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Question

Let x and y be two 2- digit number such that y is obtain by reversing the digits of x.suppose they also satisfy x2y2=m2 for some positive integer m. The value of x+y+m is.

A
88
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B
112
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C
144
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D
154
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Solution

The correct option is D 154
Xab or x=10a+b
yba or y=10b+a
Now x2y2=(10a+b)2(10b+a)2
=99(a2b2)
=32×11(a+b)(ab)..........(1)
According of Q
(a+b)(ab)=11 and ab=1
a+b=11 and ab=1
a=6,b=5
Hence
x=65
y=56 and m=33
x+y+m=154

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