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# Inequalities trick

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Inequalities trick [#permalink]  16 Mar 2010, 09:11
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I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.

Suppose you have the inequality

f(x) = (x-a)(x-b)(x-c)(x-d) < 0

Just arrange them in order as shown in the picture and draw curve starting from + from right.

now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.

Don't forget to arrange then in ascending order from left to right. a<b<c<d

So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)

If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.

For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.
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Re: Inequalities trick [#permalink]  22 Oct 2010, 05:33
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Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain.

If (x-a)(x-b)(x-c)(x-d) < 0, we can draw the points a, b, c and d on the number line.
e.g. Given (x+2)(x-1)(x-7)(x-4) < 0, draw the points -2, 1, 7 and 4 on the number line as shown.

Attachment:

doc.jpg [ 7.9 KiB | Viewed 21052 times ]

This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive.

When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x - 7) will be negative since x < 7 in this region. All other factors are still positive.

When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x - 7) and (x - 4) are negative, but negative x negative is positive... and so on till you reach the leftmost section.

Since we are looking for values of x where the expression is < 0, here the solution will be -2 < x < 1 or 4< x < 7

It should be obvious that it will also work in cases where factors are divided.
e.g. (x - a)(x - b)/(x - c)(x - d) < 0
(x + 2)(x - 1)/(x -4)(x - 7) < 0 will have exactly the same solution as above.

Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion.
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Math Forum Moderator Joined: 20 Dec 2010 Posts: 2047 Followers: 128 Kudos [?]: 919 [13] , given: 376 Re: Inequalities trick [#permalink] 11 Mar 2011, 05:49 13 This post received KUDOS 4 This post was BOOKMARKED vjsharma25 wrote: VeritasPrepKarishma wrote: vjsharma25 wrote: How you have decided on the first sign of the graph?Why it is -ve if it has three factors and +ve when four factors? Check out my post above for explanation. I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? I always struggle with this as well!!! There is a trick Bunuel suggested; (x+2)(x-1)(x-7) < 0 Here the roots are; -2,1,7 Arrange them in ascending order; -2,1,7; These are three points where the wave will alternate. The ranges are; x<-2 -2<x<1 1<x<7 x>7 Take a big value of x; say 1000; you see the inequality will be positive for that. (1000+2)(1000-1)(1000-7) is +ve. Thus the last range(x>7) is on the positive side. Graph is +ve after 7. Between 1 and 7-> -ve between -2 and 1-> +ve Before -2 -> -ve Since the inequality has the less than sign; consider only the -ve side of the graph; 1<x<7 or x<-2 is the complete range of x that satisfies the inequality. _________________ Manager Joined: 29 Sep 2008 Posts: 154 Followers: 2 Kudos [?]: 31 [8] , given: 1 Re: Inequalities trick [#permalink] 22 Oct 2010, 10:45 8 This post received KUDOS 3 This post was BOOKMARKED if = sign is included with < then <= will be there in solution like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7 in case when factors are divided then the numerator will contain = sign like for (x + 2)(x - 1)/(x -4)(x - 7) < =0 the solution will be -2 <= x <= 1 or 4< x < 7 we cant make 4<=x<=7 as it will make the solution infinite correct me if i am wrong Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4755 Location: Pune, India Followers: 1112 Kudos [?]: 5022 [4] , given: 164 Re: Inequalities trick [#permalink] 11 Mar 2011, 18:57 4 This post received KUDOS Expert's post vjsharma25 wrote: I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? Ok, look at this expression inequality: (x+2)(x-1)(x-7) < 0 Can I say the left hand side expression will always be positive for values greater than 7? (x+2) will be positive, (x - 1) will be positive and (x-7) will also be positive... so in the rightmost regions i.e. x > 7, all three factors will be positive. The expression will be positive when x > 7, it will be negative when 1 < x < 7, positive when -2 , x < 1 and negative when x < -2. We need the region where the expression is less than 0 i.e. negative. So either 1 < x < 7 or x < -2. Now let me add another factor: (x+8)(x+2)(x-1)(x-7) Can I still say that the entire expression is positive in the rightmost region i.e. x>7 because each one of the four factors is positive? Yes. So basically, your rightmost region is always positive. You go from there and assign + and - signs to the regions. Your starting point is the rightmost region. Note: Make sure that the factors are of the form (ax - b), not (b - ax)... e.g. (x+2)(x-1)(7 - x)<0 Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1') Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-' Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: Inequalities trick [#permalink]  10 Aug 2011, 16:01
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WoW - This is a cool thread with so many thing on inequalities....I have compiled it together with some of my own ideas...It should help.

1) CORE CONCEPT
@gurpreetsingh -
Suppose you have the inequality

f(x) = (x-a)(x-b)(x-c)(x-d) < 0

Arrange the NUMBERS in ascending order from left to right. a<b<c<d
Draw curve starting from + from right.

now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.

So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)

If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.

For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.

Note: Make sure that the factors are of the form (ax - b), not (b - ax)...

example -
(x+2)(x-1)(7 - x)<0

Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1')
Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-'
Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0.

2) Variation - ODD/EVEN POWER
@ulm/Karishma -
if we have even powers like (x-a)^2(x-b)
we don't need to change a sign when jump over "a".
This will be same as (x-b)

We can ignore squares BUT SHOULD consider ODD powers
example -
2.a
(x-a)^3(x-b)<0 is the same as (x-a)(x-b) <0
2.b
(x - a)(x - b)/(x - c)(x - d) < 0 ==> (x - a)(x - b)(x-c)^-1(x-d)^-1 <0
is the same as (x - a)(x - b)(x - c)(x - d) < 0

3) Variation <= in FRACTION
@mrinal2100 -
if = sign is included with < then <= will be there in solution
like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7

BUT if it is a fraction the denominator in the solution will not have = SIGN

example -
3.a
(x + 2)(x - 1)/(x -4)(x - 7) < =0
the solution will be -2 <= x <= 1 or 4< x < 7
we cant make 4<=x<=7 as it will make the solution infinite

4) Variation - ROOTS
@Karishma -
As for roots, you have to keep in mind that given \sqrt{x}, x cannot be negative.

\sqrt{x} < 10
implies 0 < \sqrt{x} < 10
Squaring, 0 < x < 100
Root questions are specific. You have to be careful. If you have a particular question in mind, send it.

Refer - inequalities-and-roots-118619.html#p959939
Some more useful tips for ROOTS....I am too lazy to consolidate

<5> THESIS -
@gmat1220 -
Once algebra teacher told me - signs alternate between the roots. I said whatever and now I know why Watching this article is a stroll down the memory lane.

I will save this future references....

Anyone wants to add ABSOLUTE VALUES....That will be a value add to this post

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Re: Inequalities trick [#permalink]  09 Sep 2013, 22:35
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karannanda wrote:
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.

Suppose you have the inequality

f(x) = (x-a)(x-b)(x-c)(x-d) < 0

Just arrange them in order as shown in the picture and draw curve starting from + from right.

now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.

Don't forget to arrange then in ascending order from left to right. a<b<c<d

So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)

If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.

For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.

