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Random Interesting Patterns Followed by Square Numbers

A number cons...

Question

A number consists of two digits. The digit in the ten's place is greater than unit place by 4. sum of square of digits of the number is 15 less than the number. Find the number.

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Solution

Let unit's digit be x.
Then ten's digit = $x + 4$
Number = $10 (x + 4) + x = 11 x + 40$
Therefore,
$x^{2} + (x + 4)^{2} + 15 = 11 x + 40$
$2 x^{2} - 3 x - 9 = 0$
$(x - 3) (2 x + 3) = 0$
$x = 3$
Hence, the number is $33 + 40 = 73$ .

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