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GMAT Club World Cup 2026 (Day 7): If a and b are integers, for how

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vasu1104

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15 Jul 2026, 00:07

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(a-b)(a+b) = 3^3 * 5 *7

total factors for 945 = (3+1)(1+1)(1+1) = 16.

now since a and b can be negative as well. so total factors = 32

choice D

Bunuel

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

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15 Jul 2026, 00:21

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So 945=a^2-b^2=(a-b)(a+b)
let x=a-b, y=a+b which makes xy=945

a=x+y/2, b=y-x/2
so a and b are integers if x and y are both odds or both even
since 945 is odd, every factor pair that * to it must both be odd making every factor pair of 945 yield integer a and b.
prime factor of 945=9*10=3^2*3*5*7=3^3*5^1*7^1
how many positive divisors (3+1)*(1+1)*(1+1)=4*2*2=16
Since x and y can both be negative, i multiply the positive factor by 2
16*2=32
ANS: D. 32

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15 Jul 2026, 00:35

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use the difference of squares

x=a-b
y = a+b
xy = 945
factorise 945 = 3^3 x 5 x 7

number of positive divisors of 945 is (3+1)(1+1)(1+1) = 16
the negative divisors of x are also 16 so total 32.

as 945 is odd both x and y are odd meaning a and b are noth integers,, thus every ordered factor pair produces one valid ordered pair. So 32.

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15 Jul 2026, 01:42

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Given: a^2 − b^2 = 945

a2 − b2 = (a − b)(a + b)

So, (a − b)(a + b) = 945

Let: x = a − b and y = a + b

Then ----> xy = 945

Also,
a = (x + y)/2
b = (y − x)/2

For a and b to be integers x and y must have the same parity. Since 945 = 3^3 × 5 × 7 is odd, all of its factor pairs are odd, so this condition is automatically satisfied.

Now counting the ordered factor pairs of 945.

Number of positive divisors of 945:
-----> (3 + 1)(1 + 1)(1 + 1)
-----> 4 × 2 × 2
-----> 16

Each positive divisor gives one positive factor pair (x, y), so there are 16 positive ordered pairs. Since both factors can also be negative, there are another 16 ordered pairs. Total ordered factor pairs = 16 + 16 = 32. Each factor pair gives exactly one ordered pair (a, b).

Answer: D.

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15 Jul 2026, 01:51

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945 = a^2-b^2
prime factorization of 945 = 3^3 * 5^1 * 7^1
hence (3+1)(1+1)(1+1) = 16 positive factors or 16 ordered positive pairs of factors or 8 unique positive pairs of factors.

a^2-b^2 = (a+b)(a-b)
let k = a+b & m=a-b
945 = k*m

hence 945 has 16 ordered positive pairs of (k,m)

similarly we count those for negative integers and double this = 16*2 = 32 total ordered pairs of (k,m)

but since we need how many ordered pairs there are for (a,b) we build relation of a,b to k,m:

k = a +b
m = a-b
add the two eqs => a = (k+m)/2
subtract the two eqs => b = (k-m)/2

since 945 is odd, k & m are both odd since odd*odd=odd

also odd+odd = even & odd-odd = even hence both (k+m) & (k-m) are even and can be divided by 2 for all values of k & m. we check this b/c we are given that a & b are both integers hence (k+m) & (k-m) must divide by 2 to give integer. this means all possible value pairs of (k,m) are valid for (a,b).

(a,b) can take 32 total valid ordered pairs. D

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15 Jul 2026, 02:32

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a^2 - b^2 = (a - b) (a + b) = 945

Let (a-b) = x and (a + b) = y. So, xy = 945

Since 945 is odd, both x and y must be odd.

Now, 945 = 3^3 * 5 * 7

No. of +ve factors = (3+1) (1+1)(1+1) = 16

There will be 16 +ve ordered factor pairs and similarly 16 -ve ones

Total ordered pairs (x,y) = 32, and each gives a unique (a,b)

Ans : D

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15 Jul 2026, 02:44

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The correct answer is D.

Lets look at the equation given.