Hi Gurpreet,
Thanks for the wonderful method.
I am trying to understand it so that i can apply it in tests.
Can you help me in applying this method to the below expression to find range of x.
x^3 – 4x^5 < 0?

I am getting the roots as -1/2, 0, 1/2 and when i plot them using this method, putting + in the rightmost region, I am not getting correct result. Not sure where i am going wrong. Can you pls help.

Before you apply the method, ensure that the factors are of the form (x - a)(x - b) etc

x^3 - 4x^5 < 0

x^3 ( 1 - 4x^2) < 0

x^3(1 - 2x) (1 + 2x) < 0

4x^3(x - 1/2)(x + 1/2) > 0 (Notice the flipped sign. We multiplied both sides by -1 to convert 1/2 - x to x - 1/2)

Now the transition points are 0, -1/2 and 1/2 so put + in the rightmost region.
The solution will be x > 1/2 or -1/2 < x< 0.

Check out these posts discussing such complications:
http://www.veritasprep.com/blog/2012/06 ... e-factors/
http://www.veritasprep.com/blog/2012/07 ... ns-part-i/
http://www.veritasprep.com/blog/2012/07 ... s-part-ii/
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Senior Manager Joined: 16 Feb 2012 Posts: 260 Concentration: Finance, Economics Followers: 4 Kudos [?]: 65 [1] , given: 110 Re: Inequalities trick [#permalink] 22 Jul 2012, 02:03 1 This post received KUDOS VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation\frac {(x+2)(x-1)}{(x-4)(x-7)} were 4\leq x \leq 7, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation. _________________ Kudos if you like the post! Failing to plan is planning to fail. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4755 Location: Pune, India Followers: 1112 Kudos [?]: 5022 [1] , given: 164 Re: Inequalities trick [#permalink] 23 Jul 2012, 02:13 1 This post received KUDOS Expert's post Stiv wrote: VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation\frac {(x+2)(x-1)}{(x-4)(x-7)} were 4\leq x \leq 7, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation. x cannot be equal to 4 or 7 because if x = 4 or x = 7, the denominator will be 0 and the expression will not be defined. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: Inequalities trick [#permalink]  18 Oct 2012, 04:17
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GMATBaumgartner wrote:
gurpreetsingh wrote:
ulm wrote:
if we have smth like (x-a)^2(x-b)
we don't need to change a sign when jump over "a".

yes even powers wont contribute to the inequality sign. But be wary of the root value of x=a

Hi Gurpreet,
Could you elaborate what exactly you meant here in highlighted text ?

Even I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above.

If the powers are even then the inequality won't be affected.

eg if u have to find the range of values of x satisfying (x-a)^2 *(x-b)(x-c) >0

just use (x-b)*(x-c) >0 because x-a raised to the power 2 will not affect the inequality sign. But just make sure x=a is taken care off , as it would make the inequality zero.
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Re: Inequalities trick [#permalink]  18 Oct 2012, 09:27
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GMATBaumgartner wrote:
Hi Gurpreet,
Could you elaborate what exactly you meant here in highlighted text ?

Even I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above.

In addition, you can check out this post:

http://www.veritasprep.com/blog/2012/07 ... s-part-ii/

I have discussed how to handle powers in it.
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 26582 Followers: 3487 Kudos [?]: 26158 [1] , given: 2705 Re: Inequalities trick [#permalink] 02 Dec 2012, 06:25 1 This post received KUDOS Expert's post GMATGURU1 wrote: Going ahead in this inequalities area of GMAT, can some one has the problem numbers of inequalities in OG 11 and OG 12? Cheers, Danny Search for hundreds of question with solutions by tags: viewforumtags.php DS questions on inequalities: search.php?search_id=tag&tag_id=184 PS questions on inequalities: search.php?search_id=tag&tag_id=189 Hardest DS inequality questions with detailed solutions: inequality-and-absolute-value-questions-from-my-collection-86939.html Hope it helps. _________________ CEO Status: Nothing comes easy: neither do I want. Joined: 12 Oct 2009 Posts: 2793 Location: Malaysia Concentration: Technology, Entrepreneurship Schools: ISB '15 (M) GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Followers: 178 Kudos [?]: 945 [1] , given: 235 Re: Inequalities trick [#permalink] 20 Dec 2012, 20:04 1 This post received KUDOS VeritasPrepKarishma wrote: The entire concept is based on positive/negative factors which means <0 or >0 is a must. If the question is not in this format, you need to bring it to this format by taking the constant to the left hand side. e.g. (x + 2)(x + 3) < 2 x^2 + 5x + 6 - 2 < 0 x^2 + 5x + 4 < 0 (x+4)(x+1) < 0 Now use the concept. Yes this is probable but it might not be possible always to group them. So in case you are unsure just follow the number plugging approach. But most of the times this trick would be very handy. _________________ Fight for your dreams :For all those who fear from Verbal- lets give it a fight Money Saved is the Money Earned Jo Bole So Nihaal , Sat Shri Akaal Support GMAT Club by putting a GMAT Club badge on your blog/Facebook Get the best GMAT Prep Resources with GMAT Club Premium Membership Gmat test review : 670-to-710-a-long-journey-without-destination-still-happy-141642.html Math Expert Joined: 02 Sep 2009 Posts: 26582 Followers: 3487 Kudos [?]: 26158 [1] , given: 2705 Re: Inequalities trick [#permalink] 11 Nov 2013, 06:51 1 This post received KUDOS Expert's post anujpadia wrote: Can you please explain the above mentioned concept in relation to the following question? Is a > O? (1) a^3 - 0 < 0 (2) 1- a^2 > 0 Can you please explain the scenario when (x-a)(x-B)(x-C)(x-d)>0? Sorry, but finding it difficult to understand. Check alternative solutions here: is-a-0-1-a-3-a-0-2-1-a-86749.html Hope this helps. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4755 Location: Pune, India Followers: 1112 Kudos [?]: 5022 [1] , given: 164 Re: Inequalities trick [#permalink] 23 Apr 2014, 03:43 1 This post received KUDOS Expert's post 1 This post was BOOKMARKED PathFinder007 wrote: I have a query. I have following question x^3 - 4x^5 < 0 I can define this as (1+2x).x^3(1-2x). now I have roots -1/2, 0, 1/2. so in case of >1/2 I will always get inequality value as <0 and in case of -1/2 and 0 I will get value as 0. So How I will define them in graph and what range I will consider for this inequality. Thanks The factors must be of the form (x - a)(x - b) .... etc x^3 - 4x^5 < 0 x^3 * (1 - 4x^2) < 0 x^3 * (1 - 2x) * (1 + 2x) < 0 x^3 * (2x - 1) * (2x + 1) > 0 (Note the sign flip because 1-2x was changed to 2x - 1) x^3 * 2(x - 1/2) *2(x + 1/2) > 0 So transition points are 0, 1/2 and -1/2. ____________ - 1/2 _____ 0 ______1/2 _________ This is what it looks like on the number line. The rightmost region is positive. We want the positive regions in the inequality. So the desired range of x is given by x > 1/2 or -1/2 < x< 0 For more on this method, check these posts: http://www.veritasprep.com/blog/2012/06 ... e-factors/ http://www.veritasprep.com/blog/2012/07 ... ns-part-i/ http://www.veritasprep.com/blog/2012/07 ... s-part-ii/ The links will give you the theory behind this method in detail. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: Inequalities trick [#permalink]  16 Mar 2010, 09:46
Can u plz explainn the backgoround of this & then the explanation.

