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14 Jul 2026, 08:00
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Difficulty:
95%
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Question Stats:
55% (01:57) correct
45%
(02:42)
wrong
based on 191
sessions
History
Date
Time
Result
Not Attempted Yet
If a and b are integers, for how many ordered pairs (a, b) is 945 = a^2 - b^2?
A. 8
B. 16
C. 24
D. 32
E. 64
A. 8
B. 16
C. 24
D. 32
E. 64
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M46-43
14 Jul 2026, 09:00
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GMAT Club Official Solution:
If a and b are integers, for how many ordered pairs (a, b) is 945 = a^2 - b^2?
A. 8
B. 16
C. 24
D. 32
E. 64
945 = a^2 - b^2 = (a - b)(a + b)
Since 945 is positive, a - b and a + b must be either both positive or both negative.
Now factor 945:
945 = 3^3 * 5 * 7
So the number of positive divisors of 945 is:
(3 + 1) * (1 + 1) * (1 + 1) = 16
For each positive divisor d of 945, we can set a - b = d and a + b = 945/d. Thus, there are 16 such ordered pairs with positive a - b and a + b.
For example:
If a - b = 1, then a + b = 945/1 = 945.
Adding the two equations:
2a = 946
a = 473
Then b = 472.
So this gives one ordered pair:
(a, b) = (473, 472)
If a - b = 3, then a + b = 315.
Adding the two equations:
2a = 318
a = 159
Then b = 156.
So this gives another ordered pair:
(a, b) = (159, 156)
Also, reversing the factor values gives a different ordered pair.
For example:
If a - b = 945, then a + b = 1.
Adding the two equations:
2a = 946
a = 473
Then b = -472.
So this gives:
(a, b) = (473, -472)
Since 945 is odd, both d and 945/d are odd, so the resulting values of a and b from each such pair will be integers. Thus, each positive divisor gives one ordered pair (a, b). Since there are 16 positive divisors, we get 16 ordered pairs from positive factor pairs.
Similarly, the negatives (-d, -945/d) give another 16 ordered pairs of values for a - b and a + b, with both values negative.
For example:
If a - b = -1, then a + b = -945.
Adding the two equations:
2a = -946
a = -473
Then b = -472.
So the negative factor pairs give another 16 ordered pairs.
Therefore, the total number of ordered pairs (a, b) is:
16 + 16 = 32
Answer: D.
If a and b are integers, for how many ordered pairs (a, b) is 945 = a^2 - b^2?
A. 8
B. 16
C. 24
D. 32
E. 64
945 = a^2 - b^2 = (a - b)(a + b)
Since 945 is positive, a - b and a + b must be either both positive or both negative.
Now factor 945:
945 = 3^3 * 5 * 7
So the number of positive divisors of 945 is:
(3 + 1) * (1 + 1) * (1 + 1) = 16
For each positive divisor d of 945, we can set a - b = d and a + b = 945/d. Thus, there are 16 such ordered pairs with positive a - b and a + b.
For example:
If a - b = 1, then a + b = 945/1 = 945.
Adding the two equations:
2a = 946
a = 473
Then b = 472.
So this gives one ordered pair:
(a, b) = (473, 472)
If a - b = 3, then a + b = 315.
Adding the two equations:
2a = 318
a = 159
Then b = 156.
So this gives another ordered pair:
(a, b) = (159, 156)
Also, reversing the factor values gives a different ordered pair.
For example:
If a - b = 945, then a + b = 1.
Adding the two equations:
2a = 946
a = 473
Then b = -472.
So this gives:
(a, b) = (473, -472)
Since 945 is odd, both d and 945/d are odd, so the resulting values of a and b from each such pair will be integers. Thus, each positive divisor gives one ordered pair (a, b). Since there are 16 positive divisors, we get 16 ordered pairs from positive factor pairs.
Similarly, the negatives (-d, -945/d) give another 16 ordered pairs of values for a - b and a + b, with both values negative.
For example:
If a - b = -1, then a + b = -945.
Adding the two equations:
2a = -946
a = -473
Then b = -472.
So the negative factor pairs give another 16 ordered pairs.
Therefore, the total number of ordered pairs (a, b) is:
16 + 16 = 32
Answer: D.
The 32 ordered pairs are:
(473, 472), (159, 156), (97, 92), (71, 64), (57, 48), (39, 24), (33, 12), (31, 4)
(31, -4), (33, -12), (39, -24), (57, -48), (71, -64), (97, -92), (159, -156), (473, -472)
(-473, -472), (-159, -156), (-97, -92), (-71, -64), (-57, -48), (-39, -24), (-33, -12), (-31, -4)
(-31, 4), (-33, 12), (-39, 24), (-57, 48), (-71, 64), (-97, 92), (-159, 156), (-473, 472)
(473, 472), (159, 156), (97, 92), (71, 64), (57, 48), (39, 24), (33, 12), (31, 4)
(31, -4), (33, -12), (39, -24), (57, -48), (71, -64), (97, -92), (159, -156), (473, -472)
(-473, -472), (-159, -156), (-97, -92), (-71, -64), (-57, -48), (-39, -24), (-33, -12), (-31, -4)
(-31, 4), (-33, 12), (-39, 24), (-57, 48), (-71, 64), (-97, 92), (-159, 156), (-473, 472)
General Discussion
14 Jul 2026, 08:16
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If a and b are integers, for how many ordered pairs (a, b) is 945 = a^2 - b^2?
A. 8
B. 16
C. 24
D. 32
E. 64
945 factors 3^3 * 5*7
4*2*2 = 16
since ordered pairs are in power of 2 , so a,b can be both positive and negative
ordered pairs possible are
16*2 = 32
OPTION D ; 32 is correct
A. 8
B. 16
C. 24
D. 32
E. 64
945 factors 3^3 * 5*7
4*2*2 = 16
since ordered pairs are in power of 2 , so a,b can be both positive and negative
ordered pairs possible are
16*2 = 32
OPTION D ; 32 is correct
14 Jul 2026, 08:26
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IMO D
Given : a^2-b^2 = 945 = (a+b)(a-b)
assuming a+b=x, a-b=y
x.y=945
Then Prime factorization of 945 = 5x7x3x3x3
No of positive divisors = (1+1)(1+1)(3+1)=16
Now x&y can be any divisors (positive or negative), so pairs will be doubled = 32
Hence for a & b, there will be 32 ordered pairs
Given : a^2-b^2 = 945 = (a+b)(a-b)
assuming a+b=x, a-b=y
x.y=945
Then Prime factorization of 945 = 5x7x3x3x3
No of positive divisors = (1+1)(1+1)(3+1)=16
Now x&y can be any divisors (positive or negative), so pairs will be doubled = 32
Hence for a & b, there will be 32 ordered pairs
Bunuel
