Difference between revisions of "2011 AMC 8 Problems/Problem 22"
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<math> \textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7 </math> | <math> \textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }7 </math> | ||
| − | ==Solution== | + | ==Solution 1== |
The first couple powers of <math>7</math> are <math>7, 49, 343, 2401, 16807.</math> As you can see, the last two digits cycle after every 4 powers. <math>7^{1}\ (\text{mod }100) \equiv 7^{5}\ (\text{mod }100) \equiv 7^{2009}\ (\text{mod }100).</math> From there, we go two more powers. The last two digits are <math>43</math> so the tens digit is <math>\boxed{\textbf{(D)}\ 4}</math> | The first couple powers of <math>7</math> are <math>7, 49, 343, 2401, 16807.</math> As you can see, the last two digits cycle after every 4 powers. <math>7^{1}\ (\text{mod }100) \equiv 7^{5}\ (\text{mod }100) \equiv 7^{2009}\ (\text{mod }100).</math> From there, we go two more powers. The last two digits are <math>43</math> so the tens digit is <math>\boxed{\textbf{(D)}\ 4}</math> | ||
| + | ==Solution 2== | ||
| + | We want the tens digit | ||
==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 10:45, 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
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. 
