# 1996 AJHSME Problems/Problem 15

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## Problem

The remainder when the product $1492\cdot 1776\cdot 1812\cdot 1996$ is divided by 5 is

$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$

## Solution 1

To determine a remainder when a number is divided by $5$, you only need to look at the last digit. If the last digit is $0$ or $5$, the remainder is $0$. If the last digit is $1$ or $6$, the remainder is $1$, and so on.

To determine the last digit of $1492\cdot 1776\cdot 1812\cdot 1996$, you only need to look at the last digit of each number in the product. Thus, we compute $2\cdot 6\cdot 2\cdot 6 = 12^2 = 144$. The last digit of the number $1492\cdot 1776\cdot 1812\cdot 1996$ is also $4$, and thus the remainder when the number is divided by $5$ is also $4$, which gives an answer of $\boxed{E}$.

## Solution 2

$1492\cdot 1776\cdot 1812\cdot 1996 \mod 5$

$\equiv 2\cdot 1\cdot 2\cdot 1 \mod 5$

$=4$, which is option $\boxed{E}$.