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

Question

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

chetan2u · Accepted Answer

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

EMPOWERgmatRichC · Answer

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:  D

GMAT assassins aren't born, they're made, 
Rich

AMahmoodi · Answer

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.

JavierS · Answer

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).

JavierS · Answer

chetan2u  thanks, I see where I went wrong. Great explanation!

RJ7X0DefiningMyX · Answer

I almost got to the correct answer, but in haste, failed to consider that b could also have a negative value, and hence, got to only 4 as the answer.

Oppenheimer1945 · Answer

b=0 (not possible)
b=±1, we get a=±10±1 (4 pairs)
b=±2, we get irrational a
b=±3, we get a=±6±3 (4 pairs)

8 pairs

Ulfat · Answer

8( b^2 )+  (a-b)^2  = 108
or 108 - 8( b^2 ) =  (a-b)^2 
Therefore, 108 - 8 b^2  =  c^2 , where c = (a-b)

Since both the terms are squares and squares are always non-negative. 
That is  b^2  and  c^2   will be from {0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100}.

108 
 - 
 8 
 x 
   
b^2   
 = 
  c^2  
 Remarks 
 
  108 
 - 
 8 
 x 
   0   
 = 
 108 
 Not possible, because  c^2 =108 is not a perfect square 
 
  108 
 - 
 8 
 x 
   1   
 = 
  100  
  Valid  
 
  108 
 - 
 8 
 x 
   4   
 = 
 76 
 Not possible, because Not possible, because  c^2 =76 is not a perfect square 
 
  108 
 - 
 8 
 x 
   9   
 = 
  36  
  Valid  
 
  108 
 - 
 8 
 x 
   16   
 = 
 -20 
 Not possible, because Not possible, because  c^2 = -20 is a negative number

For  b^2  ≥ 16, the value of  c^2  becomes negative, which is inadmissible, as  c^2  being a perfect square has to be non-negative.

Therefore  b^2  = 1 and  c^2  = 100 
=> b=± and c=±10 
=> b=± and (a-b)=±10
=> (a, b) = (11, 1), (-9, 1), (9, -1), (-11, -1)

And,  b^2  = 9 and  c^2  = 36 
=> b=±3 and c=±6 
=> b=±3 and (a-b)=±6
=> (a, b) = (9, 3), (-3, 3), (3, -3), (-9, -3)

Thus, we have total 8 ordered pairs of (a, b)

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

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108	-	8	x	\(\\ b^2\)	=	\(c^2\)	Remarks
108	-	8	x	0	=	108	Not possible, because \(c^2\)=108 is not a perfect square
108	-	8	x	1	=	100	Valid
108	-	8	x	4	=	76	Not possible, because Not possible, because \(c^2\)=76 is not a perfect square
108	-	8	x	9	=	36	Valid
108	-	8	x	16	=	-20	Not possible, because Not possible, because \(c^2\)= -20 is a negative number

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards