# More on 0.999…

#### (An archive question of the week)

In collecting questions and answers about 0.999… for the last post, there were two that were too long to include, but that dig more deeply into issues that some of the standard answers tend to gloss over. So here, I want to look at those two answers, both of which deal with how to handle arithmetic on infinitely many digits: How can you add, subtract, or multiply when there is no rightmost digit? This is not central to the main question, but enough people express uncertainty about this aspect to make it worth covering.

## Adding without a place to start

Here’s the first question, from Darryl in 1999 (appropriately enough):

What is 0.999... + 0.999...?

While discussing the 1 = 0.999... solution, a person asked what is 0.999... + 0.999...? I think it is a good question - one that I could not answer. It should be 2 but how do we show this?

In the archives at

Getting 0.99999...
http://mathforum.org/dr.math/problems/dusty4.15.98.html

you say "Can you figure out why 0.3bar + 0.3bar = 0.6bar? Because these numbers go on forever, you will need to use a little logic to add them. (The algorithm that you learned for adding numbers doesn't work very well when you can't get to the rightmost number.)"

If we copy the idea of limits, we could say that for any number n we could start adding at this rightmost point and get the 0.66...6(nth) place. And since we can do this for any n then the equation holds. But if we try this for 0.9bar + 0.9bar we seem to get something that looks less like 2, namely 1.9....8(nth). But if we replace 8 with 9, we seem to get a number between 2 and the one we have, which makes me feel as if 0.9bar + 0.9bar < 2.

The link is to an elaborated version of the 1/3 + 1/3 + 1/3 = 1 argument, which I only gave a link to last time. (The notation 0.3bar as a typable version of $$0.\overline{3}$$, comes from there.) As Doctor Derrel pointed out there, we normally add from right to left, and there is no rightmost digit to start at in 0.333… + 0.333…; fortunately, we know that it isn’t necessary to start at the right when there are no carries, so we can just add left to right. This is discussed here:

Darryl’s suggestion is to consider any n digits, where we can clearly add 0.333…3 + 0.333…3 = 0.666…6, so we get the desired result by letting n increase without bound.

But, as Darryl points out, things get trickier when you try to add 0.999… + 0.999… . With 10 decimal places, for example, we get

0.9999999999
+ 0.9999999999
--------------
1.9999999998

You get a carry into each place from the place to its right, but where can you start that? And if there’s an 8 anywhere, then we’re less than 1.999… = 2, right? The answer, ultimately, is that in fact the 8 is nowhere!

Hi Darryl, thanks for your question. It's an interesting twist on the perennial debate about whether 0.999... could really equal 1.

Since that 8 is the nth digit and you take n to infinity, there is no digit 8 in the limit.

With no rightmost place where you would have 9 + 9 = 18, every digit will be 9. But that’s just impossible to imagine. So we have to do as mathematicians do, and find a way to work around infinities. That way is to use limits formally.

Apply the definition of a limit to what you have. We can express it as a game: you pick a number epsilon, as small as you want, and I have to pick a number n such that

|2 - 1.999...98| < epsilon

where the 8 is in the nth decimal place. Since

|2 - 1.999...98| = 2*10^(-n)

I just pick

n > -log(epsilon/2)

This is no harder than in the case of 0.999...9. The 8 in the last place has no effect on the limit; it only delays somewhat the approach to the limit, as indicated by that factor of 1/2.

The idea here is that we can consider such a sum with any number n of decimal places, which will have an 8 in the last place. We can get this sum to be as close as we want to 2, just by taking enough decimal places; in effect, to get as close as we want, we just have to take enough digits to push the 8 beyond where it would make the error too large. If we want our sum to be less than $$\epsilon = 0.00000000000000000001$$ away from 2, we just need to use $$n > -\log(\epsilon/2) = -\log(0.00000000000000000001/2) = 20.3$$, so 21 digits will be enough: that is, 0.999999999999999999999 + 0.999999999999999999999 = 1.999999999999999999998.

Here is another way to approach the problem, by manipulating infinite sums. If you want to make everything explicit, you can write the sums as limits of finite sums, and get into the issue of switching the limit and the sum as you go through this process.

Let's write 0.999... as an infinite sum:

0.999... = Sum[n = 1 to infinity](9*10^-n)

Now we can add infinite sums:

0.999... + 0.999...
= Sum[n = 1 to infinity](9*10^-n)
+ Sum[n = 1 to infinity](9*10^-n)

= Sum[n = 1 to infinity](18*10^-n)

= Sum[n = 1 to infinity](10*10^-n + 8*10^-n)

= Sum[n = 1 to infinity](10^(-n+1))
+ Sum[n = 1 to infinity](8*10^-n)

= Sum[n = 0 to infinity](10^-n)
+ Sum[n = 1 to infinity](8*10^-n)

= 1 + Sum[n = 1 to infinity](10^-n)
+ Sum[n = 1 to infinity](8*10^-n)

= 1 + Sum[n = 1 to infinity]((8+1)*10^-n)

= 1 + Sum[n = 1 to infinity](9*10^-n)

= 1.999...

I think this is the result you are looking for. Since we know that 0.999... = 1, the answer is 1 + 1 = 2; but you wanted to see that unbroken string of 9's.

He is adding an infinite string of carries (1) to an infinite string of 8’s, and getting an infinite string of 9’s.

By the way, notice that the final 8 never appears in this method. That's because we're dealing with infinite sums from the start, so there is always a carry from the next digit to the right. The sum of 10^-n is the sum of the infinite number of carries.

## Multiplying without a place to start

A similar question arises in thinking about the standard method of conversion from a repeating decimal. This question is from 2000:

Induction on .999...

I have been having an argument with my physics teacher over the fact that point nine recurring (.999... or PNR) equals 1. I showed him the proof on your site and he pointed out a fact that I hadn't noticed.

How can you multiply PNR by ten if you can't get to the beginning? For example, to multiply 215 * 3, you would start off on the right, multiplying 3 by 5, then 1, then finally 2. Now how can you multiply PNR by ten if the farthest right value cannot be reached?

Thank you for any clarification you can give.

As you should recall, we convert 0.999… to a fraction by subtracting it (x) from 10x:

x = 0.9999...
10x = 9.9999...
10x - x = 9.0000...
9x = 9
x = 1

Here, we first had to multiply 0.9999… by 10. But, as in the addition case above, there is no rightmost digit at which to start the multiplication, so can we actually do that multiplication? That is Blake’s teacher’s challenge.

Doctor TWE took it up, first referring to the FAQ (which includes links to several of the answers I’ve referred to) as a source of supplemental arguments before turning to the main issue.

I wanted to comment on your (or your teacher's) idea that "you can't get to the beginning." We multiply values from right to left only as a matter of convenience, so we don't have to "backtrack" each time we carry. It is equally valid to multiply from left to right. This is often done when we want to get a quick estimate for the answer, then get more accuracy later. The LEFTMOST digits contribute the most to the answer, so to get "in the ballpark," we can multiply them first. I'll demonstrate with a concrete example, then a general argument.

We’ve discussed this left-to-right idea in Dividing Right to Left, Adding Left to Right.

Let's use your example of 215 * 3. I can multiply them as follows; first, I'll multiply the 3 by the 2 (actually, by 200) and put the answer in the hundreds place, like this:

215
*   3
---
6

Then I'll multiply the 3 by 1 and put the answer in the tens place:

215
*   3
---
6
3

Finally, I'll multiply the 3 by 5 and put the answer in the units place:

215
*   3
---
6
3
15

215
*   3
---
6
3
15
---
645

Notice that along the way, I got pretty good estimates for my final answer. After the first step my total was 600 (not a bad estimate, off by less than 10%), after the second step I had 630, and after the third step, I got 645. Going right to left, my totals after each step would be: 15 (not a very good estimate of the final answer), 45 (still not very good) and finally 645.

So multiplying left to right just requires us to modify the digits we already have to account for carries from new columns as we do them; but the effect of the new column rarely propagates very far back into our work, and never makes a large change. Each new digit makes a smaller adjustment than the one before. That will be relevant as we continue.

In the general case, consider multiplying a 3-digit value ABC by some value X, where A, B and C are digits. What we really have, then, is

(100*A + 10*B + 1*C) * X

Conventionally, we solve this starting with the units digit as:

X*C*1 + X*B*10 + X*A*100

But using the commutative property of addition, this is equal to:

X*A*100 + X*B*10 + X*C*1

showing that the order doesn't matter. If I wanted to, I could start in the middle, for example:

X*B*10 + X*A*100 + X*C*1

I'd just have to be careful not to miss any digits or do any ones twice.

