There are only three positive two-digit numbers such that each number is equal to the incomplete square + +ab of the sum of its digits. Find the smaller of two of them, being given that the second number exceeds the two of them by 50.
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Assume x is the ten's digit and y is the unit digit of the smaller number
Therefore, 10x + y = x 2 + xy + y 2 ............................................(1)
& 10(x+5) + y = [ x + 5 ] 2 + (x+5)y + y 2
or 10x + 50 + y = x 2 + 25 + 10x + xy + 5y + y 2
or 25 - 4y = x 2 + xy + y 2 ...............................................................(2)
By (1) & (2), y = 5 - 2x
Therefore, x can be 1 or 2
On putting values, x = 1
Therefore number = 13