Difference between revisions of "1983 AHSME Problems/Problem 14"
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==Solution 2== | ==Solution 2== | ||
| − | By Euler's Totient Theorem, if <math>\gcd(a, 10) = 1</math>, then <math>a^{\phi(10)} \equiv a^4 \equiv 1\ (\mathrm{mod}\ 10)</math>. So <math>3^{1001} \cdot 7^{1002} \cdot 13^{1003} \equiv 3^{1001} \cdot 7^{1002} \cdot 3^{1003} \equiv 3^{2004} \cdot 7^{1002} \equiv 1 \cdot 9 \equiv \boxed{\textbf{(E)}\ 9}\ (\mathrm{mod}\ 10)</math>. | + | By Euler's Totient Theorem, if <math>\gcd(a, 10) = 1</math>, then <math>a^{\phi(10)} \equiv a^4 \equiv 1\ (\mathrm{mod}\ 10)</math>. So <math>3^{1001} \cdot 7^{1002} \cdot 13^{1003} \equiv 3^{1001} \cdot 7^{1002} \cdot 3^{1003} \equiv 3^{2004} \cdot 7^{1002} \equiv 3^0 \cdot 7^2 \equiv 1 \cdot 9 \equiv \boxed{\textbf{(E)}\ 9}\ (\mathrm{mod}\ 10)</math>. |
-j314andrews | -j314andrews | ||
Latest revision as of 15:10, 3 July 2025
Contents
Problem
The units digit of 
is
Solution 1
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 
.
Solution 2
By Euler's Totient Theorem, if 
, then 
. So 
.
-j314andrews
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
