Difference between revisions of "1983 AHSME Problems/Problem 14"
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− | + | ==Problem== | |
− | + | The units digit of <math>3^{1001} 7^{1002} 13^{1003}</math> is | |
− | + | <math>\textbf{(A)}\ 1\qquad | |
+ | \textbf{(B)}\ 3\qquad | ||
+ | \textbf{(C)}\ 5\qquad | ||
+ | \textbf{(D)}\ 7\qquad | ||
+ | \textbf{(E)}\ 9 </math> | ||
− | + | ==Solution== | |
− | <math>3\ | + | First, we notice that <math>3^0</math> is congruent to <math>1 \ \text{(mod 10)}</math>, <math>3^1</math> is <math>3 \ \text{(mod 10)}</math>, <math>3^2</math> is <math>9 \ \text{(mod 10)}</math>, <math>3^3</math> is <math>7 \ \text{(mod 10)}</math>, <math>3^4</math> is <math>1 \ \text{(mod 10)}</math>, and so on. This turns out to be a cycle repeating every <math>4</math> terms, so <math>3^{1001}</math> is congruent to <math>3 \ \text{(mod 10)}</math>. |
+ | |||
+ | The number <math>7</math> has a similar cycle, going <math>1, 7, 9, 3, 1, ...</math> Hence we have that <math>7^{1002}</math> is congruent to <math>9 \ \text{(mod 10)}</math>. Finally, <math>13^{1003}</math> is congruent to <math>3^{1003} \equiv 7 \ \text{(mod 10)}</math>. Thus the required units digit is <math>3\cdot 9\cdot 7 \equiv 9 \ \text{(mod 10)}</math>, so the answer is <math>\boxed{\textbf{(E)}\ 9}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AHSME box|year=1983|num-b=13|num-a=15}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 00:51, 20 February 2019
Problem
The units digit of is
Solution
First, we notice that is congruent to , is , is , is , is , and so on. This turns out to be a cycle repeating every terms, so is congruent to .
The number has a similar cycle, going Hence we have that is congruent to . Finally, is congruent to . Thus the required units digit is , so the answer is .
See Also
1983 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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All AHSME Problems and Solutions |
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