What we see here is that, conceptually, we can think of the whole multiplication as being done at once; it doesn’t matter in what order we do the pieces, because the sum is commutative.

Without being able to start with the most significant digit, we could never find a value like 2*pi, because pi (like all irrationals) is an infinite non-repeating decimal. When we say 2*pi is approximately 6.2831853, we can do so because we started multiplying at the end and not the beginning.

One final note: How can we be sure that the multiplications by 10 continue as we expect (i.e. they continue shifting the digits 1 place to the left) and that the subtractions of successive digits produce zeroes infinitely? These steps can be proven using a technique called mathematical induction. That's too complex to explain here, but if you search our Ask Dr. Math archives for the word "induction" (type it without the quotes), you'll find many questions and answers about it.

Whenever we do a calculation on numbers like π in a calculator, we are working only with the leftmost digits (those the calculator can hold); but since we know that the digits we have omitted will add no more than a 1 in the rightmost digit by carrying, the answer will be as close as we need. (In fact, calculators actually work with another digit or two beyond what they display, so that such errors never even show up.)

## Proof by induction

That last teaser – that there is a formal method for actually proving infinite things without having to do infinite things – was just what Blake needed. He replied a couple days later (back then we didn’t have threaded conversations, so he didn’t know whether the same Doctor would get the message):

I recently sent you a letter regarding the proof that .999... = 1 in the FAQ. Specifically, I asked how it was possible to multiply PNR by ten when you could not get to the right-hand side of it. This was explained quite neatly.

However, the nice person who helped me out said it could be proven that one can multiply PNR by 10 by using a process of induction. I tried to search the archives, but the results I found there were not very satisfactory because it seemed to me that each induction question was actually problem-specific.

To cut short, I would like to know the induction proof for multiplying PNR by ten.

Doctor TWE first summarized the concept of induction:

Hi again Blake! Thanks for writing back!

In general, we use proof by induction whenever we want a proof that involves an infinite sequence or series, in this case .999...

Inductive proofs require two steps:

Step 1 (the basis step): Prove it for some starting value, like n = 1.

Step 2 (the inductive step): Prove that if it's true for n = k, then it is true for n = k+1.

For the inductive step, we assume that it is true for n = k, and we usually use this assumption in the proof itself.

A word of caution: because of their self-referential nature, inductive proofs are usually hard to follow; it's easy to get lost in the details, losing track of what n, k, and k+1 are supposed to be.

For other explanations and examples of mathematical induction, see

Proof by Mathematical Induction

Sum of n Odd Numbers

Inductive Misunderstanding

Here is the proof. As he’ll explain, rather than taking n as 1, 2, 3, …, he is taking n as the exponent, -1, -2, -3, …, going in the negative direction and so reversing the usual procedure.

In our case, we want to show that when multiplying the nth digit by 10, we get a 9 in the (n+1)st digit, no carry to the (n+2)nd digit, and a 0 in the nth digit (counting from the right.) We want no carry to the (n+2)nd digit so that we don't have to adjust the previous (greater place value) digits multiplied because of a "retroactive carry." We want a 0 in the nth digit so that it will not produce further carries when multiplying the next digit (smaller place value) by 10. Mathematically:

10 * (9*10^n) = 9 * 10^(n+1)

We want to show that:

a) the nth digit is 0
b) the (n+1)st digit is 9
c) the (n+2)nd digit is 0

To minimize the confusion, I'm going to use n = -1 as my basis step, and show that if it is true for n = k, then it is true for n = k-1 as my inductive step. This is the "negative" of the standard, but it will eliminate the need for using -k's and -(k+1)'s, etc. in my equations.

Step 1: Show that it is the case for n = -1

10 * [9*10^(-1)]  =  9 * 10^(-1+1)
10 * (9*0.1)  =  9*10^0
10 * 0.9  =  9*1
9  =  9

a) the -1st digit (the tenths place) is 0: True
b) the 0th digit (the units place) is 9: True
c) the 1st digit (the tens place) is 0: True

So we've proved the basis. Now for the inductive step.

Step 2: Prove that if it's true for n = k, then it's true for n = k-1.

Let's let d = the initial (k+1)st digit. This digit is 0 from conclusion (a) of the previous step. Then:

10 * (9*10^k) + d  =  9 * 10^(k+1)
9 * (10*10^k) + 0  =  9 * 10^(k+1)
9 * 10^(k+1)  =  9 * 10^(k+1)

a) the kth digit is 0: True - 9*10^(k+1) has a 0 in the kth place.
b) the (k+1)st digit is 9: True - 9*10^(k+1) has a 9 in the (k+1)st place
c) the (k+2)nd digit is 0: True - 9*10^(k+1) has a 9 in the (k+2)nd place

So, if 10 times the kth digit is 9*10^(k+1), then 10 times the (k-1)st digit is 9*10^k, and thus 10 times the (k-2)nd digit is 9*10^(k-1), and thus...

Therefore 10 * 0.999... = 9.999...

Q.E.D.

More could be done to make this foolproof:

Depending on the skepticism of the intended audience, you may have to use a similar inductive proof for the step:

9.999...
- 0.999...
--------
9.000...

For most audiences, the following will do:

Let x = 0.999...
then 9.999... = 9 + x

so 9.999... - 0.999...  =  (9 + x) - x  = 9

However, some audiences (like your teacher, perhaps) will not be satisfied that algebraic operations work on "infinite strings" (i.e. they won't be convinced that x - x = 0 when x is an infinite repeating decimal.)

I think some of the other explanations discussed last time may be more useful for certain kinds of skeptics; but it’s good to have a variety of approaches, to meet a variety of objections.

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# value of $\int_{-\infty}^{\infty}\arcsin\frac1{\cosh x}\,dx$

I want to know the value of $$I=\int_{-\infty}^{\infty}\arcsin\frac1{\cosh x}\,dx$$ The Symbolab integral calculator says that the integral diverges, but when one graphs it obvious that it converges. So what is the value?

I was thinking that I might try Feynman integration, but I can't think of the right substitution.

From the answer provided by @user10354138, we can reach $$\int\arcsin\frac1{\cosh x}dx=i\operatorname{Li}_2(i\phi)-i\operatorname{Li}_2(-i\phi)+C$$ Where $$\phi=\tan\bigg(\frac12\arcsin\frac1{\cosh x}\bigg)$$ And $$\operatorname{Li}_2(z)=\sum_{n\geq1}\frac{z^n}{n^2}$$ is the Di-logarithm.

Wolfy says it is 4 times the Catalan's constant.

One (not optimal) way to derive this is \def\sech{\operatorname{sech}} \begin{align*} \int_{-\infty}^\infty\arcsin\sech x\,\mathrm{d}x&=2\int_0^\infty\arcsin\sech x\,\mathrm{d}x\\ &=2\int_0^1\frac{\arcsin u\,\mathrm{d}u}{u\sqrt{1-u^2}}\quad(u=\sech x)\\ &=2\int_0^{\pi/2}\frac{\theta\,\mathrm{d}\theta}{\sin\theta}\quad(u=\sin\theta)\\ &=2\int_0^1\frac{2\tan^{-1}t\,\frac{2\,\mathrm{d}t}{1+t^2}}{\frac{2t}{1+t^2}}\quad(t=\tan\tfrac12\theta)\\ &=4\int_0^1\frac{\tan^{-1}t}{t}\,\mathrm{d}t\\ &=4\int_0^1\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}t^{2n}\,\mathrm{d}t\\ &=4\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}=4G\\ \end{align*}

• Dude that's dope! Thank you! – clathratus Oct 31 '18 at 3:31
• This substitution simplifies to $t=e^{-x}$. – J.G. Oct 31 '18 at 8:44
• how do you get from $\frac{\theta d\theta}{\sin\theta}$ to $$\frac{2\arctan(t)\frac{2dt}{1+t^2}}{\frac{2t}{1+t^2}}$$ With the substitution $t=\tan\frac\theta2$? – clathratus Nov 1 '18 at 23:06
• I keep getting $$\frac{\theta d\theta}{\sin\theta}=\frac{2\arctan t}{t\sqrt{t^2+1}}dt$$ – clathratus Nov 1 '18 at 23:10
• @clathratus This is the universal trigonometric substitution – user10354138 Nov 1 '18 at 23:25

