Which of the 10 digits does A Represent

Question

in the following problem? (A, S, M, and N are all different)

 

AS * A = MAN

Answer 1

Multiplying AS by A will give you a three-digit number, so A can't be too small.

A can't be 0 (since then the answer would be 0)
A can't be 1 (since then the answer would be AS)
A can't be 2 (since twenty-something times 2 isn't a three-digit number)

 

Let's think about actually multiplying AS by A. It works out like this, with another unknown number R added in:

 

  R

  AS

×  A

---------

 MAN

 

This means that A×S = RN and A² + R = MA.

 

Let's think through possible values of A knowing that.

 

If A = 3, then 9 + R = M3, so R must be 4 and M must be 1: 9 + 4 = 13. If R is 4, then 3×S = 4N. But there's no single-digit number you can multiply by 3 to get to forty-something. So, A isn't 3.

 

If A = 4, then 16 + R = M4, so R must be 8 and M must be 2: 16 + 8 = 24. If R is 8, then 4×S = 8N. But there's no single-digit number you can multiply by 4 to get eighty-something. So, A isn't 4.

 

If A = 5, then 25 + R = M5, so R must be 0 and M must be 2: 25 + 0 = 25. If R is 0, then 5×S = 0N. If S isn't one, N is a two-digit number, which can't happen. If S is one, then N is 5, but A is already 5. And if S is 0, then so is N, which can't happen. So, A isn't 5.

 

If A = 6, then 36 + R = M6, so R must be 0 and M must be 3: 36 + 0 = 36. If R is 0, then 6×S = 0N. Just like with 5, this leads to a contradiction. So, A isn't 6.

 

If A = 7, then 49 + R = M7, so R must be 8 and M must be 5: 49 + 8 = 57. If R is 8, then 7×S = 8N. But there's no single-digit number you can multiply by 7 to get eighty something. So, A isn't 7.

 

If A = 8, then 64 + R = M8, so R must be 4 and M must be 6: 64 + 4 = 68. If R is 4, then 8×S = 4N. S could be 5 or 6. If S is 5, then N is 0. This is actually consistent: 85 × 8 = 680 solves the puzzle. So, A = 8.