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Difficulty:
95%
(hard)
Question Stats:
29% (02:38) correct
71%
(02:45)
wrong
based on 488
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Date
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Not Attempted Yet
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
A. 2
B. 4
C. 6
D. 8
E. 10
17 May 2017, 20:47
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Amby02
Hi MikeMighty and AMahmoodi
The approach is correct but you are faltering 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)
D
12 Jan 2018, 13:38
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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:
GMAT assassins aren't born, they're made,
Rich
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:
GMAT assassins aren't born, they're made,
Rich