Alternatively, you can integrate by parts: \begin{align} \int \limits_{-\infty}^\infty \arcsin(\operatorname{sech}(x)) \, \mathrm{d} x &= 2 \int \limits_0^\infty \arcsin(\operatorname{sech}(x)) \, \mathrm{d} x \\ &= 2x \arcsin(\operatorname{sech}(x)) \Bigg \rvert_{x=0}^{x=\infty} - 2 \int \limits_0^\infty x \frac{- \sinh(x) \operatorname{sech}^2(x)}{\sqrt{1-\operatorname{sech}^2(x)}} \, \mathrm{d} x \\ &= 2 \int \limits_0^\infty \frac{x}{\cosh(x)} \, \mathrm{d} x = 4 \sum \limits_{n=0}^\infty (-1)^n \int \limits_0^\infty x \, \mathrm{e}^{-(2n+1) x} \, \mathrm{d} x \\ &= 4 \Gamma(2) \sum \limits_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} = 4 \mathrm{G} \, . \end{align}

• I'm a fan of the alternative approach. Thank you! – clathratus Oct 31 '18 at 14:59

Yet another alternative approach: once shown that $$\int_{\mathbb{R}}\frac{dx}{\cosh(x)^{2k+1}} = \frac{\pi \binom{2k}{k}}{4^k}\tag{1}$$ and recalled that $$\arcsin z = \sum_{k\geq 0}\frac{\binom{2k}{k}}{4^k(2k+1)}z^{2k+1} \tag{2}$$ we have the following identity: $$\int_{\mathbb{R}}\arcsin\frac{1}{\cosh x}\,dx = \pi\sum_{k\geq 0}\frac{1}{2k+1}\left[\frac{1}{4^k}\binom{2k}{k}\right]^2.\tag{3}$$ Now we may invoke a function whose Maclaurin series involves squared central binomial coefficients, namely the complete elliptic integral of the first kind $$K(x)$$, here denoted according to Mathematica's notation (the argument of $$K$$ is the elliptic modulus): $$\sum_{k\geq 0}\left[\frac{1}{4^k}\binom{2k}{k}\right]^2 x^{2k}=\frac{2}{\pi}K(x^2)\tag{4}$$ leading to: $$\int_{\mathbb{R}}\arcsin\frac{1}{\cosh x}\,dx = 2\int_{0}^{1} K(x^2)\,dx = \int_{0}^{1}\frac{K(x)}{\sqrt{x}}\,dx.\tag{5}$$ At last, we recall that both $$K(x)$$ and $$\frac{1}{\sqrt{x}}$$ have fairly simple Fourier-Legendre series expansions: $$K(x)=2\sum_{n\geq 0}\frac{P_n(2x-1)}{2n+1},\qquad \frac{1}{\sqrt{x}}=2\sum_{n\geq 0}(-1)^n P_n(2x-1)$$ and by the orthogonality of shifted Legendre polynomials

$$\int_{\mathbb{R}}\arcsin\frac{1}{\cosh x}\,dx =\int_{0}^{1}\frac{K(x)}{\sqrt{x}}\,dx = 4\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^2} = 4G.\tag{6}$$

• This is really cool. I'll have to look at this one a little more in order to understand it better... But this is great. Thanks! – clathratus Oct 31 '18 at 18:37
• You seem to link any special function with any other special function. The link between elliptic integral and Legendre function is pretty smart. +1 – Paramanand Singh Nov 1 '18 at 16:41
• How do we show $$\int_{\Bbb R}\frac{\mathrm dx}{\cosh(x)^{2k+1}}=\frac{\pi{2k\choose k}}{4^k}$$ It looks a lot like a Beta integral – clathratus Dec 21 '18 at 0:56
• @clathratus: it is, indeed, it is enough to set $\frac{1}{\cosh(x)}=u$. – Jack D'Aurizio Dec 21 '18 at 10:48


# Applying Gauss' Divergence Theorem to given integral

We are given the following two integrals:

$$\iint\limits_S D_n f\:dS$$ and $$\iiint\limits_B \nabla \cdot (\nabla f) \: dV$$

where $$S$$ is the portion of the sphere $$x^2+y^2+z^2=a^2$$ in the first octant, $$n$$ is the unit normal vector to $$S$$ at $$(x,y,z)$$ and $$f(x,y,z) = ln(x^2+y^2+z^2)$$

For the first integral, we have: $$D_n f = \nabla f \cdot n$$ and $$n = \frac{1}{a}$$

Since $$\nabla f = \: <\frac{2x}{x^2+y^2+z^2}, \frac{2y}{x^2+y^2+z^2}, \frac{2z}{x^2+y^2+z^2}>$$, this gives $$D_n f = \nabla f \cdot n = \frac{2}{a}$$.

So the integral becomes

$$\iint\limits_S D_nf \: dS = \iint\limits_S \frac{2}{a} \: dS = \frac{2}{a} \iint\limits_S dS = \frac{2}{a} \cdot A(S) = \frac{2}{a} \cdot 4\pi a^2 \cdot \frac{1}{8} = \pi a$$ ($$\frac{1}{8}$$ since the sphere is in the first octant)

For the second integral, we have $$\nabla \cdot (\nabla f) = \frac{2}{a^2}$$ (since $$x^2 + y^2 + z^2 = a^2$$). So the integral becomes $$\iiint\limits_B \nabla \cdot (\nabla f) \: dV = \frac{2}{a^2} \iiint\limits_B dV = \frac{2}{a^2} \cdot V(B) = \frac{2}{a^2} \cdot \frac{4}{3} \pi a^3 \cdot \frac{1}{8} = \frac{1}{3} \pi a$$

However, according to Gauss' Divergence theorem, these two integrals should be equal to each other right? Since we have $$\iiint\limits_B \nabla \cdot (\nabla f) \: dV = \iint\limits_S \nabla f \cdot dS = \iint\limits_S \nabla f \cdot n \: dS = \iint\limits_S D_n f \: dS$$

So how is it possible that I get two different answers, even though the Divergence theorem shows that the two integrals should be the same?

• You're not counting the flat parts of the surface which are quarter circles in the coordinate planes. – B. Goddard Mar 29 at 11:09
• Oh I see, so if I understand correctly, is it true that the results of the first two integrals are correct, but the application of Gauss' theorem isn't? Because the surface S is not the same as the boundary surface of B, since the boundary surface contains extra flat parts? – Stallmp Mar 29 at 11:14

To apply divergence theorem, you must have a closed surface. So we close the surface by placing $$3$$ quarter disks in plane $$x = 0, y = 0, z = 0$$.

Please note that when you are doing volume integral, you cannot equate $$\nabla \cdot (\nabla f) = \displaystyle \frac{2}{x^2+y^2+z^2} = \frac{2}{a^2}$$. It should rather be,

$$\nabla \cdot (\nabla f) = \displaystyle \frac{2}{x^2+y^2+z^2} = \frac{2}{\rho^2}$$

So the integral becomes,

$$\displaystyle \int_0^{\pi/2} \int_0^{\pi/2} \int_0^a \frac{2}{\rho^2} \ \cdot\rho^2 \cdot \sin \phi \ d\rho \ d\phi \ d\theta = \pi a$$.

Now to find flux through $$S$$, we must subtract flux through planar surfaces $$x = 0, y = 0, z = 0$$ but in this case, they are simply zero.

• simply because $x^2+y^2+z^2 = a^2$ only on the surface of the sphere, not inside it. We are talking volume integral which includes all points inside the sphere too and so we write $x^2+y^2+z^2 \leq a^2$. – Math Lover Mar 29 at 11:31
• Yes you are right about surface $S$. What you get from divergence theorem will include flux through closed surface which is spherical surface $S$ and $3$ planar surfaces as mentioned in my answer. We are supposed to calculate that separately and subtract from the result that we get from divergence theorem to obtain flux only through spherical surface $S$. But in this case they are zero. Do you see why or do you need an explanation? – Math Lover Mar 29 at 11:39
• Yes this is indeed what I thought, thank you very much for the explanation! I see why they are zero as well when calculating the flux through the planar surfaces. – Stallmp Mar 29 at 11:45
• Yes you are right, we need to show they sum to zero. Note that $\iint\limits_{S_1} D_n f \: dS = \iint\limits_{S_2} D_n f \: dS = \iint\limits_{S_3} D_n f \: dS = 0$ – Math Lover Mar 29 at 12:34
• – Math Lover Mar 29 at 12:35


# Method for Extracting the $n$th root of a number?

I was wondering if there exists a reliable way to extract the $$n$$th root of a given number. For example, if you have a large perfect square such as 10,404, how would I go about taking its square root by hand?

And what about cube roots? Is there a way to find the cube root of any number by hand?

If there is a way to take the cube and square roots of any number, is there a technique that can be applied to extract the cube or square root of polynomials? What about any root?

I think that, if there is a way to take the $$n$$th root of any number of polynomial, I would benefit from learning it, since I consider being able to work with roots important to my algebra foundation.

