Question

Consider ab, a two-digit number, and its square cde, a three-digit number. For how many values of ab, is (a + b)2 = c + d + e?
a, b, c, d and e are positive integers.​

Answer 1

Answer:

Ideally, ABC * AB = ABCD should make you think of 1s and 0s. The digits needn't be unique so the first thing that comes to mind is 100*10 = 1000. But since CA and CD are given as two digit numbers, next you should think of 101*10 = 1010 or any such number. This satisfies all conditions so we can analyse using this example. 

101 * 10 = 1010

AD = 10 - a factor of 1010

CD = 10 - a factor of 1010

CA = 11 - not a factor of 1010

Now the only thing left to figure is whether it is possible that C1 is a factor of 10C0 for some value of C.

10C0 = 1000 + C0 = 999 + C1

C1 is divisible by C1 but we don't know whether 999 is divisible by C1.

999 = 3^3*37

It has no two digit factor ending in 1 so it will not be divisible by C1.

Answer (C)