Thanks
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Re: Inequalities trick [#permalink]  19 Mar 2010, 11:48
I have applied this trick and it seemed to be quite useful.
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Re: Inequalities trick [#permalink]  19 Mar 2010, 11:59
ttks10 wrote:
Can u plz explainn the backgoround of this & then the explanation.

Thanks

i m sorry i dont have any background for it, you just re-read it again and try to implement whenever you get such question and I will help you out in any issue.

sidhu4u wrote:
I have applied this trick and it seemed to be quite useful.

Nice to hear this....good luck.
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Re: Inequalities trick [#permalink]  17 Oct 2010, 15:14
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.

So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) < 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)

I don't understand this part alone. Can you please explain?
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Re: Inequalities trick [#permalink]  17 Oct 2010, 16:19
Dreamy wrote:
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.

So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) < 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)

I don't understand this part alone. Can you please explain?

Suppose you have the inequality

f(x) = (x-a)(x-b)(x-c)(x-d) < 0 you will consider the curve with -ve inside it.. check the attached image.

f(x) = (x-a)(x-b)(x-c)(x-d) > 0 consider the +ve of the curve
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# What arithmetic should I memorize?

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Updated on: 19 Oct 2018, 13:30
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Improve Your Speed - GMAT Arithmetic to Memorize

I find myself constantly using long division and multiplication for simple things like 13 x 11.
What are some good arithmetic calcs to memorize? I created a 20x20 multiplication table but this seems like a bit much right?

Perfect squares up to 100? $$\sqrt{3}$$? $$\sqrt{5}$$?

I am not really talking about formulas to memorize since you should definitely memorize things like nCr, nPr, sum of all #'s in an evenly spaced set, etc.

Edit: Moderator note: Attached are files we gathered in this post from several other sources in addition to the author. Also see additional resources in posts below to help with improve your speed and reduce mistakes

Target Test Prep 15-page PDF summarizes ALL formulas you have to know

-
Attachments

GMAT Math Compendium.xls [855 KiB]

Geometry Formulas_template.pdf [44.28 KiB]

Geometry Formulas.pdf [72.21 KiB]

Arithmetic Fractions and Percentages.doc [32.5 KiB]

Originally posted by topher on 25 Jun 2009, 20:39.
Last edited by bb on 19 Oct 2018, 13:30, edited 7 times in total.
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Re: What arithmetic should I memorize?  [#permalink]

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25 Jun 2009, 21:18
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Fantastic question

This file should give you an idea where you are lacking.
I think you should definitely know the squares from 0-10 and preferrably from 10 to 20 as well.
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Arithmetic Fractions and Percentages.doc [32.5 KiB]

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Re: What arithmetic should I memorize?  [#permalink]

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21 Oct 2009, 23:59
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As far as common squares are concerned, I only remember the ones with 0 or 5 in the units digit. For the latter category, I use the following process:

For example: 65*65

1) Always write down 25 as this is always the last two digits of the result:
...25

2) Multiply (non-units digits) times (non-units digits + 1)
6 * (6+1) = 42

2c) Combine:
4225

This way I always have the important squares handy... very useful for estimations!

Any other math shortcuts? Anyone

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Re: What arithmetic should I memorize?  [#permalink]

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26 Jul 2009, 09:36
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Hi all,

I created a template that gather all the geometry formulas that appear on the GMAT (if i missed any please tell!). I use it every week in order to ensure that I remember all of them.

I think it could be helpful for some of you so I decided to share.

The TEMPLATE is the one for practice.

Good luck!
Attachments

Geometry Formulas_template.pdf [44.28 KiB]

Geometry Formulas.pdf [72.21 KiB]

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26 Jul 2009, 19:33
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Great resource

Thank you!

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Re: What arithmetic should I memorize?  [#permalink]

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Updated on: 26 Jan 2010, 12:23
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My fellow GMAT clubbers,

I'd like to share the excel spreadsheet in which I have been compiling all knowledge and shortcuts relevant for tackling the GMAT quant section.

It is organized by tabs. Each tab covers a different area.

Examples:
- Tab "NP. Primes" covers: Number Properties - Primes
- Tab "NP. Powers-Rts" covers: Number Properties - Patterns of powers and roots
- Tab "G. PTriples" covers: Geometry - Phytagorean triplets patterns
- Tab "G. Ci-Sq (Pi)" covers: Geometry - Relantionships between the measures of inscribed/circumscribed circles & squares
- Tab "WT. Prob" covers: Wort Translations - Probability (dice/coins)

Finally, three of the tabs are summaries made of the Man guides "Word Translations", "Number Properties" and "Geometry".

Hope it helps somebody.

Cheers,
Attachments

GMAT Math Compendium.xls [855 KiB]

Originally posted by powerka on 08 Sep 2009, 16:51.
Last edited by powerka on 26 Jan 2010, 12:23, edited 3 times in total.
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Re: What arithmetic should I memorize?  [#permalink]

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31 Oct 2009, 08:56
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I'm new here, and try to go through all the topics - this site is a treasure!

Maybe not the right place to share my experience with "multiplication for simple things like 13 x 11", but anyway...if you need to multiply any two-digits number by 11, just sum those digits and put the result in between. For example,
13x11 -> 1+3=4 -> 143 is the result.
Or, 36x11 -> 3+6=9 -> the result is 36x11=396.

It really saves time.
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Re: What arithmetic should I memorize?  [#permalink]

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31 Oct 2009, 19:21
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Shelen wrote:
I'm new here, and try to go through all the topics - this site is a treasure!

Maybe not the right place to share my experience with "multiplication for simple things like 13 x 11", but anyway...if you need to multiply any two-digits number by 11, just sum those digits and put the result in between. For example,
13x11 -> 1+3=4 -> 143 is the result.
Or, 36x11 -> 3+6=9 -> the result is 36x11=396.

It really saves time.

And if it equals more than 10, add a 1 to the first digit

EG

68*11 -> 6+8=14 -> 6 14 8 -> 748
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Re: What arithmetic should I memorize?  [#permalink]

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07 Jan 2010, 08:47
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Sorry, but on page 2 there is a misprint:
in Arc Measure (Intersecting Secants/Tangents) It must be the difference between arcs measures of AD and BD.
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Re: What arithmetic should I memorize?  [#permalink]

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07 Jan 2010, 16:13
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Also, see this post for some good theory: math-number-theory-88376.html
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Re: What arithmetic should I memorize?  [#permalink]

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11 Feb 2010, 09:34
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Thanks for sharing the Doc bb...@Shelen & @jeckll...Thanks for the tip but I would like to add some more.