Thanks

• For the square root, there is a method similar to the usual division by a number. If we know that the number is a perfect square we can restrict the possibilities (ending digits , residue mod 9). For cube roots and higher , I am not aware of an efficient method by hand. Feb 25 at 18:58
• For square roots we can use the algorithm described here under "Digit-by-digit calculation" / "Decimal (base 10)" Feb 25 at 19:03
• @StefanOctavian I've seen similar algorithms for higher roots done (somehow you take 3-digits at a time for cube roots.) But they stopped teaching the square-root extraction the year before I would have had to learn it. Feb 25 at 20:38

If you would like to compute $$\sqrt[p]{a}$$, the following iterative process will do it to whatever accuracy you please:

First, make a guess and call it $$x_0$$. Then perform this iteration:

$$x_{n+1} = \frac{(p-1)x_n^p+a}{px^{k-1}}.$$

For instance, if you want $$\sqrt[5]{11}$$, guess that $$x_0=2$$. The iteration looks like

$$x_{n+1} = \frac{4x_n^5+11}{5x_n^4}.$$

So we have $$x_1 = \frac{4\cdot2^5+11}{5\cdot 2^4} = \frac{139}{80}=1.7375.$$

$$x_2=\frac{4\cdot1.7375^5+11}{5\cdot 1.7375^4} = 1.631392308.$$

This number to the 5th power is $$11.55558806$$, so we're getting close.

$$x_3=\frac{4\cdot1.631392308^5+11}{5\cdot 1.631392308^4} = 1.615704970.$$

And $$1.615704970^5 = 11.0105\ldots.$$

This is an implementation of Newton's Method on the polynomial $$x^p-a$$.


Related Rates:

Gas is escaping from a spherical balloon at the rate of 2ft^3/min. How fast is the surface area shrinking (ds/dt) when the radius is 12ft? (A sphere of radius r has volume v=4/3 pi r^3 and surface area S=4pi r^2.)

Remember that ds/dt = ds/dr X dr/dt

Step 1: Find ds/dr

I have no clue where to start. I was going to set 2 equal to 4 pi r^2 and then take the derivative but I really have no idea.

Step 2: Find dr/dr. (HInt dv/dt= dv/dr X dr/dt

Would I set 2 equal to volume and then take derivative?

Step 3: Find ds/dt?

Step 4: Evaluate ds/dt when the radius is 12ft.
After I find my equation I would just plug in 12 correct?

1. 👍
2. 👎
3. 👁
1. S = 4pi*r^2
dS/dr = 4pi*2r = 8pi*r

V = (4/3)pi*r^3
dV/dr = 4pi*r^2
dV/dt = dV/dr * dr/dt
2 = 4pi*r^2 * dr/dt
dr/dt = 1/(2pi*r^2)

dS/dt = dS/dr * dr/dt
= 8pi*r * 1/(2pi*r^2)
= 4/r
If r = 12, just plug the value to the last equation

1. 👍
2. 👎
2. How did you get 12?

1. 👍
2. 👎
3. Nevermind I meant how did you get 4/r but I figured it out. Thank You!!

1. 👍
2. 👎
4. Is this correct for the last part?

(8)(3.14)(12) X (1/(2)(3.14)(12^2) = 4/12

1. 👍
2. 👎
5. Yes

1. 👍
2. 👎
6. Ok do I actually have to do the calculations though?

1. 👍
2. 👎
7. If you are not given the value, that means you have to write the steps I told you until the very last equation.
If you are given the value (of r), you don't have to put the value in every equation. Just follow my steps until the last equation, then plug the value after that. It's easier that way.

1. 👍
2. 👎
8. OK thank you!

1. 👍
2. 👎
9. Simplify the answer (4/12) into 1/3

1. 👍
2. 👎

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# Relation between $\bigcap_{i \in I}A_i$ and $\bigcap_{i=1}^{n}A_i$

Let I be a nonempty set and a family of sets such that every element of the family is a subset of U.

$\mathcal F = \{A_i | i \in I\}$

I understand the meaning of this operation:

$$\bigcap_{i \in I}A_i$$

That's the intersection of all the elements of the sets of the family $\mathcal F$.

But I don't truly understand what this other operation means and what's the relation between the above one. $$\bigcap_{i=1}^{n}A_i$$

I understand that this operation is also an intersection but what I am trying to understand is the relation between $\bigcap_{i \in I}A_i$ and $\bigcap_{i=1}^{n}A_i$

$$\bigcap_{i=1}^{n}A_i \subseteq\bigcap_{i \in I}A_i$$

Is the relation above true?

• In the second case we have $I = \{ 1,2,\ldots, n \}$. – Mauro ALLEGRANZA Jun 23 '18 at 15:28

The notation $$\bigcap _{i=1}^nA_i$$ denotes the same as $$\bigcap_{i\in I}A_i$$ for the special case $I=\{1,2,\ldots, n\}$.
• So $\bigcap_{i\in I}A_i$ is the general case and $\bigcap _{i=1}^nA_i$ is a specific case for which we give concrete values to i? – adriana634 Jun 23 '18 at 15:40
• @adriana634 Yes. One situation where we see both notations used simultaneously is when $I=\mathbb{N}$; then "$\bigcap_{i=1}^nA_i$" means "$\bigcap_{i\in \{1,2,..., n\}}A_i$," which is generally a superset of $\bigcap_{i\in I}A_i$. – Noah Schweber Jun 23 '18 at 15:42
• @Noah Schweber But for example $\bigcap_{i=1}^{5}A_i$ means $I=\{1,2,\ldots, 5\}$, in this example $\bigcap_{i\in I}A_i$ is a superset of $\bigcap_{i=1}^{5}A_i$? I don't understand why you said $\bigcap_{i=1}^nA_i$ is a superset of $\bigcap_{i\in I}A_i$. – adriana634 Jun 23 '18 at 16:00
• @adriana634 No. The intersection over a bigger index set yields a smaller set. E.g. if we take $I=\mathbb{N}$ and $A_i=\{i, i+1, i+2, ...\}$ for $i\in I$, then $\bigcap_{i=1}^5A_i=\{5, 6,7,...\}$ but $\bigcap_{i\in I}A_i=\emptyset$. More abstractly: $$I\supseteq J\implies \bigcap_{i\in I}A_i\subseteq \bigcap_{i\in J}A_i.$$ – Noah Schweber Jun 23 '18 at 16:02


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# In the addition shown above, A, B, C, and D represent

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In the addition shown above, A, B, C, and D represent  [#permalink]

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15 Oct 2013, 07:17
7
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Question Stats:

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ABC
+BCB
CDD

In the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?

A. 8
B. 10
C. 12
D. 14
E. 18
Magoosh GMAT Instructor
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Posts: 4488
Re: In the addition shown above, A, B, C, and D represent  [#permalink]

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15 Oct 2013, 10:47
17
12
saintforlife wrote:
ABC
+BCB
CDD

In the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?

A. 8
B. 10
C. 12
D. 14
E. 18

I'm happy to help with this.
First, look at the one's column. C + B produces a one's unit of D, and we don't know whether anything carries to the ten's place.
But now, look at the ten's place --- B + C again produces a digit of D --- that tells us definitively that nothing carried, and that B + C = D. B & C have a single digit number as a sum.
Now, C = A + B, because again, in the hundreds column, nothing carries here.

Notice that, since C = A + B and D = B + C, B must be smaller than both B and C. We want to make the product of A & B large, so we want to make those individual digits large. Well, if we make B large, then that makes C large, and then B + C would quickly become more than a one-digit sum, which is not allowed. Think about it this way. Let's just assume D = 9, the maximum value.
D = 9 = B + C = B + (A + B) = A + 2B
We want to pick A & B such that A + 2B = 9 and A*B is a maximum. It makes sense that B would be smaller.
Try A = 7, B = 1. Then A + 2B = 9 and A*B = 7
Try A = 5, B = 2. Then A + 2B = 9 and A*B = 10
Try A = 3, B = 3. Then A + 2B = 9 and A*B = 9
Try A = 1, B = 4. Then A + 2B = 9 and A*B = 4
Indeed, as B gets bigger, the product gets less. This seems to imply that the biggest possible product is 10. This corresponds to A = 5, B = 2, C = 7, and D = 9, and the original addition problem becomes
527
+272
799
Thus, the maximum product is 10, and answer = (B).

Does all this make sense?
Mike
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Re: In the addition shown above, A, B, C, and D represent  [#permalink]

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06 Dec 2013, 10:19
8
1
saintforlife wrote:
How do we do all that in less than 2 mins, is my only question :)

Posted from my mobile device

This problem can be solved much faster if you start with the options. As Mike has wonderfully explained, none of the digits will have a carry over.

