#math trivia #107 solution

Assuming that a is non-zero, the square root must be a three-digit number.  Let xyz be the digits of a number N, so that

N = 100*x + 10*y + z

The square of N is then

N^2 = 10000*x^2 + 1000*(2*x*y) + 100*(2*x*z+y^2) + 10*(2*y*z) + z^2.

When y = 0, the equation simplifies to:

N^2 = 10000*x^2 + 100*(2*x*z) + z^2.

To get the pattern aabbc, we need:

  • x = 1, 2, or 3, so that x^2 is a single digit
  • 2*x*z has the same 10’s digit as x^2
  • 2*x*z has the same 1’s digit as z^2

Let’s consider the three values of x in turn.

If x = 1, then we need 2*x*z to be between 10 and 19, which means that z must be between 5 and 9.  The possible values of 2*x*z are thus 10, 12, 14, 16, and 18.  In the last three cases, 2*x*z has the same 1’s digits as z^2, corresponding to the values

107^2 = 11449
108^2 = 11664
109^2 = 11881

If x = 2, then we need 2*x*z to be between 20 and 29, which means that z must be between 5 and 7.  The possible values of 2*x*z are thus 20, 24, and 28.  None of these cases matches z^2.

If x = 3, z must be either 5 or 6.  None of these cases matches z^2 either.

So, the other numbers (so far) are 108 and 109.

Are there any with y > 0?  A quick spreadsheet calculation reveals two more:  235^2 = 55225 and 315^2 = 99225.

If we allow a be zero, then we can also consider 1^1 = 1, 2^2 = 4, 3^2 = 9, 15^2 = 225, and 21^2 = 441.

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