2008 AMC 12A Problem 15

Attempt Problem 15 of the 2008 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 12A solutions, or check the answer key.

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15.

Let

k = 2008^2 + 2^{2008}.

What is the units digit of

k^2 + 2^k?

0

2

4

6

8

The units digit of

2^n

cycles

2, 4, 8, 6

with period

4

Find the units digit of

k

first, then note that

k

is a multiple of

4

The units digit of

2008^2

4.

Since

2008

is a multiple of

4,

the units digit of

2^{2008}

6.

Thus

k

has units digit

0,

and so does

k^2.

Both

2008^2

and

2^{2008}

are multiples of

4,

k

is a multiple of

4.

Therefore the units digit of

2^k

6.

The units digit of

k^2 + 2^k

is then

0 + 6 = 6.

Thus, D is the correct answer.

Problem 15 in other years