Option E is $$18 = 9 * 2$$(no other case is possible). Now$$9+2>=10$$. Therefore, it will have a carry over digit 1. So reject this option.

Option D is $$14 = 7 * 2$$(no other case is possible). Since $$C=A+B$$, therefore $$C= 9$$. Since $$C=9$$ and $$B$$ is a non zero digit, $$C+B>=10$$. Therefore, it will have a carry over digit 1. So reject this option.

Option C is 12. This will have 2 cases-
$$Case 1--> 12 = 4 * 3$$

Here $$C= 4+3=7$$. Now $$7+4>=10$$ and [$$7+3>=10$$. Therefore, both cases will have a carry over digit 1. So reject this option.

(OR) $$Case 2--> 6 * 2$$.

Here $$C= 6+2=8$$. Now $$8+2>=10$$ and $$8+6>=10$$. Therefore, both cases will have a carry over digit 1. So reject this option.

Option B is $$10 = 5 * 2$$. If you follow the same set of steps you'll notice that there will not be any carry over digit for the case $$C=7, A=5, B=2$$. So right answer!

##### General Discussion
Manager
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Re: In the addition shown above, A, B, C, and D represent  [#permalink]

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23 Oct 2013, 15:44
2
How do we do all that in less than 2 mins, is my only question

Posted from my mobile device
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4488
Re: In the addition shown above, A, B, C, and D represent  [#permalink]

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23 Oct 2013, 15:54
6
saintforlife wrote:
How do we do all that in less than 2 mins, is my only question

Dear Saint For Life,
Problems such as this are quite out-of-the-box. To some extent, the GMAT intends us to handle many of the other questions in 90 secs or less, so that we have a bit of a time-cushion when we run into one of these oddball questions.

Having said that, the more familiar you are with number properties, the faster it will go. In this problem, it took me quite some time to write out everything in verbal form, but I saw things relatively quickly. As you practice seeing patterns, you will see them more quickly, even if they are hard to explain to someone else. You may find this post helpful:
http://magoosh.com/gmat/2013/how-to-do- ... th-faster/

Mike
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Re: In the addition shown above, A, B, C, and D represent  [#permalink]

### Show Tags

23 Oct 2013, 23:13
5
2
saintforlife wrote:
ABC
+BCB
CDD

In the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?

A. 8
B. 10
C. 12
D. 14
E. 18

Similar questions to practice:
tough-tricky-set-of-problems-85211.html#p638336

Hope it helps
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Re: In the addition shown above, A, B, C, and D represent the  [#permalink]

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26 Oct 2013, 21:47
2
... got E as an answer. However, it took me 6 minutes, since I plugged in the answer options.
Is there any faster way, or am I just too slow in back-solving?

TirthankarP wrote:

In the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?
A) 8
B) 10
C) 12
D) 14
E) 18

My explanation:

According to the question stem C+B=D and D<10, since the second addition (2nd column) is C+B=D again (if there were a carry, the second column's result would not be D again), and A+B=C
Now I break up the answer options:
A) A*B=8, so A and B -> 2 and 4 or 1 and 8. A+B=2+4=6=C, C+B=6+2=8=D, which is smaller than 10, thus the answer is not "out", but there could be a larger possible value of A*B
B) A*B=10, so A and B -> 2 and 5. A+B=2+5=7=C, C+B=7+2=9=D, which is smaller than 10, thus the is also possible. Let's look for the next options.
For C-E:
C) A*B=12, so A and B -> 6 and or 4 and 3.
D) A*B=14, so A and -> 7 and 2.
E) A*B=18, so A and B -> 9 and 2 or 3 and 6.
Do the same process with those numbers and you will find that all will yield a sum of C+B>10, thus the constraint of D<10 is not satisfied. The answer options are not possible.
Because 10>8 answer option B) is the largest possible.
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Re: In the addition shown above, A, B, C, and D represent the  [#permalink]

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27 Oct 2013, 00:04
Chiranjeevee wrote:
TirthankarP wrote:
Attachment:

In the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?
A) 8
B) 10
C) 12
D) 14
E) 18

I do a thorough search before posting any new question. But this time I couldn't find this link
Forum moderator may delete this thread
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Collection of some good questions on Number System

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In the addition shown above, A, B, C, and D represent  [#permalink]

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01 Jul 2015, 07:02
HKHR wrote:
saintforlife wrote:
How do we do all that in less than 2 mins, is my only question :)

Posted from my mobile device

This problem can be solved much faster if you start with the options. As Mike has wonderfully explained, none of the digits will have a carry over.

Option E is $$18 = 9 * 2$$(no other case is possible). Now$$9+2>=10$$. Therefore, it will have a carry over digit 1. So reject this option.

Option D is $$14 = 7 * 2$$(no other case is possible). Since $$C=A+B$$, therefore $$C= 9$$. Since $$C=9$$ and $$B$$ is a non zero digit, $$C+B>=10$$. Therefore, it will have a carry over digit 1. So reject this option.

Option C is 12. This will have 2 cases-
$$Case 1--> 12 = 4 * 3$$

Here $$C= 4+3=7$$. Now $$7+4>=10$$ and [$$7+3>=10$$. Therefore, both cases will have a carry over digit 1. So reject this option.

(OR) $$Case 2--> 6 * 2$$.

Here $$C= 6+2=8$$. Now $$8+2>=10$$ and $$8+6>=10$$. Therefore, both cases will have a carry over digit 1. So reject this option.

Option B is $$10 = 5 * 2$$. If you follow the same set of steps you'll notice that there will not be any carry over digit for the case $$C=7, A=5, B=2$$. So right answer!

Wonderful explanation. Just thought to add some more from my side. As we see only B satisfies all the necessary requirement(not to carry over any digits).

In option A, taking A=4 and B=2 also satisfies the condition as it is something like
426
+262
-------
688

But since we need maximum value well go for option B
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Re: In the addition shown above, A, B, C, and D represent  [#permalink]

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30 Nov 2017, 18:03
Dear Mike, I was wondering why are we maximising D since it no where mentioned to do so. We have to maximise A*B right so if we choose C=1 and B=5 and A=4 i think we staisfy all conditions and our product of A*B (5*4) comes to be 20. Dont know where i am going wrong. Kindly help
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Posts: 4488
Re: In the addition shown above, A, B, C, and D represent  [#permalink]

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01 Dec 2017, 13:54
1
1
Sarthak.bhatt wrote:
Dear Mike, I was wondering why are we maximising D since it no where mentioned to do so. We have to maximise A*B right so if we choose C=1 and B=5 and A=4 i think we staisfy all conditions and our product of A*B (5*4) comes to be 20. Dont know where i am going wrong. Kindly help

Dear Sarthak.bhatt,

I'm happy to respond.

With all due respect, my friend, you are not interpreting the question correctly. There is absolutely no multiplication happening in this question. Here's the prompt again:

ABC
+BCB
CDD

In the addition shown above, A, B, C, and D represent
the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?

Thus, for example if C = 1, B = 5, and A = 4, then ABC would not be the product of those three numbers but instead the single three-digit number 451. With those numbers, the problem would be

451
+515
966

These choices satisfy the equation.

The problem is completely different from the way you were conceptualizing it, so the strategy is completely different from what it would be in the problem you had in mind.

Does this make sense?

Mike
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Re: In the addition shown above, A, B, C, and D represent  [#permalink]

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26 May 2018, 03:22
mikemcgarry. Can you please explain Sarthak's query as i have the same query but unfortunately i didn't get from the reply you gave to him.
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Re: In the addition shown above, A, B, C, and D represent  [#permalink]

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15 Jun 2018, 15:24
Do we not consider negative numbers in this problem at all?
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Joined: 02 Sep 2009
Posts: 50727
Re: In the addition shown above, A, B, C, and D represent  [#permalink]

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15 Jun 2018, 21:55
rohithtv89 wrote:
Do we not consider negative numbers in this problem at all?

No. We are given that A, B, C, and D represent the nonzero digits. There are 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
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Re: In the addition shown above, A, B, C, and D represent &nbs [#permalink] 15 Jun 2018, 21:55
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# A photography dealer ordered 60 Model X cameras to be sold

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A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit ##### Most Helpful Expert Reply Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8195 Location: Pune, India Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 24 Jun 2012, 22:20 61 38 pgmat wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras?

