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 <math>\boxed{77}</math>
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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>
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<i>-bronzetruck2016<i>
  
 
== See also ==
 
== See also ==

Latest revision as of 18:21, 19 April 2021


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 $17^{2017}\mod4$, 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 (ProblemsAnswer KeyResources)
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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