Difference between revisions of "2008 AMC 12A Problems/Problem 15"
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So, <math>k^2 \equiv 0 \pmod{10}</math>. Since <math>k = 2008^2+2^{2008}</math> is a multiple of four and the units digit of powers of two repeat in cycles of four, <math>2^k \equiv 2^4 \equiv 6 \pmod{10}</math>. | So, <math>k^2 \equiv 0 \pmod{10}</math>. Since <math>k = 2008^2+2^{2008}</math> is a multiple of four and the units digit of powers of two repeat in cycles of four, <math>2^k \equiv 2^4 \equiv 6 \pmod{10}</math>. | ||
− | Therefore, <math>k^2+2^k \equiv 0+6 \equiv 6 \pmod{10}</math>. So the units digit is <math>6 \Rightarrow D</math>. | + | Therefore, <math>k^2+2^k \equiv 0+6 \equiv 6 \pmod{10}</math>. So the units digit is <math>6 \Rightarrow \boxed{D}</math>. |
==See Also== | ==See Also== |
Revision as of 16:40, 13 January 2018
- The following problem is from both the 2008 AMC 12A #15 and 2008 AMC 10A #24, so both problems redirect to this page.
Problem
Let . What is the units digit of ?
Solution
.
So, . Since is a multiple of four and the units digit of powers of two repeat in cycles of four, .
Therefore, . So the units digit is .
See Also
2008 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 AMC 12 Problems and Solutions |
2008 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.