Difference between revisions of "2011 AMC 8 Problems/Problem 22"
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==Solution 2== | ==Solution 2== | ||
We want the tens digit | We want the tens digit | ||
− | So, we take <math>7^{ | + | So, we take <math>7^{2011}\ (\text{mod }100)</math>. That is congruent to <math>7^{11}\ (\text{mod}100)</math>. From here, it is an easy bash, 7, 49, 43, 01, 07, 49, 43, 01, 07, 49, 43. So the answer is 4 |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=21|num-a=23}} | {{AMC8 box|year=2011|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:50, 12 November 2019
Contents
Problem
What is the tens digit of ?
Solution 1
The first couple powers of are As you can see, the last two digits cycle after every 4 powers. From there, we go two more powers. The last two digits are so the tens digit is
Solution 2
We want the tens digit So, we take . That is congruent to . From here, it is an easy bash, 7, 49, 43, 01, 07, 49, 43, 01, 07, 49, 43. So the answer is 4
See Also
2011 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
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