What if u need to multiply 3 digits or 4 didgits by 11...the procedure is same as u mentioned but it would be done like below

For 3 digits
133x11 --> 1 1+3 3+3 3=1463
And for 4 digits
1243x11 --> 1 1+2 2+4 4+3 3=13673

And for 5 digits
15453x11 --> 1 1+5 5+4 4+5 5+3 3=169983
and so on.
you can plug other numbers in and check it out.

Hope it helps!
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Re: What arithmetic should I memorize?  [#permalink]

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11 Feb 2010, 21:41
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Guys! I just found another way of checking whether a number is divisible by 8 or not ( the rule is same but another approach or route ). It's for a number with more than two digits.

Let's take 1936
1) First of all check whether last two digits of the number are divisible by 4 or not.
For 1936, we do this way 36/4=9

2) If it is divisible by 4 then add the quotient to the 3rd last digit of the number and if the sum of them is divisible by 2 then the whole number is divisible by 8.

--> 9 (quotient)+ 9 ( 3rd digit from right)= 18, and -->18/2=9
So the whole number is divisible by 8.

Once you understand it and do a little practice, you'll find it easy and fast.
**You can try other numbers to see whether it is true or not
Hope it helps!
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Re: What arithmetic should I memorize?  [#permalink]

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14 Feb 2010, 05:12
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AtifS wrote:
Thanks for sharing the Doc bb...@Shelen & @jeckll...Thanks for the tip but I would like to add some more.

What if u need to multiply 3 digits or 4 didgits by 11...the procedure is same as u mentioned but it would be done like below

For 3 digits
133x11 --> 1 1+3 3+3 3=1463
And for 4 digits
1243x11 --> 1 1+2 2+4 4+3 3=13673

And for 5 digits
15453x11 --> 1 1+5 5+4 4+5 5+3 3=169983
and so on.
you can plug other numbers in and check it out.

Hope it helps!

welldone dude,
just a small addition which i tried and will help to avoid confusion:

start writing the answer from right hand side in case if the addition of two no. exceeds 10, and add it to consecutive no. on left hand side. try out!!
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Re: What arithmetic should I memorize?  [#permalink]

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14 Feb 2010, 05:21
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AtifS wrote:
Guys! I just found another way of checking whether a number is divisible by 8 or not ( the rule is same but another approach or route ). It's for a number with more than two digits.

Let's take 1936
1) First of all check whether last two digits of the number are divisible by 4 or not.
For 1936, we do this way 36/4=9

2) If it is divisible by 4 then add the quotient to the 3rd last digit of the number and if the sum of them is divisible by 2 then the whole number is divisible by 8.

--> 9 (quotient)+ 9 ( 3rd digit from right)= 18, and -->18/2=9
So the whole number is divisible by 8.

Once you understand it and do a little practice, you'll find it easy and fast.
**You can try other numbers to see whether it is true or not
Hope it helps!

hi,
i feel the process is bit complicated as it doesn't give the value of quotient, it just tells you whether no. is divisible by 8.
here is another tric, if the no. formed by last three digits is divisible by 8 then the whole no. is divisible by 8.

953360 is divisible by 8 since 360 is divisible by 8,
529418: not divisible as 418 is not divisible by 8.

plug in different values and try.
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Re: What arithmetic should I memorize?  [#permalink]

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17 Feb 2010, 18:53
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RE:
What if u need to multiply 3 digits or 4 didgits by 11...the procedure is same as u mentioned but it would be done like below

For 3 digits
133x11 --> 1 1+3 3+3 3=1463
And for 4 digits
1243x11 --> 1 1+2 2+4 4+3 3=13673

And for 5 digits
15453x11 --> 1 1+5 5+4 4+5 5+3 3=169983
and so on.

Hi guys,
When multiplying by 11 I find it much easier to multiply the # by 10 (or add 0) and then add the original # to it.
For example, 1234x11 = 12340 + 1234

When dividing by 11 and the following condition is satisfied one could factor the # as follows:
671/11 = 61x11/11 = 61 because in 671, 6+1=7 which is the # in the middle of 671. This works for 3digit #'s
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Re: What arithmetic should I memorize?  [#permalink]

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Updated on: 05 Oct 2010, 10:03
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Now, what does this mean:

Area of Isosceles Triangle = 1/2[(leg)^2]
(from Geometry Formulas.pdf)

I haven't been able to give myself any convincing explanation of this formula...

Originally posted by rishabhsingla on 11 Sep 2010, 10:27.
Last edited by rishabhsingla on 05 Oct 2010, 10:03, edited 1 time in total.
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Re: What arithmetic should I memorize?  [#permalink]

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11 Sep 2010, 10:42
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rishabhsingla wrote:
Now, what does this mean:

Area of Isosceles Triangle = 1/2[(leg)^2]
(from Geometry Formulas.pdf)

I haven't been able to give myself any convincing explanation of this formula...

Well that's because it's not true.

It should be area of isosceles right triangle = 1/2*(leg)^2, simply because area of right triangle equals to 1/2*leg1*leg2, and since in isosceles right triangle leg1=leg2, then this formula becomes area=1/2*(leg)^2.
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Re: What arithmetic should I memorize?  [#permalink]

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20 Apr 2011, 18:47
Thanks for this memory sheet, very useful!

However I believe some formulas, details might be added...

Angles of a polygons (n-2)* 180, angles inscribed in a circle (compared to central angle there a section about it in the VERITAS Prep book)

Also I did not see two real IMPORTANT triangles sides ratio:

3:4:5
5:12:13

Good you put the formula for the equilateral triangle (which is not given in the OG and allow you to gain so much time)

I found formula I did not know! Thank you for sharing
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23 Apr 2011, 12:59
1
I'd also add one more element to Coordinate geometry. The coordinates of a point P on line joining two points A(x1,y1) and B(x2,y2) such that PA:PB is in the ration m:n would be ((mx2+nx1)/(m+n),(my2+ny1)/(m+n)).
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04 Jan 2012, 15:59
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These are a few math tricks my brother used when studying for the MCAT. Hope they help!
(attached a word document)

1. A nice math trick is multiplying two integers that have multiple digits relatively quickly. It does not apply to all integers the following has to be met:
TENS DIGIT in both integers HAS TO BE SAME
ONES DIGIT in each integer HAS TO ADD UP TO 10
Also, if the ones digits are 1 and 9 you just write 09.

Example 34x36:
Step 1: Add one to tens place then multiply (1+3)x3 = 12
Step 2: Ones place 4x6 = 24
Step 3: Place Steps 1 & 2 = 1224

Example 1 & 9 in ones place: 21x29
Step 1: Add one to tens place then multiply (1+2)x2 = 6
Step 2: Ones place 1x9= 09
Step 3: Place Steps 1 & 2 together = 609

2. This one is for people having difficulty memorizing a few square root numbers!
sqrt 1 = 1 (as in 1/1 = New Year's Day)
sqrt 2 = 1.4 (as in 2/14 = Valentine's Day)
sqrt 3 = 1.7 (as in 3/17 = St. Patrick's Day)

3. The next math trick is the Babylonian Method it can be useful when estimating square roots.
First guess roughly what you think it would be

Step 1: For a number less than 1 guess bigger.
For a number greater than 1 guess smaller.
Step 2: Divide your guess into the square root number.
Step 4: Divide by 2

Example sqrt(.78):
Step 1: sqrt of .78 < 1 so guess = .85
Step 2. .78/.85 = ~.9
Step 3: (.85+.9) = 1.75
Step 4: 1.75/2 = ~.88 And this should be your answer (or close enough)

If you guess really wildly just use the answer from your first guess and run through the process again. You can do it in seconds once you get good at it.

