Cool Multiplications

in 10! or 10 square last digit is 10 so no matter whichever digit you multiply with it is 0. So, one need not find the other two last digits. Ans : 0

The last digit of $A$ and $B$ is 0,because $10!$ and $10^{2}$ contains a 10 and multiplying by 10 adds a 0 to a number , so $a = 0,b = 0$ .Since anything multiplied by 0 is 0, $a \times b \times c = 0$ .

As $A$ already contains at least product of $10^3$ , thus, the value of $A$ ends with $000$ . Thus, after being multiplied by $B$ , it can only increase the number of $0's$ behind. Thus, $C$ ends with $000$ .

Hence, $a$ x $b$ x $c = 0$ x $0$ x $0$ = $\boxed{0}$

$A$ can be rewritten as $10!\cdot 10!$ and hence $A \times B = (10!)^{3} \times x$ . (Let's say $x = 1! \times 2! \times \ldots \times 9!$ , we just want the last three digits). Well, we know $(10!)^{3}$ ends with six $0$ 's, so multiplying it with $x$ will give more than six $0$ 's. That's enough to conclude that the last three digits are $\overline{000}$ . Thus, $a \times b \times c = 0 \times 0 \times 0 = \boxed {0}$ .