Difference between revisions of "2017 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1"
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== Solution== | == Solution== | ||
− | In this problem, we will use the Chinese remainder theorem. This question is asking us to find <math>2017^{2017}\mod 100=17^{2017}\mod 100</math>. By the Chinese remainder theorem, we can find <math>17^{2017}\mod 25</math> and <math>17^{2017}\mod 4</math>, and then "combine" them. <math>17^{2017}\mod4\equiv1^{2017}\mod4\equiv1\mod4</math>. To find <math>17^{2017}\mod4</math>, we'll look for a pattern. The pattern is <math>17, 14, 13, 21, 7, 19, 23, 16, 22, -1, -17, -14, -13, -21, -7, -19, -23, -16, -22, 1</math>. Since there are 20 terms, and <math>2017\equiv17\mod20</math>, we have that <math>17^{2017}\mod25\equiv2\mod25</math>. Now, we want to find a number that's <math>2\mod25</math> and <math>1\mod4</math>. We find this number by clever guess and check. We know our number is odd, so it is <math>27\mod50</math>. First, we try <math>27</math>, but that doesn't work. Next, we try <math>77</math>, and that works! So, our answer is < | + | In this problem, we will use the Chinese remainder theorem. This question is asking us to find <math>2017^{2017}\mod 100=17^{2017}\mod 100</math>. By the Chinese remainder theorem, we can find <math>17^{2017}\mod 25</math> and <math>17^{2017}\mod 4</math>, and then "combine" them. <math>17^{2017}\mod4\equiv1^{2017}\mod4\equiv1\mod4</math>. To find <math>17^{2017}\mod4</math>, we'll look for a pattern. The pattern is <math>17, 14, 13, 21, 7, 19, 23, 16, 22, -1, -17, -14, -13, -21, -7, -19, -23, -16, -22, 1</math>. Since there are 20 terms, and <math>2017\equiv17\mod20</math>, we have that <math>17^{2017}\mod25\equiv2\mod25</math>. Now, we want to find a number that's <math>2\mod25</math> and <math>1\mod4</math>. We find this number by clever guess and check. We know our number is odd, so it is <math>27\mod50</math>. First, we try <math>27</math>, but that doesn't work. Next, we try <math>77</math>, and that works! So, our answer is <cmath>\boxed{77}</cmath> |
+ | <i>-bronzetruck2016<i> | ||
== See also == | == See also == |
Latest revision as of 18:21, 19 April 2021
Problem
What are the last two digits of ?
Solution
In this problem, we will use the Chinese remainder theorem. This question is asking us to find . By the Chinese remainder theorem, we can find and , and then "combine" them. . To find , we'll look for a pattern. The pattern is . Since there are 20 terms, and , we have that . Now, we want to find a number that's and . We find this number by clever guess and check. We know our number is odd, so it is . First, we try , but that doesn't work. Next, we try , and that works! So, our answer is -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 |