Wild Guess Example: sqrt 70 = ?
Step 1 sqrt of 70>1 so guess 10
Step 2: 70/10= 7
Step 3: 7+10= 17
Step 4: 17/2 = 8.5
*8.5x8.5 = 72.25 Still off (10 kind of a wild guess, so repeat process with the new answer from step 4)
Step 1: guess = 8.5
Step 2: 70/8.5= 8.2
Step 3: 8.2+8.5= 16.7
Step 4: 16.7/2 = ~8.4
8.4*8.4 = 70.6

4. If two numbers (both even or odd) are close together and their average is an integer, then this method can be used.
Need to recognize that:
x^2 - y^2 = (x + y)(x - y) and vice-versa (x + y)(x - y) = x^2 - y^2
Example 1
48*52 = (50-2)(50+2) = 50^2 - 2^2 = 2496.

Example 2
3^2 - 2^2 = 3 + 2
4^2 - 3^2 = 4 + 3
5^2 - 4^2 = 5 + 4

5. When x and y are consecutive integers, then (x - y) = 1.

It's useful for calculating large squares:

Example: 71^2 = ?
71^2 - 70^2 = 71 + 70
so 71^2 = 70^2 + (141)
= 4900 + 141 = 5041

Example: 53^2 = ?
53^2 – 52^2 = (53 + 52) + (52 + 51) + (51 + 50)
53^2 = 50^2 + (105) + (103) + (101)
= 2500 + 309 = 2809

0.79^2
80^2 - 79^2 = 80 + 79
so 80^2 - (80 + 79) = 79^2
6400 - 159 = 6241 so 0.79^2 = 0.6241
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# Inversion Mapping is Automorphism iff Group is Abelian

## Theorem

Let $\struct {G, \circ}$ be a group.

Let $\iota: G \to G$ be the inversion mapping on $G$, defined as:

$\forall g \in G: \map \iota g = g^{-1}$

Then $\iota$ is an automorphism if and only if $G$ is abelian.

## Proof

From Inversion Mapping is Permutation, $\iota$ is a permutation.

It remains to be shown that $\iota$ has the morphism property if and only if $G$ is abelian.

### Sufficient Condition

Suppose $\iota$ is an automorphism.

Then:

 $\displaystyle \forall x, y \in G: \map \iota {x \circ y}$ $=$ $\displaystyle \map \iota x \circ \map \iota y$ Definition of Morphism Property $\displaystyle \leadsto \ \$ $\displaystyle \paren {x \circ y}^{-1}$ $=$ $\displaystyle x^{-1} \circ y^{-1}$ Definition of $\iota$

Thus from Inverse of Commuting Pair, $x$ commutes with $y$.

This holds for all $x, y \in G$.

So $\struct {G, \circ}$ is abelian by definition.

$\Box$

### Necessary Condition

Let $\struct {G, \circ}$ be abelian.

 $\displaystyle \forall x, y \in G: \paren {x \circ y}^{-1}$ $=$ $\displaystyle x^{-1} \circ y^{-1}$ Inverse of Commuting Pair $\displaystyle \leadsto \ \$ $\displaystyle \map \iota {x \circ y}$ $=$ $\displaystyle \map \iota x \circ \map \iota y$ Definition of $\iota$

Thus $\iota$ has the morphism property and is therefore an automorphism.

$\blacksquare$

## Sources

except this source requests only that the morphism property is demonstrated, and not the bijectivity.
except this source proves only the necessary condition.