A. 7% loss
B. 13% loss
C. 7% profit
D. 13% profit
E. 15% profit

Use weighted avgs for a quick solution:

On 54 cameras, the dealer made a 20% profit and on 6 cameras, he made a 50% loss. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1

Avg Profit/Loss = (.2*9 + (-.5)*1)/10 = 1.3/10 = .13 = 13% profit

For more on weighted avgs, check: http://www.veritasprep.com/blog/2011/03 ... -averages/
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Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! ##### Most Helpful Community Reply Manager Status: mba here i come! Joined: 07 Aug 2011 Posts: 230 A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags Updated on: 25 Feb 2016, 08:28 98 27 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. suppose total cameras = 10 and each costs$10
total profit = $2*9 -$5*1 = $13 or 13% (with$100 cost)
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Originally posted by MBAhereIcome on 23 Jun 2012, 07:31.
Last edited by MBAhereIcome on 25 Feb 2016, 08:28, edited 1 time in total.
##### General Discussion
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Posts: 48037
Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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06 May 2012, 11:07
25
1
24
pgmat wrote:
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit Total cost 60*($250/1.2)=50*250;

# of cameras sold is 60-6=54 total revenue is 54*250;
# of cameras returned is 6 total refund 6*(250/1.2)*0.5;
So, total income 54*250+ 6*(250/1.2)*0.5

The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%

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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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22 Jun 2012, 14:14
Bunuel wrote:
pgmat wrote:
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit Total cost$60*250/1.2=50*250;

# of cameras sold is 60-6=54 total revenue is 54*250;
# of cameras returned is 6 total refund 6*(250/1.2)*0.5;
So, total income 54*250+ 6*(250/1.2)*0.5

The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%

Dear Bunuel,
Ths is a doubt i face with %ages, in ths quest too.
could you pls tell where i m going wrong?
when 20% mark up for initial cost is given , how to calculate it?
250 - 20/100*250 or to take if 120 is 250 then how much is 100?
how have you arrived at 1.2?
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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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23 Jun 2012, 04:06
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2
kashishh wrote:
Bunuel wrote:
pgmat wrote:
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit Total cost 60*$250/1.2=50*250;

# of cameras sold is 60-6=54 total revenue is 54*250;
# of cameras returned is 6 total refund 6*(250/1.2)*0.5;
So, total income 54*250+ 6*(250/1.2)*0.5

The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%

Dear Bunuel,
Ths is a doubt i face with %ages, in ths quest too.
could you pls tell where i m going wrong?
when 20% mark up for initial cost is given , how to calculate it?
250 - 20/100*250 or to take if 120 is 250 then how much is 100?
how have you arrived at 1.2?

If it's given that the selling price is $250 and the markup over the initial cost is 20%, then: {Cost}+0.2{Cost}=1.2*{Cost}={Selling price} --> 1.2*{Cost}=$250 --> {Cost}=$250/1.2. Hope it's clear. _________________ Intern Joined: 28 Feb 2012 Posts: 21 GMAT 1: 700 Q48 V39 WE: Information Technology (Computer Software) Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 24 Jun 2012, 02:06 1 MBAhereIcome wrote: 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. suppose total cameras = 10, out of which 9 gave 20% profit suppose 100 was the cost of each. total profit 100*10=1000 profit from 9 = 20*9 = 180 loss from 1 = 50*1 = 50 total profit = 180-50 = 130 so, 130 profit from a cost of 1000. that is 13% profit. Your method seems easy, but I kind of got lost with this - 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. Could you please explain, how you deduced this? Thanks. Manager Joined: 02 Jun 2011 Posts: 132 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 24 Jun 2012, 12:26 1 2 gmatDeep wrote: MBAhereIcome wrote: 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. suppose total cameras = 10, out of which 9 gave 20% profit suppose 100 was the cost of each. total profit 100*10=1000 profit from 9 = 20*9 = 180 loss from 1 = 50*1 = 50 total profit = 180-50 = 130 so, 130 profit from a cost of 1000. that is 13% profit. Your method seems easy, but I kind of got lost with this - 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. Could you please explain, how you deduced this? Thanks. yess.. 54:6 = 9:1 mark up is of 20% = profit earned will be 20% VP Joined: 02 Jul 2012 Posts: 1192 Location: India Concentration: Strategy GMAT 1: 740 Q49 V42 GPA: 3.8 WE: Engineering (Energy and Utilities) Re: Of the cameras ordered, 6 were never sold and were returned [#permalink] ### Show Tags 16 Nov 2012, 23:15 15 4 carcass wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, whichrepresents a 20 percent markup over the dealer's initialcost for each camera.
Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer's initial cost for the 60 cameras?

(A) 7% loss

(B) 13% loss

(C) 7% profit

(D) 13% profit

(E) 15% profit

have fun. is not difficult to conceptualize but is difficult because if you are not organized and clear you are get in trouble.

I would not show the OA but the rules are stringent. either way try on your own

Cost to dealer for 60 cameras = $$\frac{250}{1.2}*60$$ = $12,500 Revenue for 54 cameras = 54*250 =$13,500

Revenue from 6 cameras = $$\frac{250}{2.4}*6$$ = $625 Total Revenue =$14,125

Profit percent = $$\frac{14125-12500}{12500}$$ = 13%

Otherwise, If one were short on time and I had to guess, One can eliminate A,B & C by looking at the problem itself and improve my chances of guessing the right answer to 50% between D & E.

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Re: Of the cameras ordered, 6 were never sold and were returned  [#permalink]

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16 Nov 2012, 23:27
1
60 cameras were ordered to be sold at a 20% markup for $250 each => Dealer's cost for each camera =$208
=> Cost of all 60 cameras = $60*208 6 cameras were not sold and returned for a refund of 50% of the dealer's cost => The amount the dealer got in returns = 50% of 6*208 =$624
The rest of the cameras (60-6 = 54 in number) were sold
=> Revenue made from selling the cameras = $54 * 250 For the 60 cameras, dealer's profit = 100* [54*(250-208) + (624-1248)]/[(60*200)] =100* (14100-12000)/12000 = ~13% (the error arises because 250/1.2 is not exactly 208 but 208.33) Option (D) _________________ GyanOne | Top MBA Rankings and MBA Admissions Blog Top MBA Admissions Consulting | Top MiM Admissions Consulting Premium MBA Essay Review|Best MBA Interview Preparation|Exclusive GMAT coaching Get a FREE Detailed MBA Profile Evaluation | Call us now +91 98998 31738 Director Status: Done with formalities.. and back.. Joined: 15 Sep 2012 Posts: 616 Location: India Concentration: Strategy, General Management Schools: Olin - Wash U - Class of 2015 WE: Information Technology (Computer Software) Re: Of the cameras ordered, 6 were never sold and were returned [#permalink] ### Show Tags 16 Nov 2012, 23:48 24 13 carcass wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, whichrepresents a 20 percent markup over the dealer's initialcost for each camera.
Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer's initial cost for the 60 cameras?
(A) 7% loss
(B) 13% loss
(C) 7% profit
(D) 13% profit
(E) 15% profit
have fun. is not difficult to conceptualize but is difficult because if you are not organized and clear you are get in trouble.
I would not show the OA but the rules are stringent. either way try on your own

actually, if one could visualize this problem properly, it could be solved in less than 30 secs without getting into any dirty calculations.
Here is my 30 sec approach:
Notice that it is actually a wighted ratio problem, where 9 parts earned 20% profit and 1 part earned 50% loss. (54 for profit, 6 for loss).
Hence overall profit/loss = $$(9*20 -1*50) /(9+1) = 13$$
Ans D it is!
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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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24 Nov 2012, 11:16
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Hi,

For those using normal way to solve this problem, I think that we can avoid huge calculation by just replace Cost by letter C and put it in the final formula:
Income: 54*1.2C + 6*0.5C
Cost: 60*C
Then percent =(Income-cost)/cost = (54*1.2C + 6*0.5C - 60*C)/60*C. We can see that we can eliminate C from the fraction --> percent = (54*1.2 + 6*0.5 - 60)/60 = (54*1.2 + 6*0.5)/60 - 1. Here, we can see clearly the weighted average part of the problem or we can simply calculate the fraction to reach the result.
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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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27 May 2013, 12:12
Bunuel wrote:
pgmat wrote:
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit Total cost 60*($250/1.2)=50*250;

# of cameras sold is 60-6=54 total revenue is 54*250;
# of cameras returned is 6 total refund 6*(250/1.2)*0.5;
So, total income 54*250+ 6*(250/1.2)*0.5

The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%

this must be kind of awkward question but i got no option other than asking you.

