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Question

# How many two digit positive integers N have the property that the sum of N and number obtained by reversing the order of the digits of N is a perfect square.

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Solution

## Let the two digits be x and y. N=10x+y; Let M be the number obtained on reversing the digits. M=10y+x; N+M=10x+y+10y+x= =11x+11y=11 (x+y); N+M is a perfect square; i.e. 11 (x+y) is a perfect square; i.e. 11 divides the perfect square only 121 is divisible by 11. Hence M+N=121 i.e. 11 (x+y)=121 x+y=11 and x, y are singel digit integers. The integer solutions to the above equation are (2,9), (3,8), (4,7), (5,6), (6,5), (7,4), (8,3), (9,2). Hence there are eight such two digits numbers such that the sum of the number and its reverse is a perfect square. They are 29, 38, 47, 56, 65, 74, 83, 92.

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