945= a^2-b^2= (a-b) (a+b)
Let, x= a-b and y= a+b
so, x.y= 945

Solving for a and b,
Adding both x and y, we get a= (x+y)/2 and b = (y-x)/2
Hence, we now know that x+y and x-y are even.

The factors of 945: 3^3*5^1*7^1
Total positive factors= (3+1)*(1+1)*(1+1)= 4*2*2= 16.

There are negative pairs as ewll, thus 16*2= 32 pairs.

Hence the answer.

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15 Jul 2026, 02:46

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Bunuel

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

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(a - b)(a + b) = 945

Now: 945 = 3^3 * 5 * 7

This gives us the total number of positive divisors of 945: (3 + 1)(1 + 1)(1 + 1) = 16

Given that 945 is odd, it means both factors are odd and we can take all divisor pairs.

Thus, we have 16 positive factor pairs.

For every factor pair, we get: one solution of (a, b) one of (-a, -b)

Hence, total ordered pairs: 16 * 2 = 32

option D

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15 Jul 2026, 02:58

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945 = a^2 - b^2 = (a+b)(a-b)

Let x = a+b and y = a-b. Hence, xy = 945. So we need to find all the product combinations that result in 945
Factorize 945 = 3*3*3*5*7 = 3^3 * 5^1 * 7^1

Total number of factors = (3+1)(1+1)(1+1) = 16.

Possible combinations of (x,y)
(1,945), (3,315), (5,189), (7,135), (9,105), (15,63), (21,45), (27,35), (35,27), (45,21), (63,15), (105,9), (135,7), (189,5), (315,3), (945,1) = 16 pairs. Becasue there are 16 different factors, there can only be 16 ordered pairs of x and y.

The above pairs can be in negatives as well ((-1,-945), (-3,-315)...) which will give me another 16 pairs. For each of these pairs, there will be one unique value of a and b (as shown in a few examples below) and hence there will be 32 ordered pairs of a and b in total.

Ex1. x = 945, y = 1. a = 473 and b = 472. Hence x = (473+472), y = (473-472) and (473+472)(473-472) = 945
Ex2. x = 315, y = 3. a = 159 and b = 156. Hence x = (159+156), y = (159-156) and (159+156)(159-156) = 945

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15 Jul 2026, 03:43

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a^2 - b^2 = (a-b)(a+b)
945 = 5*189 = 5*3*63 = 5*3*7*3*3

Using the above data I used a trial and error method - Total possible ways to right 945 as multiplication of 2 numbers:
1) 1*945 = 473 = a, 472 = b
2) 3*315 = 159 = a 156 = b
3) 5*189 = 97 = a 92 =b
4) 7*135 = 71 = a 64= b
5) 9*105 = 57 = a 48=b
6) 615*63 = 39 = a 24 =b
7) 21*45 = 33 = a 12=b
8) 27*25 = 31 = a 4 = b

Hence total positive pairs is 8.

But I took more time in solving the same through above approach

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15 Jul 2026, 03:49

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a2 − b2 = (a+b)(a−b) = 945
945 = 33 × 5 × 7
no. of factor pairs of 945 = (3+1)(1+1)(1+1) = 16 divisors ⇒ 8 +ve pairs.
(a+b) and (a−b) must both be odd. all factor pairs work.
8 +ve pairs + 8 −ve pairs = 16 factor pairs.
each factor pair gives 1 integer (a,b):
a = (x+y)/2, b = (x−y)/2
ans = 16 (b)

Bunuel

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

This question was provided by GMAT Club
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15 Jul 2026, 04:16

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Here's my solution for this question

Attachments

IMG_20260715_164621.jpg [ 83.43 KiB | Viewed 111 times ]

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15 Jul 2026, 04:27

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a^2-b^2 = 945
(a+b)(a-b) = 945
So lets find the total factors for 945
3^3*5*7
So (3+1)*(1+1)*(1+1) = 16
This tells us that there are 16 positive devisor so there would be 16 ordered pairs.
But we haven't been told that a and b are positive so if we consider the negative pairs as well.
It would be 16*2 = 32.
Answer is D.

