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 \pmod 4</math> , and <math>4</math> for <math>N=0 \pmod 4</math>. So the answer is <math>\boxed{75}</math>. | ||
== See also == | == See also == |
Latest revision as of 14:34, 28 May 2021
Problem
(a) Let . For how many between and inclusive is a multiple of ?
(b) For how many between and inclusive is a multiple of 5?
Solution
(a) Looking at the units digits, we need the units digit of to be either or . We know that will always have a units digit of . Looking at , however, cycles every four powers with units digits and in that order. We see that we can only get a units digit of if we have as a units digit for , and there is no way to get as a units digit. Therefore, our answer is because the four units digits cycle times in the integers to .
(b) Similarly, cycles every four powers with units digits and in that order. And cycles every two powers with units digits and . Together the units digit of their sum is for , and for . So the answer is .
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 |