20% markup over the dealer’s initial cost for each camera- I made a wrong assumption , but reducing 20%from 250 & got 200 as an initial cost for each. please explain that "Total cost 60*($250/1.2)=50*250;" Math Expert Joined: 02 Sep 2009 Posts: 48037 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 27 May 2013, 12:19 3 1 FTGNGU wrote: Bunuel wrote: pgmat wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras?

A. 7% loss
B. 13% loss
C. 7% profit
D. 13% profit
E. 15% profit

Total cost 60*($250/1.2)=50*250; # of cameras sold is 60-6=54 total revenue is 54*250; # of cameras returned is 6 total refund 6*(250/1.2)*0.5; So, total income 54*250+ 6*(250/1.2)*0.5 The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13% Answer: D. this must be kind of awkward question but i got no option other than asking you. 20% markup over the dealer’s initial cost for each camera- I made a wrong assumption , but reducing 20%from 250 & got 200 as an initial cost for each. please explain that "Total cost 60*($250/1.2)=50*250;"

(Cost per unit) + 0.2*(Cost per unit) = $250 1.2*(Cost per unit) =$250
(Cost per unit) = $250/1.2 Total cost for 60 units = 60*(Cost per unit) = 60*($250/1.2) = 50*250.

Hope it's clear.
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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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07 Apr 2014, 19:23
4
Because the answers are in percentages, I thought not to worry about the $amounts and just focus on the relationships: Let C be the total cost of all 60 cameras. Originally, the dealer thought to sell all these 60 cameras at 20% profit or for (1.2)C. However, he sold only (60-6)=54 or 90% of cameras at this price. So revenue from these cameras = (0.9)(1.2)C = (1.08)C For the remaining 10%, he got a refund of 50% of cost or (0.5)C. So total refund = (0.1)(0.5)C = (0.05)C Therefore, total revenue in terms of original cost = (1.08 + 0.05)C = 1.13C or 13% profit. So D is the correct ans. Intern Status: 1st attempt: Can I do it? Joined: 09 Jun 2013 Posts: 1 Location: India Concentration: Healthcare, Marketing GMAT Date: 05-30-2014 GPA: 2.99 WE: Pharmaceuticals (Pharmaceuticals and Biotech) Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 27 May 2014, 10:33 2 1 {(54*20) + (6*(-50))}/ 60 This gives 13% profit as answer. 20% proft on 54 cameras and 50% loss on 6 cameras. Use weighted averages here. Rest of the information in the question is misguiding. Intern Joined: 09 Feb 2013 Posts: 26 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 29 Jun 2014, 23:52 My soln is : Dealer's per unit initial cost =250 -20/100*250 = 200 Total initial cost = 60*200=12000 Total profit = (54*250+6*100) - 12000 = 2100 Profit % as a percent of dealer's initial cost = 2100/12000*100 = 17.5 % Can you please tell me where I went wrong ? Posted from my mobile device Math Expert Joined: 02 Sep 2009 Posts: 48037 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 30 Jun 2014, 02:50 kshitij89 wrote: My soln is : Dealer's per unit initial cost =250 -20/100*250 = 200 Total initial cost = 60*200=12000 Total profit = (54*250+6*100) - 12000 = 2100 Profit % as a percent of dealer's initial cost = 2100/12000*100 = 17.5 % Can you please tell me where I went wrong ? Posted from my mobile device The cost per unit is not 0.8*250 = 200, it's 250/1.2 = ~208. Markup is calculated on the cost value (check here: http://gmatclub.com/forum/a-photography ... l#p1229596). Hope it helps. _________________ Manager Joined: 28 Dec 2013 Posts: 68 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 21 Aug 2014, 08:48 VeritasPrepKarishma wrote: pgmat wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras?

A. 7% loss
B. 13% loss
C. 7% profit
D. 13% profit
E. 15% profit

Use weighted avgs for a quick solution:

On 54 cameras, the dealer made a 20% profit and on 6 cameras, he made a 50% loss. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1

Avg Profit/Loss = (.2*9 + (-.5)*1)/10 = 1.3/10 = .13 = 13% profit

For more on weighted avgs, check: http://www.veritasprep.com/blog/2011/03 ... -averages/

why are we dividing by 10?
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Re: A photography dealer ordered 60 Model X cameras to be sold  [#permalink]

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24 Aug 2014, 21:54
1
sagnik242 wrote:
VeritasPrepKarishma wrote:
pgmat wrote:
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit Use weighted avgs for a quick solution: On 54 cameras, the dealer made a 20% profit and on 6 cameras, he made a 50% loss. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1 Avg Profit/Loss = (.2*9 + (-.5)*1)/10 = 1.3/10 = .13 = 13% profit For more on weighted avgs, check: http://www.veritasprep.com/blog/2011/03 ... -averages/ why are we dividing by 10? Weighted Average Formula: Cavg = (C1*w1 + C2*w2)/(w1 + w2) C1 and C2 represent the quantity which we want to average so they will be profit/loss here. C1 = 20% = .2 C2 = -50% (loss) = -.5 The weights, given by w1 and w2, are the cost prices. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1 So w1 =9 and w2 = 1 Avg Profit/Loss = (.2*9 + (-.5)*1)/(9 + 1) = (.2*9 + (-.5)*1)/10 Get more details on this concept from the link given in my post above. _________________ Karishma Veritas Prep GMAT Instructor Save up to$1,000 on GMAT prep through 8/20! Learn more here >

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Re: A photography dealer ordered 60 Model X cameras to be sold &nbs [#permalink] 24 Aug 2014, 21:54

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# If x is an integer then x(x-1)(x-k) must be evenly divisible

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If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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31 Jan 2012, 15:49
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If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT

A. -4
B. -2
C. -1
D. 2
E. 5

[Reveal] Spoiler:
The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?

Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.
[Reveal] Spoiler: OA

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31 Jan 2012, 16:06
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enigma123 wrote:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT

A)-4
B)-2
C)-1
D) 2
E) 5

The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?

Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.

We have the product of 3 integers: (x-1)x(x-k).

Note that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x – 1), and (x – k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.

Next, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.

30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.

Hope it helps.
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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31 Jan 2012, 16:09
Thanks very much for a thorough explanation.
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Hi, there. I'm happy to help with this.

The rule that the product of three consecutive integers is a good start, but not the be all and end all.

number = x(x – 1)(x – k)

So far, we have integer x and one less that it (x - 1), so we could go down one more, or up one from x --

k = 2 ----> x(x – 1)(x – 2)

k = -1 ----> x(x – 1)(x + 1)

Now, we don't know which of the three factors are divisible by 3 -- x, or (x - 1), or the (x - k). If it's either of the first two, then we're golden, and k doesn't matter. But pretend that neither x nor (x - 1) is divisible by 3, then we are dependent on that last factor. Well, if (x - 2) is a multiple of three, we should be able to add or subtract three and still get a multiple of three.

(x - 2) - 3 = (x - 5)

(x - 5) - 3 = (x - 8)

(x - 2) + 3 = (x + 1), which we have already

(x + 1) + 3 = (x + 4)

(x + 3) + 3 = (x + 7)

So, for divisibility purposes, (x - 8), (x - 5), (x - 2), (x + 1), (x + 4), (x + 7) are all equivalent -- if any one of them is a multiple of three, all the others are. (You can check that the difference between any two is a multiple of 3.) BTW, if they are not divisible by three, then they all would have equal remainders if divided by three.

Back to the question:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A)-4
B)-2
C)-1
D) 2
E) 5

All of those choices give us a term on our list except for (B) -2.

BTW, notice all the answer choices are spaced apart by three except for (B).

Does that make sense? Please do not hesitate to ask if you have any questions.

Mike
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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31 Jan 2012, 16:18
Thanks Mike. Really appreciate your solution too.
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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31 Jan 2012, 16:34
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This is my 1st post finally thought of jumping in instead of just being an observer
I attacked this problem in a simple way. As it states it is divisible by 3
that means both x & (x-1) cannot be a multiple of 3 otherwise whatever the value of k it will be still divisible by 3
so plugging in number i chose 5 in this case you can establish answer is -2 does not fit...

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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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31 Jan 2012, 16:37
Thanks and welcome to GMAT Club Azim. I am sure you will have a great experience from a very helpful community.
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04 Oct 2012, 18:59
Bunuel wrote:
enigma123 wrote:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT

A)-4
B)-2
C)-1
D) 2
E) 5

The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?

Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.

We have the product of 3 integers: (x-1)x(x-k).

Note that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x – 1), and (x – k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.

Next, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.

30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.

Hope it helps.

Bunuel,

Would this approach work for all integer divisors (NOT the 30 sec approach)?
Say the divisor is 4 and the choices had four terms instead of three, e.g. (x-1)(x-2)(x+k)(x+1)?

