If \(x^2-y^2=45\), then
\((x+y)(x-y)=45\)

\((x+y)=a\): one factor of 45
\((x-y)=b\): another factor of 45

Factors of 45
\((a)*(b)=45\)
\(1*45=45\)
\(3*15=45\)
\(5*9=45\)

The first set of factors will not work
\(x+y=45\)
\(x-y=1\) Add
\(2x=46\)
\(x=23\)

Second set of factors
\(x+y=15\)
\(x-y=3\) Add
\(2x=18\)
\(x=9\)

\(x-y=3\)
\(9-y=3\)
\(y=6\)

One pair, xy and yx: 96 and 69
\((x^2-y^2)=(9^2-6^2)=(81-36)=45\)

Third set of factors
\(x+y=9\)
\(x-y=5\) Add
\(2x=14\)
\(x=7\)

\(x-y=5\)
\(7-y=5\)
\(y=2\)

Another pair: 72 and 27
\((x^2-y^2)=(7^2-2^2)=(49-4)=45\)

There are no more factor sets.

xy and yx are
96 and 69
72 and 27

Number of pairs: 2

Answer C
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\(1≤(x,y)≤9…2≤(x+y)≤18\)
\(x^2-y^2=45…(x+y)(x-y)=45…factor.pairs(45)=(45,1);(15,3);(9,5)\)
\((x+y)(x-y)=(45)(1)…(x+y)=45=invalid…(x+y≤18)\)
\((x+y)(x-y)=(15)(3)…(x+y)=15…(x-y)=3…2x=18…x=9…y=6\)
\((x+y)(x-y)=(9)(5)…(x+y)=9…(x-y)=5…2x=14…x=7…y=2\)
\((xy,yx)=(96,69);(72,72)=2.pairs\)

Answer (C)

Given: xy and yx are a pair of reversed two digit positive integers.

Asked: If x^2-y^2=45, how many such pairs are there?

(x-y)(x+y) = 45 = 9*5 = 15*3

Case 1:
x-y = 5
x+y =9
(x,y) = (7,2)

Case 2:
x-y = 3
x+y = 15
(x,y) = (9,6)

xy = {72,27,96,69}
Pairs = 2

IMO C
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