The ABC Mystery

Clearly, if $B = \{0,1,2,3\}$ , this forces $C = B$ and $A = 0$ , which cannot happen by the given condition. From here, we have

$B = \{4,5,6,7,8,9\}$ $B^2 = \{16, 25, 36, 49, 64, 81\}$

Since we do not want $B^2 \equiv C \bmod 10$ , where $B = C$ , this rules out $B = 5$ and $B = 6$ . Squaring $B \geq 4$ carries out the arbitrary tens digit of $B^2$ for the following sum:

(\text{Arbitrary digit of \$B^2\$}) + B^2 = AB

which in modular arithmetic

(\text{Arbitrary digit of \$B^2\$}) + B^2 \equiv B \bmod 10

In other words, we are finding the digits for $A, B, C$ that satisfy the above condition. Thus, since $81 + 8 \equiv 8 + 1 \equiv 9 \bmod 10$ , $B = 9$ , $C = 1$ and $A = 8$ .

The following digits for $B$ don't work

• If $B = 4$ , $1 + 6 = 7 \neq B$ .
• If $B = 7$ , $4 + 9 \equiv 3 \bmod 10$ , which doesn't match $B$ .
• If $B = 8$ , $6 + 4 \equiv 0 \bmod 10$ , which also doesn't match $B$ .

$99 \times 9 = 891$ , so the last digit is $\boxed { 1 }$ .