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A right triangle has sides of a, b, and 11, respectively,

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Re: A right triangle has sides of a, b, and 11, respectively, [#permalink]

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New post 24 Jan 2017, 05:42
neeraj609 wrote:
anceer wrote:
A right triangle has sides of a, b, and 11, respectively, where a and b are both integers. What is the value of (a + b)?

A. 15
B. 57
C. 93
D. 109
E. 121


So if a and b both are integers, then a+b should also be integer. Now we are asked to find the value of (a+b)^2. I started POE. None of the values is a perfect square (which one square root result into a integer, which should be the value of base a+b) EXCEPT 121.

I am pretty sure it cant be that simple, what am i missing here :)


How do you figure that we are looking for (a + b)^2?
We are looking for the value of (a + b) only.
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Re: A right triangle has sides of a, b, and 11, respectively, [#permalink]

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New post 24 Jan 2017, 07:07
VeritasPrepKarishma wrote:
neeraj609 wrote:
anceer wrote:
A right triangle has sides of a, b, and 11, respectively, where a and b are both integers. What is the value of (a + b)?

A. 15
B. 57
C. 93
D. 109
E. 121


So if a and b both are integers, then a+b should also be integer. Now we are asked to find the value of (a+b)^2. I started POE. None of the values is a perfect square (which one square root result into a integer, which should be the value of base a+b) EXCEPT 121.

I am pretty sure it cant be that simple, what am i missing here :)


How do you figure that we are looking for (a + b)^2?
We are looking for the value of (a + b) only.


Actually i reading (a+b)? wrongly as (a+b)^2. Thanks alot!

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Re: A right triangle has sides of a, b, and 11, respectively, [#permalink]

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New post 24 Jan 2017, 08:38
A right triangle has sides of a, b, and 11, respectively, where a and b are both integers. What is the value of (a + b)?


Case 1: \(a^2 + b^2 = 11^2 = 121\) (unit digit 1)

Perfect squares end with any one of the following: 0; 1; 4; 9; 6; 5.
To have unit digit 1, we have to use either of the combos: 0 and 1; 6 and 5. But none satisfies. X

Case 2: \(a^2 + 121 = b^2\)

Subcase (a)
=> \(a^2 = b^2 - 121\)
=> \(a^2 = (b+11) (b-11)\)
Since, \((b+11)\) is not equal to \((b-11)\)and also \((b+11) > (b-11)\)therefore
\(b-11 = 1\) and \(b+11 = a^2\)
\(b=12\) and \(a^2=23\) (a is not a integer) X

Subcase (b)

=> \(121 = b^2 - a^2\)
=> \((b-a) (b+a) = 121\)
Therefore, \(b-a=1\) and \(b+a=121\)

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Re: A right triangle has sides of a, b, and 11, respectively, [#permalink]

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New post 12 Sep 2017, 09:30
GMATinsight wrote:
anceer wrote:
A right triangle has sides of a, b, and 11, respectively, where a and b are both integers. What is the value of (a + b)?

A. 15
B. 57
C. 93
D. 109
E. 121


The trick for this question is

If the side of a right angle triangle is a prime number then other two sides will be

Second Side = [(Prime)^2 + 1]/2

Third Side = [(Prime)^2 - 1]/2


i.e. for One side = 11

Second Side = (11^2 -1)/2 = (121-1)/2 = 60
Third Side = (11^2 +1)/2 = (121+1)/2 = 61

I hope this helps!

I personally find it an UNSUITABLE question for GMAT... GMAT Doesn't expect such tricks from students.


Bookmarking this post and making a mental note. Definitely worth knowing!
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Re: A right triangle has sides of a, b, and 11, respectively,   [#permalink] 12 Sep 2017, 09:30

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