2008 AMC 10A Problem 24

Attempt Problem 24 of the 2008 AMC 10A 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 10A solutions, or check the answer key.

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24.

Let

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

What is the units digit of

k^2 + 2^k?

0

2

4

6

8

Find the units digit of

k

first

2^{2008}

ends in

6

and

2008^2

ends in

4;

then determine

k \bmod 4

to handle

2^k

The units digit of

2^n

cycles

2, 4, 8, 6,

2^{2008}

ends in

6.

Also

2008^2

ends in

4.

Thus

k

ends in

0,

k^2

ends in

0.

Both

2008^2

and

2^{2008}

are multiples of

4,

k \equiv 0 \pmod 4,

which makes

2^k

end in

6.

The units digit of

k^2 + 2^k

0 + 6 = 6.

Thus, the correct answer is D.

Problem 24 in other years