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Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 2"
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== Solution ==
 
== Solution ==
 
(a) Looking at the units digits, we need the units digit of <math>Q_n</math> to be either <math>0</math> or <math>5</math>.  We know that <math>1^n</math> will always have a units digit of <math>1</math>.  Looking at <math>2^n</math>, however, cycles every four powers with units digits <math>2, 4, 8,</math> and <math>6</math> in that order.  We see that we can only get a units digit of <math>5</math> if we have <math>4</math> as a units digit for <math>2^n</math>, and there is no way to get <math>0</math> as a units digit.  Therefore, our answer is <math>\boxed{25}</math> because the four units digits cycle <math>25</math> times in the integers <math>1</math> to <math>100</math>.
 
(a) Looking at the units digits, we need the units digit of <math>Q_n</math> to be either <math>0</math> or <math>5</math>.  We know that <math>1^n</math> will always have a units digit of <math>1</math>.  Looking at <math>2^n</math>, however, cycles every four powers with units digits <math>2, 4, 8,</math> and <math>6</math> in that order.  We see that we can only get a units digit of <math>5</math> if we have <math>4</math> as a units digit for <math>2^n</math>, and there is no way to get <math>0</math> as a units digit.  Therefore, our answer is <math>\boxed{25}</math> because the four units digits cycle <math>25</math> times in the integers <math>1</math> to <math>100</math>.
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(b) Similarly, <math>3^n</math> cycles every four powers with units digits <math>3, 9, 7,</math> and <math>1</math> in that order. And <math>4^n</math> cycles every two powers with units digits <math>4</math> and  <math>6</math>. Together the units digit of their sum is <math>0</math> for <math>n=1,2,3</math> Mod(<math>4</math>) , and <math>4</math> for <math>N=0</math> Mod(<math>4</math>). So the answer is <math>\boxed{75}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 00:58, 13 January 2019

Problem

(a) Let $Q_n=1^n+2^n$. For how many $n$ between $1$ and $100$ inclusive is $Q_n$ a multiple of $5$?

(b) For how many $n$ between $1$ and $100$ inclusive is $R_n=1^n+2^n+3^n+4^n$ a multiple of 5?


Solution

(a) Looking at the units digits, we need the units digit of $Q_n$ to be either $0$ or $5$. We know that $1^n$ will always have a units digit of $1$. Looking at $2^n$, however, cycles every four powers with units digits $2, 4, 8,$ and $6$ in that order. We see that we can only get a units digit of $5$ if we have $4$ as a units digit for $2^n$, and there is no way to get $0$ as a units digit. Therefore, our answer is $\boxed{25}$ because the four units digits cycle $25$ times in the integers $1$ to $100$.


(b) Similarly, $3^n$ cycles every four powers with units digits $3, 9, 7,$ and $1$ in that order. And $4^n$ cycles every two powers with units digits $4$ and $6$. Together the units digit of their sum is $0$ for $n=1,2,3$ Mod($4$) , and $4$ for $N=0$ Mod($4$). So the answer is $\boxed{75}$.

See also

2009 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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