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

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Solution

Let units digit $x$ and tens digit be $y$
Number $= 10 y + x$
Reverse number $= 10 x + y$
Sum $= 11 x + 11 y$
$= 11 (x + y)$
For sum to be square
$x + y = 11$
$∴$ Number of ways $=^{7 + 2 - 1} C_{2 - 1} =^{8} C_{1} = 8$

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

Let units digit x and tens digit be yNumber=10y+xReverse number=10x+ySum =11x+11y=11(x+y)For sum to be squarex+y=11∴ Number of ways=7+2−1C2−1=8C1=8

Let units digit $x$ and tens digit be $y$
Number $= 10 y + x$
Reverse number $= 10 x + y$
Sum $= 11 x + 11 y$
$= 11 (x + y)$
For sum to be square
$x + y = 11$
$∴$ Number of ways $=^{7 + 2 - 1} C_{2 - 1} =^{8} C_{1} = 8$