The number of permutations, permutations, of seating these five people in five chairs is five factorial. Suppose we are given a total of n distinct objects and want to select r of them. Counting Problem - discrete math. So, in Mathematics we use more precise language: When the order doesn't matter, it is a Combination. Example. Hence, the total number of permutation is $6 \times 6 = 36$ Combinations. way to permute the numbers. The formulas for each are very similar, there is just an extra $$k!$$ in the denominator of $${n \choose k}\text{. Permutations and Combinations 00:37. The number of all combinations of n things, taken r at a time is − This selection of subsets is called a permutation when the order of selection is a factor, a combination when order is not a factor. Discrete Mathematics Permutations and Combinations? Active today. This is particularly true for some probability problems. There is a group of 15 women and 10 men. Combinations and Permutations. How many ways are there to form the committee if there must be more women than men? There are \binom{9}{4} ways. I'm stuggling to get my head around this question. }$$ That extra $$k!$$ accounts for the fact that $${n \choose k}$$ does not distinguish between the different orders that the $$k$$ objects can appear in. In the Match of the Day’s goal of the month competition, you had to pick the top 3 goals out of 10. Our 1000+ Discrete Mathematics questions and answers focuses on all areas of Discrete Mathematics subject covering 100+ topics in Discrete Mathematics. This number of permutations is huge. See more ideas about discrete mathematics, mathematics, permutations and combinations. Each computer must complete its own tasks in order. We say $$P(n,k)$$ counts permutations, and $${n \choose k}$$ counts combinations. Math 3336 Section 6. with full confidence. 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In this section you can learn and practice Aptitude Questions based on "Permutation and Combination" and improve your skills in order to face the interview, competitive examination and various entrance test (CAT, GATE, GRE, MAT, Bank Exam, Railway Exam etc.) Permutations are specific selections of elements within a set where the order in which the elements are arranged is important, while combinations involve the selection of elements without regard for order. \end{equation*} We know that we have them all listed above —there are 3 choices for which letter we put first, then 2 choices for which letter comes next, which leaves only 1 choice for the last letter. Related Pages Permutations Permutations and Combinations Counting Methods Factorial Lessons Probability. For example, suppose we are arranging the letters A, B and C. (Denoted by n P r or n r or P (n, r)) : Let us consider the problem of finding the number of ways in which the first r rankings are secured by n students in a class. I understand how to calculate normal permutations and combinations but the fact that there are 4 unique types, each with different quantities is confusing. After, each computer sends its output to a shared fourth computer. Permutations and Combinations involve counting the number of different selections possible from a set of objects given certain restrictions and conditions. The information that determines the ordering is called the key. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Ask Question Asked 3 years, 9 months ago. Jan 20, 2018 - Explore deepak mahajan's board "combination" on Pinterest. Discrete Mathematics (c)Marcin Sydow Productand SumRule Inclusion-Exclusion Principle Pigeonhole Principle Permutations Generalised Permutations andCombi-nations Combinatorial Proof Binomial Coeﬃcients Countingthenumberoffunctions Thesetofallfunctionsf : X !Y isdenotedasYX The numberofdiﬀerentfunctionsf : X !Y isgivenbythe expression jYX = jXj. Permutations and combinations, the various ways in which objects from a set may be selected, generally without replacement, to form subsets. "724" won't work, nor will "247". A combination is selection of some given elements in which order does not matter. An arrangement of objects in which the order is not important is called a combination. Search. Aymara G. Numerade Educator 05:14. Of the statements in these questions, 17 are true. A permutation is a (possible) rearrangement of objects. In this article, we will learn about the Introduction permutation group, and the types of permutation in discrete mathematics. It has to be exactly 4-7-2. It's not clear to me if I am using permutation or combination for this question. That is for each permutation chances that 4 and 5 are adjacent are 4/10, hence the result becomes $5!*(1-4/10)$. Problem 1 List all the permutations of {a, b, c}. Submitted by Prerana Jain, on August 17, 2018 Permutation Group. View Permutations, Combinations and Discrete Probability.pdf from MATH 307 at Massachusetts Institute of Technology. Permutations and Combinations with overcounting. We have already covered this in a previous video. 3) There are 5! In Section 2.1 we investigated the most basic concept in combinatorics, namely, the rule of products. 1. Clarissa N. ... A professor writes 40 discrete mathematics true/false questions. Permutations and combinations are part of a branch of mathematics called combinatorics, which involves studying finite, discrete structures. It is of paramount importance to keep this fundamental rule in mind. I have come up with an answer of my own (10,155,600), however I wanted to check with the community and see if I have come up with the correct answer. Throughout mathematics and statistics, we need to know how to count. (n – r)! permutations with vowels next to each other. - 2*6! … "The combination to the safe is 472". One should spend 1 hour daily for 2-3 months to learn and assimilate Discrete Mathematics comprehensively. so u can hv 6! It is denoted by P (n, r) P (n, r) = Five factorial, which is equal to five times four times three times two times one, which, of course, is equal to, let's see, 20 times six, which is equal to 120. Courses. We say $$P(n,k)$$ counts permutations, and $${n \choose k}$$ counts combinations. When we select the data or objects from a certain group, it is said to be permutations, whereas the order in which they are represented is called combination. What Is Combination In Math? = 6$ways. A permutation of X is a one-one function from X onto X. Permutation and combination are the ways to represent a group of objects by selecting them in a set and forming subsets. permutations for EA glued together. The number of ordered arrangements of r objects taken from n unlike objects is: n P r = n! Combinations with Repetition HARD example. Discrete Math - Permutation, Combination. Thread starter BACONATOR; Start date Sep 14, 2008; Tags combinations permutations; Home. = 3600. total unrestricted permutations = 7! Since the order is important, it is the permutation formula which we use. ANS: 3600 Permutations; Combinations; Combinatorial Proofs; Permutations. permutations . 4) Firstly choose 4 numbers. He has in stock 3 identical amethysts, 4 identical diamonds, 5 identical emeralds and 7 identical rubies. There are$\binom{5}{2}=10$types of such pairs. discrete-mathematics permutations combinations. We now look to distinguish between permutations and combinations. Search for courses, skills, and videos. CS311H: Discrete Mathematics Permutations and Combinations Instructor: Is l Dillig Instructor: Is l Dillig, CS311H: Discrete Mathematics Permutations and Combinations 1/26 Permutations I Apermutationof a set of distinct objects is anordered arrangement of these objects I No object can be selected more than once I Order of arrangement matters I Example: S = fa;b;cg. Between them a committee of 6 people is chosen. permutation and combinations discrete mathematics 0 In how many ways can a board of five people that includes the male Head of School be formed if it must include at least two men and two women? thus 2*6! For example, there are 6 permutations of the letters a, b, c: \begin{equation*} abc, ~~ acb, ~~ bac, ~~bca, ~~ cab, ~~ cba. Main content. … }\) That extra $$k!$$ accounts for the fact that $${n \choose k}$$ does not distinguish between the different orders that the $$k$$ objects can appear in. Any arrangement of any r ≤ n of these objects in a given order is called an r-permutation or a permutation of n object taken r at a time. [Discrete Mathematics] Permutation Practice - Duration: 14:41. The formulas for each are very similar, there is just an extra $$k!$$ in the denominator of \({n \choose k}\text{. A penny is tossed 60 times yielding 45 heads and 15 tails. An ordered arrangement of r elements of a set is called an r- permutations. This unit covers methods for counting how many possible outcomes there are in various situations. Definition: A permutation of a set of distinct objects is an ordered arrangement of these objects. Feb 2008 13 0. Permutations. University Math Help. Viewed 2 times 0$\begingroup$I need help with the following problem: There are three computers A, B, and C. Computer A has 10 tasks, Computer B has 15 tasks, and Computer C has 20 tasks. . When the order does matter it is a Permutation. It defines the various ways to arrange a certain group of data. These topics are chosen from a collection of most authoritative and best reference books on Discrete Mathematics. We'll learn about factorial, permutations, and combinations. Viewed 240 times 0$\begingroup$So for this question, would it be 2C1 or 2P1 multiplied by 6P3 or 6C3 and why? This topic is an introduction to counting methods used in Discrete Mathematics. Solution to this Discrete Math practice problem is … How many ways are there to assign scores to the problems if the sum of the scores is 100 and each questions is worth at least 5 points? thus the required permutations = 7! Thank you for your help! We would expect that each key would give a different permutation of the names. 14 Solutions Manual of Elements of Discrete Mathematics CHAPTER TWO PERMUTATIONS, Forums. The remaining 3 vacant places will be filled up by 3 vowels in$^3P_{3} = 3! TheTrevTutor 87,951 views. There are 10 questions on a discrete mathematics final exam. In Section 2.2 we saw a subclass of rule-of-products problems, permutations, and we derived a formula as a computational aid to assist us. If the questions can be positioned in any order, how many different answer keys are possible? A permutation is an ordered arrangement. Discrete Math B. BACONATOR. Active 3 years, 9 months ago. A jeweller wants to choose 3 gemstones to set in a ring that he is making. If you're seeing this message, it means we're having trouble loading external resources on our website. BASIC CONCEPTS OF PERMUTATIONS AND COMBINATIONS ... 5.4 BUSINESS MATHEMATICS Number of Permutations when r objects are chosen out of n different objects. X. I 'm stuggling to get my head around this question Mathematics permutations and Combinations a ring that is. Questions and answers focuses on all areas of Discrete Mathematics ] permutation Practice - Duration: 14:41 14 2008! If you 're seeing this message, it is the permutation formula which we.! 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# max(xy) and min(xy)

#### Albert

##### Well-known member
x,y are integers and

$x^2-xy+2y^2=116$

find max(xy) and min(xy)

