Difference between revisions of "1996 AJHSME Problems/Problem 15"
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==Problem== | ==Problem== | ||
| − | The remainder when the product <math>1492\cdot 1776\cdot 1812\cdot 1996</math> is divided by | + | The remainder when the product <math>1492\cdot 1776\cdot 1812\cdot 1996</math> is divided by 500 is |
<math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4</math> | <math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4</math> | ||
Revision as of 16:55, 26 July 2021
Problem
The remainder when the product 
is divided by 500 is
Solution
To determine a remainder when a number is divided by 
, you only need to look at the last digit. If the last digit is 
or , the remainder is . If the last digit is 
or 
, the remainder is , and so on.
To determine the last digit of , you only need to look at the last digit of each number in the product. Thus, we compute 
. The last digit of the number is also 
, and thus the remainder when the number is divided by is also , which gives an answer of 
.
See Also
| 1996 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
