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If a, b are integers, and (a-b)^2 + 8b^2=108, what is the number of th

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If a, b are integers, and (a-b)^2 + 8b^2=108, what is the number of th [#permalink]

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If a, b are integers, and \((a-b)^2 + 8b^2=108\), what is the number of the ordered pairs (a,b)?

A. 2
B. 4
C. 6
D. 8
E. 10
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 May 2017, 22:40, edited 2 times in total.
Renamed the topic and edited the question.
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Re: If a, b are integers, and (a-b)^2 + 8b^2=108, what is the number of th [#permalink]

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New post 18 Feb 2017, 02:26
amberbajaj02 wrote:
If a, b are integers, and (a-b)^2 + 8b^2=108, what is the number of the ordered pairs (a,b)?

A. 2
B. 4
C. 6
D. 8
E. 10




Hi,

this is my approach,
start with the part that has more limit, so I start with 8b^2. b^2 could be 1, 4, 9, but not greater because if so then 8b^2 would be 128 that is greater than 108.
therefore we have that 8b^2 is one of 8*1,8*4=32 or 8*9=72, in each case (a-b)^2 is 108-8=100, 108-32=76 or 108-72=36 respectively, but (a-b)^2 can not be 100 or 76 because it's a complete square. therefore b^2=9 & (a-b)^2=36 so we have b=+or- 3 and a-b=+or -6 we get a=+or_6+(+or-3) that is 4 number for a and 2 number for b. which is 8 pairs.
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Re: If a, b are integers, and (a-b)^2 + 8b^2=108, what is the number of th [#permalink]

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New post 17 May 2017, 14:43
AMahmoodi wrote:
amberbajaj02 wrote:
If a, b are integers, and (a-b)^2 + 8b^2=108, what is the number of the ordered pairs (a,b)?

A. 2
B. 4
C. 6
D. 8
E. 10




Hi,

this is my approach,
start with the part that has more limit, so I start with 8b^2. b^2 could be 1, 4, 9, but not greater because if so then 8b^2 would be 128 that is greater than 108.
therefore we have that 8b^2 is one of 8*1,8*4=32 or 8*9=72, in each case (a-b)^2 is 108-8=100, 108-32=76 or 108-72=36 respectively, but (a-b)^2 can not be 100 or 76 because it's a complete square. therefore b^2=9 & (a-b)^2=36 so we have b=+or- 3 and a-b=+or -6 we geta=+or_6+(+or-3) that is 4 number for a and 2 number for b. which is 8 pairs.



The highlighted part is incorrect. a could be +3 or -3, +9 or -9, while b could +3 or -3. So, four options for a and two for b (8 combinations).
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Re: If a, b are integers, and (a-b)^2 + 8b^2=108, what is the number of th [#permalink]

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New post 17 May 2017, 19:47
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Amby02 wrote:
If a, b are integers, and (a-b)^2 + 8b^2=108, what is the number of the ordered pairs (a,b)?

A. 2
B. 4
C. 6
D. 8
E. 10



Hi MikeMighty and AMahmoodi

The approach is correct but you are faultering at two places.
1) you are looking for ordered pairs a,b so you cannot find combinations by simply multiplying. Each value of a and b, combined, should satisfy equation.
2) 100 is also a perfect square..

now the solution
Let's look at 8b^2.... B can take value as 0,1,2,3...
When b=4, 8b^2 becomes 128, more than Right Hand side..

When we subtract 108-8b^2, it should give us a perfect square as (a-b)^2 is a perfect square.
1) b=0,
\(108-8b^2=108..\)... NOT a perfect square.... OUT

2) b=1 or -1,
\(108-8*1^2=100... So.. (a-b)^2=100=10^2..\)...
# a-b=10....b=1,a=11 ; b=-1,a=9….....SO 2 ordered pairs
#a-b=-10...Again 2 ordered pairs
So 4 pairs here.

3) b= 2 or -2
\(108-8*2^2=76\)..... NOT a perfect square.... OUT

4) b=3 or -3..
\(108-8*3^2=36=6^2 ..or ..(-6)^2\)
#a-b=6.......2 ordered pairs
#a-b=-6..... Again 2 ordered pairs
So 4 pairs

Overall ordered pairs =4+4=8..

Just for info these are
a-b=10………(11,1);(9,-1)
a-b=-10………(-9,1);(-11,-1)
a-b=6………(9,3);(3,-3)
a-b=-6………(-3,3);(-9,-3)

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Re: If a, b are integers, and (a-b)^2 + 8b^2=108, what is the number of th [#permalink]

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New post 17 May 2017, 22:38
chetan2u thanks, I see where I went wrong. Great explanation!
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Re: If a, b are integers, and (a-b)^2 + 8b^2=108, what is the number of th [#permalink]

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New post 12 Jan 2018, 12:38
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Hi All,

While this question 'looks' complex, it can be organized into some simple ideas; those ideas - along with a bit of BRUTE FORCE arithmetic - can get you to the solution without too much trouble.

We're told that (A - B)^2 + 8(B^2) = 108 and that A and B are both INTEGERS.

In simple terms:

(A perfect square) + 8(A perfect square) = 108

Since perfect squares cannot be negative, and that second perfect square is multiplied by EIGHT, there can't be that many solutions to this equation. Let's focus on that second perfect square....If the first perfect square was 0, then we'd have....

0 + 8(a perfect square) = 108
(a perfect square) = 108/8 = 13.5

Since B is an INTEGER, we know that this equation is NOT possible - but we can use it to determine that that perfect square is less than 13.5. Thus, we really only have to think about 0, 1, 4 and 9 for that second perfect square. Let's start 'big' and work our way down...

IF.... B = 3 or -3, then we have...
(A-B)^2 + 72 = 108
(A-B)^2 = 36
(A-B) = +6 or -6

Thus, we could have FOUR solutions here:
A = 9, B = 3
A = -3, B = 3
A = -9, B = -3
A = 3, B = -3

IF.... B = 2 or -2, then we have...
(A-B)^2 + 32 = 108
(A-B)^2 = 76

76 is NOT a perfect square though, so there are no solutions here.

IF.... B = 1 or -1, then we have...
(A-B)^2 + 8 = 108
(A-B)^2 = 100
(A-B) = +10 or -10

Thus, we could have FOUR solutions here:
A = 11, B = 1
A = -9, B = 1
A = 9, B = -1
A = -11, B = -1

IF.... B = 0, then we have...
(A-0)^2 + 0 = 108
(A)^2 = 108

108 is NOT a perfect square though, so there are no solutions here.

Total solutions = four + four = 8

Final Answer:
[Reveal] Spoiler:
D


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Re: If a, b are integers, and (a-b)^2 + 8b^2=108, what is the number of th   [#permalink] 12 Jan 2018, 12:38
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