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Re: How many pairs of integer (a, b) are possible such that a^2 – b^2 = 28 [#permalink]
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(a+b)(a-b) =1×288
2×144
3×96
4× 72
6× 48
8× 36
9× 32
12×24
16×18
9 pairs are there out of which one even and one odd is not possible, so 6 pairs are possible.

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How many pairs of integer (a, b) are possible such that a^2 – b^2 = 28 [#permalink]
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Quote:
How many pairs of integer (a, b) are possible such that a^2 – b^2 = 288?

A. 6
B. 12
C. 24
D. 32
E. 48


totalf(288)=2^5*3^2=6*3=18 factors

1 × 288: a+b=288, a-b=1, 2a=289, a=fraction
2 × 144: a+b=144, a-b=2, 2a=146, a&b=integer
3 × 96: fraction
9 × 32: fraction

factors that a&b are not integers (fractions): 6
total integer pairs a&b: 18-6=12+negative pairs=24

ans (C)

Originally posted by exc4libur on 18 May 2020, 06:57.
Last edited by exc4libur on 19 May 2020, 07:09, edited 1 time in total.
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How many pairs of integer (a, b) are possible such that a^2 – b^2 = 28 [#permalink]
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How many pairs of integer (a, b) are possible such that \(a^2 – b^2 = 288\)?

A. 6
B. 12
C. 24
D. 32
E. 48
As a and b both are integers, both can be negative.
\(a^2 – b^2 = 288\)
\((a – b)(a + b) = 288\)
So, (integer – integer)(integer + integer) = 288
integer*integer = 288(both positive and negative integers)

\(288 = 2^5*3^2\). Hence 288 has (5+1)(2+1) = 18 factors (both even and odd included)
Also, 288 an be represented as
Case I: even * even = even
Case II: odd * odd = odd
Case III: even * odd = even
Case IV: odd * even = even

Only first case satisfies because in the last three
even integer - odd integer = odd integer and even integer + odd integer = odd integer

Now, 288 has number o odd integers = 2+1 = 3 (power of odd prime factor + 1 viz. \(3^0, 3^1\) and \(3^2\))
These three would form three pairs belonging to invalid cases II to IV.

Pairs of even factors left = \(\frac{18 - 6}{2}\) = 6 (2,144 | 4,72 | 6,48 | 8,36 | 12,24 | 16,18)
There are equal number of pairs of negative even factors(++, +-, -+, --).

Total pairs = 6 + 6 + 6 + 6 = 24

Answer C.

Originally posted by unraveled on 18 May 2020, 07:47.
Last edited by unraveled on 19 May 2020, 01:18, edited 1 time in total.
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Re: How many pairs of integer (a, b) are possible such that a^2 – b^2 = 28 [#permalink]
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How many pairs of integer (a, b) are possible such that a^2 – b^2 = 288?

A. 6
B. 12
C. 24
D. 32
E. 48
288 has 3*6=18 factors, so 9 pairs
(a-b)(a+b)=288
if a-b=9, a+b=32..a,b not possible
so removing odd, even factor pairs, left with 6 such pairs
a+b=12, a-b=24, a=18, b=-6
a^2-b^2=288,
so a is squared, a can be +ve or -ve...so a=18,-18...b=6,-6
1 factor pair gives 4 such pairs,
so 6 fatcor => 24
Ans C
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How many pairs of integer (a, b) are possible such that a^2 – b^2 = 28 [#permalink]
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Bunuel wrote:

Competition Mode Question



How many pairs of integer (a, b) are possible such that a^2 – b^2 = 288?

A. 6
B. 12
C. 24
D. 32
E. 48


Are You Up For the Challenge: 700 Level Questions


Asked: How many pairs of integer (a, b) are possible such that a^2 – b^2 = 288?

\(a^2 - b^2 = 288\)
\((a+b) (a-b) = 288 = 2*2*2*2*2*3*3 = 2^5*3^2\)
a =1/2 {(a+b) + (a-b)}
b =1/2 {(a+b) - (a-b)}
For a & b to be integers, (a+b) & (a-b) should be both even; Since both odd is not possible
Let (a+b) = 2x ; (a-b) = 2y
\(4xy = 288 = 2^5*3^2\)
\(xy = 2^3*3^2\)
Number of solutions = Number of factors /2 = 4*3/2 = 6
Since a & b can take negative values, number of integer solutions = 4*6 = 24

IMO C
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Re: How many pairs of integer (a, b) are possible such that a^2 – b^2 = 28 [#permalink]
nick1816 wrote:
\(a^2-b^2 = 288\)

\((a+b)(a-b)= 288\)

\(288=2^5*3^2\)

We'll get integral solutions od (a,b) when (a+b) and (a-b) are even. (both can't be odd as their product is even)

Number of ways to write 288 as a product of 2 positive even numbers
\(2x*2y = 288\)
\(x*y=2^3*3^2\)

total unordered solutions of (x,y)=\(\frac{(3+1)*(2+1)}{2}=6\)

Hence positive integral solutions of \(a^2-b^2 = 288\) is equal to 6

Since we require integral solutions, (a,b) can be (++), (-+), (+-) and (--).

integral solutions of \(a^2-b^2 = 288\) is equal 6*4 = 24


Hey nick1816
I didn't understand the following part:
total unordered solutions of (x,y)=\(\frac{(3+1)*(2+1)}{2}=6\)
what is the meaning of unordered solution and how did you find the same ?
the numerator seems to be what we do to find out the number of factors
you divided it by 2 because you want to cancel out the possible arrangements
But I didn't understand the logic
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Re: How many pairs of integer (a, b) are possible such that a^2 b^2 = 28 [#permalink]
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