Difference between revisions of "1996 AJHSME Problems/Problem 15"
Talkinaway (talk | contribs) (Created page with "==Problem== The remainder when the product <math>1492\cdot 1776\cdot 1812\cdot 1996</math> is divided by 5 is <math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qqu...") |
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<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> | ||
| − | ==Solution== | + | ==Solution 1== |
To determine a remainder when a number is divided by <math>5</math>, you only need to look at the last digit. If the last digit is <math>0</math> or <math>5</math>, the remainder is <math>0</math>. If the last digit is <math>1</math> or <math>6</math>, the remainder is <math>1</math>, and so on. | To determine a remainder when a number is divided by <math>5</math>, you only need to look at the last digit. If the last digit is <math>0</math> or <math>5</math>, the remainder is <math>0</math>. If the last digit is <math>1</math> or <math>6</math>, the remainder is <math>1</math>, and so on. | ||
| − | To determine the last digit of <math>1492\cdot 1776\cdot 1812\cdot 1996</math>, you only need to look at the last digit of each number in the product. Thus, we compute <math>2\cdot 6\cdot 2\cdot 6 = 12^2 = 144</math>. The last digit of the number <math>1492\cdot 1776\cdot 1812\cdot 1996</math> is <math>4</math>, and thus the remainder when the number is divided by <math>5</math> is also <math>4</math>, which gives an answer of <math>\boxed{E}</math>. | + | To determine the last digit of <math>1492\cdot 1776\cdot 1812\cdot 1996</math>, you only need to look at the last digit of each number in the product. Thus, we compute <math>2\cdot 6\cdot 2\cdot 6 = 12^2 = 144</math>. The last digit of the number <math>1492\cdot 1776\cdot 1812\cdot 1996</math> is also <math>4</math>, and thus the remainder when the number is divided by <math>5</math> is also <math>4</math>, which gives an answer of <math>\boxed{E}</math>. |
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| + | ==Solution 2== | ||
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| + | <math>1492\cdot 1776\cdot 1812\cdot 1996 \mod 5</math> | ||
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| + | <math>\equiv 2\cdot 1\cdot 2\cdot 1 \mod 5</math> | ||
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| + | <math>=4</math>, which is option <math>\boxed{E}</math>. | ||
==See Also== | ==See Also== | ||
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* [[AJHSME Problems and Solutions]] | * [[AJHSME Problems and Solutions]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 10:29, 27 June 2023
Contents
Problem
The remainder when the product 
is divided by 5 is
Solution 1
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 
.
Solution 2
, which is option .
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. 
