1996 AJHSME Problems/Problem 15

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 $\boxed{E}$ .

1996 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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1996 AJHSME Problems/Problem 15

Contents

Problem

Solution 1

Solution 2

See Also