Difference between revisions of "1983 AHSME Problems/Problem 14"
Sevenoptimus (talk | contribs) (Rewrote solution for clarity) |
Sevenoptimus (talk | contribs) m (Added box at the bottom) |
||
Line 14: | Line 14: | ||
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>. | 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 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.