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2008 AMC 12A Problems/Problem 15

Revision as of 19:58, 1 August 2021 by Graceandmommymath (talk | contribs) (→‎Note)
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 $k={2008}^{2}+{2}^{2008}$. What is the units digit of $k^2+2^k$?

$\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8$

Solution

$k \equiv 2008^2 + 2^{2008} \equiv 8^2 + 2^4 \equiv 4+6 \equiv 0 \pmod{10}$.

So, $k^2 \equiv 0 \pmod{10}$. Since $k = 2008^2+2^{2008}$ is a multiple of four and the units digit of powers of two repeat in cycles of four, $2^k \equiv 2^4 \equiv 6 \pmod{10}$.

Therefore, $k^2+2^k \equiv 0+6 \equiv 6 \pmod{10}$. So the units digit is $6 \Rightarrow \boxed{D}$.

Note

Another way to get $k \equiv 0 \pmod{10}$ is to find the cycles of the last digit.

For $2008^2$, we need only be concerned with the last digit $8$ since the other digits do not affect the last digit. Since $8^{2} = 64$, the last digit of $2008^2$ is $4$.

For $2^{2008}$, note that the last digit cycles through the pattern ${2, 4, 8, 6}$. (You can try to see this by calculating the first powers of $2$.)

Since $2008$ is a multiple of $4$, the last digit of $2^{2008}$ is evidently $6.$

Continue as follows.

~mathboy282

Mathboy282, That is actually what solution 1 is explaining in the first sentence but I think yours is a more detailed and comprehensible explanation.

Solution 2 (Video solution)

Video: https://youtu.be/Ib-onAecb1I

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

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