Bunuel

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

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15 Jul 2026, 06:01

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945 = 3^3*5*7

(a+b)(a-b) = 945
So we need to identify pairs (a,b)
Lets consider scenario 1: a+b = 7*5, a-b = 3^3
We can identify a and b from here
Also we can create 3 more cases from here - one reverse, one both negative, one both negative reverse.

So for each scenario, 4 cases can be created

Let’s count different scenarios now based on first term = (3,rest), (5, rest), (7,rest), ... (3*5, ), (3*7, ), (5*7, ), (3*5*7, ), (3^2*5*7, )

Total 8 scenarios
So 8*4 =32 is the answer (D)

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15 Jul 2026, 06:03

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a^2-b^2 can be expressed as (a+b)(a-b)=945
Factorize 945=3^3x5x7 all of which are odd
Total factors will therefore be:
(3+1)(1+1)(1+1)= 16 total pairs
Since both positive and negative factors would yield the same results then we can safely conclude that the total factors are:
16+16= 32
Ans D

Bunuel

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

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15 Jul 2026, 06:29

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We know that
a and b are integers
a^2 - b^2 = 945

We need to find possible ordered pairs of (a,b) which are valid

Now,
a^2 - b^2 = 945
(a+b)*(a-b) = 945

=> a+b and a-b must be odd since 945 is odd

Let us say,
a + b = x
a - b = y
=> x and y are odd

=> a = x + y / 2
=> b = x - y / 2

For this to be valid, x+y and x-y should be even
=> Only possible when x and y are both even or odd
But we already know above that x and y are odd
=> x and y are odd

Now,
945 = 3^3 * 5^1 * 7^1

Total number of factors = (3+1)*(1+1)*(1+1) = 4*2*2 = 16
=> 8 positive factor pairs possible

a and b can be negative too
=> 8 negative factor pairs possible

Total ordered pairs (a,b) = 8+8 = 16

B. 16

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15 Jul 2026, 06:37

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a^2-b^2 = (a-b)*(a+b) = x*y

x = a-b
y = a+b

Solving for a and b:
a=(x+y)/2
b=(y-x)/2

x and y must be both even or both odd so a and b are integers (their sum and difference are even)

945 = 3^3 * 5 * 7 = x*y
all its factors are odd so we can choose any combination of them: choose one factor x of 945 and y=945/f

number of factors: 4*2*2=16
choosing -x and -y also gives us another ordered pair -> another 16

16+16=32

IMO D

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15 Jul 2026, 06:59

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945 = a^2 - b^2 = (a-b)*(a+b)
945 = 3^3*5*7

For example:
(a-b)=1, (a+b)=945
(a-b)=3, (a+b)=3^2*5*7
...

The question is if all this equations do that a and b are integers:

a-b=m
a+b=n

2a=m+n, a=(m+n)/2
2b=n-m, b=(n-m)/2

a and b are integers if m and n have same parity and, as all factors of 945 are odd, then a and b are always integers choosing any factor of 945.

factors=(3+1)*(1+1)*(1+1)=16
As 945 is positive, it is possible to choose the negative of each ordered pair and obtain a new ordered pair.

16*2=32

Answer D

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15 Jul 2026, 07:02

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IMO 32
(a+b)(a-b) = 945
945 = 3*3*3*5*7
let a+b =x and a-b =y
now x. y = 945
also since x=a+b and y = a-b
we can equate them and get = => a = x+y /2 b = y-x/2 == om trial method we find
therefore 32

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15 Jul 2026, 07:20

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a^2-b^2=945
(a+b)(a-b)=945

a and b must be integers (positive or negative)
(a+b) and (a-b) are integers with same parity and, as 945 is odd, then (a+b) and (a-b) are both odd.

945=7*5*3^3
number of factors of 945 = 2*2*4=16

choose factor=(a+b) and 945/factor=(a-b) and solve for a and b to obtain the integers: 16 solutions
factors of 945 are odd, so any combination of factors results in a and b integers.

it can be chosen too -factor=(a+b) and -945/factor=(a-b) because minus*minus=plus: 16 solutions

16+16=32

The answer is D