Difference between revisions of "2008 AMC 12A Problems/Problem 15"

Revision as of 19:50, 4 June 2021

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

Solution 2 (Video solution)

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

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.

@@ Line 10: / Line 10: @@
 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>.
+===Note===
+Another way to get <math>k \equiv 0 \pmod{10}</math> is to find the cycles of the last digit.
+For <math>2008^2</math>, we need only be concerned with the last digit <math>8</math> since the other digits do not affect the last digit. Since <math>8^{2} = 64</math>, the last digit of <math>2008^2</math> is <math>4</math>.
+For <math>2^{2008}</math>, note that the last digit cycles through the pattern <math>{2, 4, 8, 6}</math>. (You can try to see this by calculating the first powers of <math>2</math>.)
+Since <math>2008</math> is a multiple of <math>4</math>, the last digit of <math>2^{2008}</math> is evidently <math>6.</math>
+Continue as follows.
+~mathboy282
+==Solution 2 (Video solution)==
+Video: https://youtu.be/Ib-onAecb1I
 ==See Also==

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