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15 Jul 2026, 07:23
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a^2-b^2=945,
.: (a+b)(a-b)=(3^3)*5*7
.: No of factors possible for 945 is= 4*2*2=16
That means, 8 are the larger factors and other 8 are their smaller counter part.
We can say no of (a+b)s =8 & other 8 factors are the respective (a-b)s.
Since, 945 is an odd number, no factor of 945 can be even.
Therefore, for all the (a+b) & (a-b) pairs will get integer a & b s.
because, a+b=odd
a-b=odd
------------
.:2a=even, .: a=integer. .:b=integer.
That means total 8 possible positive pairs.
Now the negative counter parts will also come into count.
So 8*4=32 no of ordered pair will be there.
.: D is the correct answer.
.: (a+b)(a-b)=(3^3)*5*7
.: No of factors possible for 945 is= 4*2*2=16
That means, 8 are the larger factors and other 8 are their smaller counter part.
We can say no of (a+b)s =8 & other 8 factors are the respective (a-b)s.
Since, 945 is an odd number, no factor of 945 can be even.
Therefore, for all the (a+b) & (a-b) pairs will get integer a & b s.
because, a+b=odd
a-b=odd
------------
.:2a=even, .: a=integer. .:b=integer.
That means total 8 possible positive pairs.
Now the negative counter parts will also come into count.
So 8*4=32 no of ordered pair will be there.
.: D is the correct answer.
Bunuel
15 Jul 2026, 07:40
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We break down 945 into prime factors = 3^3 * 5 * 7
a^2-b^2 = (a+b)*(a-b)
Comparing both sides of the equation we can divide 3^3*5*7 in two multiplicands and assing one to (a+b) and another to (a-b).
Then solve for a and b an obtain each time an ordered pair.
But we have to be sure than a and b are integers, not fractions.
(a+b) = k
(a-b) = l
a=(k+l)/2
b=(k-l)/2
a and b are integers if k and l are both odd of both even.
All the factors of 945 are odd, then the two multiplicands are odd and a and b are always integers.
factors of 945 = 4*2*2 = 16
But, for example, if the two multiplicands are 1 and 945, then -1 and -945 is a valid solution two. Add 16 more ordered pairs
16 + 16 = 32 ordered pairs
The correct answer is D
a^2-b^2 = (a+b)*(a-b)
Comparing both sides of the equation we can divide 3^3*5*7 in two multiplicands and assing one to (a+b) and another to (a-b).
Then solve for a and b an obtain each time an ordered pair.
But we have to be sure than a and b are integers, not fractions.
(a+b) = k
(a-b) = l
a=(k+l)/2
b=(k-l)/2
a and b are integers if k and l are both odd of both even.
All the factors of 945 are odd, then the two multiplicands are odd and a and b are always integers.
factors of 945 = 4*2*2 = 16
But, for example, if the two multiplicands are 1 and 945, then -1 and -945 is a valid solution two. Add 16 more ordered pairs
16 + 16 = 32 ordered pairs
The correct answer is D
sunshineeee
15 Jul 2026, 07:47
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a^2 - b^2 = (a+b) (a-b) = 945
If we put x=(a+b) and y=(a-b) -> x*y = 945
Factor of 945 = 3^4 * 5 * 7
For 3^4 -> 3^0 to 3^4 = 4 option
For 5 -> 5^0 and 5^1 = 2 option
For 7 -> 7^0 and 7^1 = 2 option
Total 4*2*2 = 16 option
Since 945 can be 2 models: minus * minus and plus * plus
the result should be 16*2 = 32 (answer D)
If we put x=(a+b) and y=(a-b) -> x*y = 945
Factor of 945 = 3^4 * 5 * 7
For 3^4 -> 3^0 to 3^4 = 4 option
For 5 -> 5^0 and 5^1 = 2 option
For 7 -> 7^0 and 7^1 = 2 option
Total 4*2*2 = 16 option
Since 945 can be 2 models: minus * minus and plus * plus
the result should be 16*2 = 32 (answer D)
Bunuel