##### Active member
\begin{align*} x^2-xy+2y^2&=116\\ x^2-2\sqrt{2}xy+(\sqrt{2}y)^2+2\sqrt{2}xy-xy&=116\\ (x-\sqrt{2}y)^2&=116+(1-2\sqrt{2})xy\\ \end{align*}

since, $(x-\sqrt{2}y)^2\geq 0$,
\begin{align*} 116+(1-2\sqrt 2)xy &\geq 0\\ (2\sqrt 2-1)xy &\leq 116\\ xy &\leq \frac{116}{2\sqrt 2 -1} \qquad (2\sqrt 2-1 > 0) \end{align*}

also,
\begin{align*} x^2-xy+2y^2&=116\\ x^2+2\sqrt{2}xy+(\sqrt{2}y)^2-2\sqrt{2}xy-xy&=116\\ (x+\sqrt{2}y)^2&=116+(1+2\sqrt{2})xy\\ \end{align*}

with, $(x+\sqrt{2}y)^2\geq 0$,

\begin{align*} 116+(1+2\sqrt 2)xy &\geq 0\\ (2\sqrt 2+1)xy &\geq -116\\ xy &\geq \frac{-116}{2\sqrt 2 +1} \end{align*}

eventually we have,

$$\frac{-116}{2\sqrt 2 +1} \leq xy \leq \frac{116}{2\sqrt 2 -1}$$

$$-30.36 \leq xy \leq 63.73$$

but this is for $x,y \in R$ I cannot figure out it for integers

Last edited:

#### Opalg

##### MHB Oldtimer
Staff member
Continuing BAdhi's analysis, the max integer value is at most 63, and the min integer value is at least -30. But you can achieve each of those values, by taking $(x,y) = (9,7)$ or $(-9,-7)$ for the max, and $(x,y) = (6,-5)$ or $(-6,5)$ for the min.

I found those points by graphing, using MHB's Desmos grapher (click on it to see the detail):
[graph]35ihvotanp[/graph]
It would be interesting to see a more analytical solution.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I rewrote the problem with $w=xy \Rightarrow y=\frac w x$ to get:

$x^2-w+2\frac {w^2} {x^2}=116$

Feeding it to Wolfram gives a nice graph and all 12 integer solutions, with the minimum of -30 and the maximum of 63.

Likewise, it would be nice to see a more analytical solution.

#### Albert

##### Well-known member
I rewrote the problem with $w=xy \Rightarrow y=\frac w x$ to get:

$x^2-w+2\frac {w^2} {x^2}=116$

Feeding it to Wolfram gives a nice graph and all 12 integer solutions, with the minimum of -30 and the maximum of 63.

Likewise, it would be nice to see a more analytical solution.
yes, it could be solved to use a more analytical solution,try it ! it is not hard !

I will give the solution soon

#### Albert

##### Well-known member
$x^2-xy+2y^2=116$
we rerrange it and get
$x^2-xy+2y^2-116=0----(1)$
solving for x ,since x,y are integers the determinant:
$7y^2-464 \leq 0$
ant it must be a perfect square
$\therefore -8 \leq y\leq 8$
furthermore if we replace x , y with -x and -y the equation remain unchanged
so we only have to put y=0,1,2,3,4,5,6,7,8 to
(1) and get the corresponding x
by taking for (x,y)=(9,7) or (-9,-7) the max(xy)=63
(x,y)=(6,-5) or (-6,5) the min(xy)=-30

#### anemone

##### MHB POTW Director
Staff member
Since $$\displaystyle x^2-xy+2y^2-116=0$$ are defined over the real integers, we know that the discriminant of the equation (if we solve it for x) will be greater than or equal to zero.

Thus,

$$\displaystyle (-y)^2-4(1)(2y^2-116) \ge 0$$

$$\displaystyle 464-7y^2 \ge 0$$

$$\displaystyle (\sqrt{464}+\sqrt{7}y)(\sqrt{464}-\sqrt{7}y) \ge 0$$ $$\displaystyle \rightarrow {-8.1416 \le y \le 8.1416}$$

But y will be an integer, thus, we know that $$\displaystyle {-8 \le y \le 8}$$ must be true.

From the given equation $$\displaystyle x^2-xy+2y^2=116$$

We know that we can manipulate the RHS of the equation by doing the following:
$$\displaystyle x^2-xy+2y^2=114+2(1)$$ which implies $$\displaystyle y=\pm1$$

This gives $$\displaystyle x^2-x(\pm 1)=114$$ but this leads to non-integer solutions for x.

We repeat the process from $$\displaystyle y=2$$ to $$\displaystyle y=8$$ (we will cover the negative values of y as well because of the fact that $$\displaystyle y^2=1, 4, 9, 14, 25, 36, 49, 64$$ would cover the y values from $$\displaystyle y=\pm1, \pm2, \pm3, \pm4, \pm5, \pm6, \pm7, \pm8$$) and we find that all of the integer solutions to the equations are $$\displaystyle (6, -5), (-11, -5), (11, 5), (-6, 5), (2, -7), (-9, -7), (9, 7), (-2, 7), (-2, -8), (-6, -8), (6, 8),$$ and $$\displaystyle (2, 8)$$ which gives us the range of $$\displaystyle xy$$ as $$\displaystyle -30 \le xy \le 63$$.

(Edit:I now notice that this solution is similar to that posted by Albert, which I did not see until after making my post....I am sorry!)

Last edited:


Question

Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle $\theta$ below the horizontal. The baseline from which the ball is served is 11.9 m from the net, which is 0.91 m high. What is the angle $\theta$ such that the ball just crosses the net? Will the ball land in the service box, which has an outermost service line that is 6.40 m from the net?

$\theta = 6.1^\circ$

Yes, the ball will land within the service box.

Solution Video


Fomin, "Elements of the theory of functions and functional analysis" , L.D. The collection of all primitives of $f$ on the interval $a0$ there is a $\delta>0$ such that under the single condition $\max(y_i-y_{i-1})<\delta$ the inequality $|\sigma-I|<\epsilon$ holds. Boros, G. and Moll, V. Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals. J. Diestel, J.J. Uhl jr., "Vector measures" . Yes, finding a definite integral can be thought of as finding the area under a curve (where area above the x-axis counts as positive, and area below the x-axis counts as negative). Stover, Christopher and Weisstein, Eric W. Deeply thinking an antiderivative of f(x) is just any function whose derivative is f(x). For instance, the Riemann integral is based on Kaplan, W. Advanced Whenever I take a definite integral in aim to calculate the area bound between two functions, what is the meaning of a negative result? The integral symbol is U+222B ∫ INTEGRAL in Unicode and \int in LaTeX.In HTML, it is written as ∫ (hexadecimal), ∫ and ∫ (named entity).. A definite integral is an integral int_a^bf(x)dx (1) with upper and lower limits. Does it simly mean that the said area is under the the x - axis, in the negative domain of the axis? The case of arbitrary functions was studied by B. Riemann (1853). Introduction to Integration. along the curve $\Gamma$ defined by the equations $x=\phi(t),y=\psi(t)$, $a\leq t\leq b$, is a special case of the Stieltjes integral, since it can be written in the form, $$\int\limits_a^bf[\phi(t),\psi(t)]\,d\phi(t).$$, A further generalization of the notion of the integral is obtained by integration over an arbitrary set in a space of any number of variables. u d v = u v-? Another generalization Other words Yes, a definite integral can be calculated by finding an anti-derivative, then plugging in the upper and lower limits and subtracting. Pesin, "Classical and modern integration theories" , Acad. Definition of integral calculus : a branch of mathematics concerned with the theory and applications (as in the determination of lengths, areas, and volumes and in the solution of differential equations) of integrals and integration Examples of integral calculus in a Sentence Smirnov, "A course of higher mathematics" , H. Lebesgue, "Leçons sur l'intégration et la récherche des fonctions primitives" , Gauthier-Villars (1928), E. Hewitt, K.R. In mathematics, an integral assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data. of a differential form over the boundary of some orientable 1993. where $U$ is a set function on $M$ (its measure in a particular case) and the points belong to the set $M$ over which the integration proceeds. 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What is a more natural example of the following?

