Difference between revisions of "2011 AMC 8 Problems/Problem 22"
Firebolt360 (talk | contribs) (→Solution) |
Firebolt360 (talk | contribs) |
||
| Line 8: | Line 8: | ||
==Solution 2== | ==Solution 2== | ||
We want the tens digit | We want the tens digit | ||
| + | So, we take <math>7^{2009}\ (\text{mod }100)</math>. That is congruent to <math>7^9\ (\text{mod}100)</math>. From here, it is an easy bash, 7, 49, 43, 01, 07, 49, 43, 01, 07 | ||
==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 10:49, 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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
