2017 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1

Problem

What are the last two digits of $2017^{2017}$ ?

Solution

In this problem, we will use the Chinese remainder theorem. This question is asking us to find $2017^{2017}\mod 100=17^{2017}\mod 100$ . By the Chinese remainder theorem, we can find $17^{2017}\mod 25$ and $17^{2017}\mod 4$ , and then "combine" them. $17^{2017}\mod4\equiv1^{2017}\mod4\equiv1\mod4$ . To find $17^{2017}\mod25$ , we'll look for a pattern. The pattern is $17, 14, 13, 21, 7, 19, 23, 16, 22, -1, -17, -14, -13, -21, -7, -19, -23, -16, -22, 1$ . Since there are 20 terms, and $2017\equiv17\mod20$ , we have that $17^{2017}\mod25\equiv2\mod25$ . Now, we want to find a number that's $2\mod25$ and $1\mod4$ . We find this number by clever guess and check. We know our number is odd, so it is $27\mod50$ . First, we try $27$ , but that doesn't work. Next, we try $77$ , and that works! So, our answer is $\boxed{77}$ -bronzetruck2016

See also

2017 UNM-PNM Contest II (Problems • Answer Key • Resources)

Preceded by
First question Followed by
Problem 2

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10

All UNM-PNM Problems and Solutions

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2017 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1

Problem

Solution

See also

2017 UNM-PNM Contest II (Problems • Answer Key • Resources)
Preceded by First question	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNM-PNM Problems and Solutions