The following is an exercise from Jacobson's Basic Algebra I:

Let $S=\{1,2,\cdots\}$. Give an example of two maps $\alpha, \beta$ of $S$ into $S$ such that $\alpha \beta=1_{S}$ but $\beta \alpha\ne 1_{S}$. Can this happen if $\alpha$ is bijective?

I have a fairly good answer for this exercise, I think.

An Example

Let $\alpha(s)=|s-2|$ and $\beta(s)=s+2$ for $s \in S$. Thus, $(\alpha \circ \beta)(s)=|(s+2)-2|=|s|=s$, which implies $\alpha \beta=1_{S}$.

However, $(\beta \circ \alpha)(s)=|s-2|+2$. We see that $(\beta \circ \alpha)(1)=3$, hence $\beta \alpha\ne 1_{S}$.

Consequences of $\alpha$ Being Bijective

If $\alpha$ is bijective, there exists an inverse map $\gamma$ such that $\gamma \alpha=\alpha \gamma=1_{S}$. Since $\alpha \beta=1_{S}$, we have that $\alpha \beta=\alpha \gamma$. Composing both sides of the equation with $\gamma$, we have that $\gamma(\alpha \beta)=\gamma(\alpha \gamma)$. Next, using the law of associativity, we have that $(\gamma \alpha)\beta=(\gamma \alpha)\gamma$. Since $\gamma \alpha=1_{S}$, we thus have $1_{S}\beta=1_{S}\gamma$. Hence, $\beta=\gamma$.

Since we previously stated that $\gamma \alpha=1_{S}$ and we've shown that $\gamma=\beta$, it must be so that $\beta\alpha=1_{S}$. $\square$

My question

Are there more natural examples of $\alpha$ and $\beta$? Furthermore, is my proof accurate? I ask the second question only as I see it directly pertains to the first.

P.S. By natural, I mean elegant or insightful.

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Your solution looks good. Perhaps the "most natural" example would be for $\alpha(n) = n-1$ if $n > 1$ and $\alpha(1) = 1$. (Think of $\alpha$ as the "shift one unit left -- if possible" function.) Then take $\beta$ to be the "shift right one unit" function, $\beta(n) = n+1$. Then $\alpha \beta = \text{id}_S$, but $\beta \alpha \neq \text{id}_S$. – Michael Joyce Oct 21 '12 at 20:39
@MichaelJoyce, looking at your thoughts as well as N.S.'s, I see that my issue was entirely due to myself. It was because I saw piecewise functions as unnatural that I struggled with finding an example. I also, after looking at some of the awesome maps presented by Jacobson, assumed there was an expectation for some marvelous geometric map. I see now that that's probably not the case. – 000 Oct 21 '12 at 20:42
There's nothing wrong with coming with a solution that is not "aesthetically optimal" at first. The aesthetic solutions usually come later after you've had more time to fully internalize everything about a problem. But there can be as much learning as finding a non-standard solution as there is to finding a very elegant one. BTW, I would consider the example I presented as "geometric" in so far as it can be visualized as pulling a string of beads in a one-dimensional universe. :) – Michael Joyce Oct 21 '12 at 20:47

$\alpha(x)=2x$ and $\beta(x) = \lfloor \frac{x+1}{2} \rfloor$ is probably more natural.
More generarily, given any $\alpha$ which is 1-1 but not bijective function, you can always find a $\beta$ so that $\beta \alpha =1_S$. But since $\alpha$ is not bijective, $\alpha \beta \neq 1_S$.
You may need $\beta(x) = \lceil \frac{x}{2} \rceil$ since $S$ starts at $1$ – Henry Oct 21 '12 at 20:57


# In base 10

After writing a solution to this problem I focused on product of digits and got the following result. Also I have a challenge also for all those who found this interesting try if you can do it!

Let $f_{10}(x)$ be defined as the product of digits of $x$ when written in base $10$ for example $f_{10}(279)=2 \times 7 \times 9=126$. Then, $\boxed{\sum_{i=10^{n-1}}^{10^{n}}f(i)=45^{n}}$

Proof:

First, we have to see the following lemma:

LEMMA : $\sum_{n=10^{q-1}}^{10^{q}}f(n)=45(\sum_{i=10^{q-2}}^{10^{q-1}}f(i))$

Proof:

Let $S_n=\sum_{n=10^{q-1}}^{10^{q}}f(n)$

We can remove numbers from $10^{q-1}$ to $1111...$ ($q$ $1's$) as the numbers between have a $0$ in them

Similarly we have to remove numbers from $2000...$ to $2111...$ ($q-1$ $0's$ and 1 ), from $3000...$ to $3111...$ ($q-1$ $0's$ and 1 ) and so on.

$S_n=(1\times1\times1...\times1 )+(1\times1\times1...\times2)...+(9\times9\times9...\times9)$

We can take leading digits common, reducing a single digit from each number

$S_n=1((1\times1...\times1)+(1\times1...2)...+(9\times9...\times9))+2((1\times1...\times1)+(1\times1...2)...+(9\times9...\times9))...9(1\times1...\times1)+9(1\times1...2)...+9(9\times9...\times9))$

Now we can take $\sum_{n=10^{q-2}}^{10^{q-1}}$ by include numbers like $100...01$ as $f(100...01)=0$ so it makes no change

$S_n=(\sum_{n=1}^{9})(\sum_{i=10^{q-2}}^{10^{q-1}}f(i))=45(\sum_{i=10^{q-2}}^{10^{q-1}}f(i))$

Using the lemma we can get

$\sum_{i=10^{n-1}}^{10^{n}}f(i)=45(\sum_{i=10^{n-2}}^{10^{n-3}}f(i))$

If you apply it again and again

$\sum_{i=10^{n-1}}^{10^{n}}f(i)=(45)(45)...(\sum_{i=10^{n-n}}^{10^{1}}f(i))=(45)(45)...(45)=\boxed{45^n}$ Hence proved

Challenge:

• I have proved this for base $10$ you can try for any other base or you may prove it for any base $b$

Note by Zakir Husain
3 days, 14 hours ago

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## Comments

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I think your argument can be adapted to any base, b. The sum would be (b(b-1)/2)^n. Proof could be by induction. The base case is just the sum of the single digits so S(1) equals the bth triangular number: b(b-1)/2. Assuming S(n)=(b(b-1)/2)^n, then by your leading digits argument, S(n+1)=1(S(n))+2(S(n))+...+b(S(n))=(b(b-1)/2)(b(b-1)/2)^n=(b(b-1)/2)^(n+1) completing the proof by induction.

- 15 hours ago

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Good efforts!

- 12 hours ago

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