Thanks,

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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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04 Oct 2012, 23:10
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enigma123 wrote:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT

A. -4
B. -2
C. -1
D. 2
E. 5

The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?

Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.

Since this is a multiple choice GMAT question, you can pick a particular value for x such that neither x, nor x-1 is divisible by 3 and start checking the answers.
In the given situation, choose for example x = 2 and check when 2 - k is not divisible by 3.
(A) 2 - (-4) = 6 NO
(B) 2 - (-2) = 4 BINGO!

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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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05 Oct 2012, 03:37
I too solved it using a value for x =2, but I am not sure if is it better to solve using value for such questions or otherwise.

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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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05 Oct 2012, 03:56
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Avantika5 wrote:
I too solved it using a value for x =2, but I am not sure if is it better to solve using value for such questions or otherwise.

On the GMAT, is definitely the fastest way to solve it. Being a multiple choice question, you can be sure that there is a unique correct answer. And in the given situation, the only issue is to choose for x values such that neither x, nor x - 1 is divisible by 3.
It won't harm to understand and know to use the properties of consecutive integers presented in the other posts . They can be useful any time.
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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06 Oct 2012, 01:10
Thanks EvaJager, I was always thought it is not good way to solve and I should learn the better way.
Thanks a lot...

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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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06 Oct 2012, 02:10
Avantika5 wrote:
Thanks EvaJager, I was always thought it is not good way to solve and I should learn the better way.
Thanks a lot...

Better is a relative word...Mathematicians always try to prove and justify everything in a formal, logical way.
But GMAT is not testing mathematical abilities per se. If they wanted so, the questions would have been open and not multiple choice.
Have a flexible mind, think out of the box. GMAT is not a contest for the most beautiful, elegant, mathematical solution...
Get the correct answer as quickly as possible, and go to the next question without any feeling of guilt...:O)

Though, as I said, try to understand the other properties of the integer numbers, they can become handy and also, because they are so beautiful! Isn't Mathematics wonderful?
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Last edited by EvaJager on 06 Oct 2012, 02:46, edited 1 time in total.

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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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06 Oct 2012, 02:25
1. k=-4, x+4=y so x=y-4. Now ----- (y-4)(y-5)(y). Gives number divisible by 3 in all cases.
2. k=-2, , x+2=y so x=y-2. Now ----- (y-2)(y-3)(y). Not applicable when y is 5 or when x is 7
3. k=-1, , x+1=y so x=y-1. Now ----- (y-1)(y-2)(y). Consecutive 3 integers. Divisible by 3
4. k=2, , x-2=y so x=y+2. Now ----- (y+2)(y+1)(y). Consecutive 3 integers. Divisible by 3
5. k=5, , x-5=y so x=y+5. Now ----- (y+5)(y+4)(y). Consecutive 3 integers. Divisible by 3
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16 Feb 2013, 12:34
mikemcgarry wrote:
Hi, there. I'm happy to help with this.

The rule that the product of three consecutive integers is a good start, but not the be all and end all.

number = x(x – 1)(x – k)

So far, we have integer x and one less that it (x - 1), so we could go down one more, or up one from x --

k = 2 ----> x(x – 1)(x – 2)

k = -1 ----> x(x – 1)(x + 1)

Now, we don't know which of the three factors are divisible by 3 -- x, or (x - 1), or the (x - k). If it's either of the first two, then we're golden, and k doesn't matter. But pretend that neither x nor (x - 1) is divisible by 3, then we are dependent on that last factor. Well, if (x - 2) is a multiple of three, we should be able to add or subtract three and still get a multiple of three.

(x - 2) - 3 = (x - 5)

(x - 5) - 3 = (x - 8)

(x - 2) + 3 = (x + 1), which we have already

(x + 1) + 3 = (x + 4)

(x + 3) + 3 = (x + 7)

So, for divisibility purposes, (x - 8), (x - 5), (x - 2), (x + 1), (x + 4), (x + 7) are all equivalent -- if any one of them is a multiple of three, all the others are. (You can check that the difference between any two is a multiple of 3.) BTW, if they are not divisible by three, then they all would have equal remainders if divided by three.

Back to the question:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A)-4
B)-2
C)-1
D) 2
E) 5

All of those choices give us a term on our list except for (B) -2.

BTW, notice all the answer choices are spaced apart by three except for (B).

Does that make sense? Please do not hesitate to ask if you have any questions.

Mike

Great posts everyone, including you Mike, who I am now quoting, but I would like to add that..

The first two integers for each answer choice cannot be divisible by 3 because all the answers choices would then be correct, thereby making the question invalid. Therefore, the last integer (x+k) must be divisible by 3, which helps in the theoretical approach...

It appears that was left out

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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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17 Oct 2013, 15:25
enigma123 wrote:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT

A. -4
B. -2
C. -1
D. 2
E. 5

The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?

Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.

Hi enigma123. Thank you for the nice question. But you might wanna put a spoiler on this statement The OA is B so that others can solve the question in real conditions. Thanks

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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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I tried the problem in a different way:

for any number to be divisible by 3, the sum of the integers in the number should be a factor of 3.

Taking the sum of x, (x-1) and (x-k) we have:

x+x-1+x-k = 3x-1-k

now looking at the choices

k = -4 => sum = 3x+3 --> divisible by 3
k = -2 => sum = 3x+1 --> not divisible by 3
k = -1 => sum = 3x --> divisible by 3
k = 2 => sum = 3x-3 --> divisible by 3
k = 5 ==> sum = 3x-6 --> divisible by 3

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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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17 Jun 2015, 02:13
Hi there,

The answer is right, however my explanation is quite differently:

1. x(x-1)(x-k)
2. (x^2-1) (x-k)
3. x^3-x^2k-x^2+xk
4. x-xk

Plugg-in x=1
a. 1-1(-4) = 1-(-4) = 5
b. 1-1(-2) = 1-(-2) = 3
c. 1-1(-1) = 1-(-1) = 2
d. 1-1(2) = 1-(2) = -1
e. 1-1(5) = 1-(5) = -4

Am I correct in my explanation?

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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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17 Jun 2015, 05:49
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Expert's post
santorasantu wrote:
Quote:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT

A. -4
B. -2
C. -1
D. 2
E. 5

I tried the problem in a different way:

for any number to be divisible by 3, the sum of the integers in the number should be a factor of 3.

Taking the sum of x, (x-1) and (x-k) we have:

x+x-1+x-k = 3x-1-k

now looking at the choices

k = -4 => sum = 3x+3 --> divisible by 3
k = -2 => sum = 3x+1 --> not divisible by 3
k = -1 => sum = 3x --> divisible by 3
k = 2 => sum = 3x-3 --> divisible by 3
k = 5 ==> sum = 3x-6 --> divisible by 3

Just refining the highlighted language and presenting another view to del with this problem

CONCEPT1:For any number to be divisible by 3, the sum of the Digits of the Number should be a Multiple of 3.

CONCEPT2: Product of any three consecutive Integers always include one muliple of 3 hence product of any three consecutive Integers is always a Multiple of 3

CONCEPT3: If a Number "w" is a Multiple of 3 then any number at a difference of 3 or multiple of 3 from "w" will also be a multiple of 3 i.e. If w is a multiple of 3 then (w+3), (w-3), (w+6), (w-6) etc. will all be Multiples of 3

Here, I see that x(x – 1) is a product of two consecutive Integers but if another Number next to them is obtained then x(x – 1)(x – k) will certainly be a multiple of 3 [as per Concept2 mentioned above]

for x(x – 1)(x – k) to be a product of 3 consecutive integers,

(x – k) should be either (x - 2) i.e. k=2
OR
(x – k) should be either (x - 5) i.e. k=5 [Using Concept3]
OR
(x – k) should be either (x + 1) i.e. k=-1
OR
(x – k) should be either (x + 4) i.e. k=-4 [Using Concept3]

This Eliminates options A, C D and E

[Reveal] Spoiler:
B

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If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]

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04 Nov 2016, 04:23
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enigma123 wrote:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT

A. -4
B. -2
C. -1
D. 2
E. 5

[Reveal] Spoiler:
The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?

Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.

A little bit faster approach.

We know from the start that product of $$3$$ consecutive integers is divisible by $$3$$, because it’s indeed a multiple of $$3$$.

So we have $$x(x+1)(x+2)$$ is definitely divisible by 3 and k=2 is our anchor point.

Now we can use principles of modular arithmetic to generate multiples of $$3$$. We can either add or subtract $$3$$ from our anchor number. And we got following string of numbers:

… -4, -1, 2, 5, 7 …

As you can see in this progression only answer B (-2) is absent.

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