1989 AMC 12 Problem 1

Attempt Problem 1 of the 1989 AMC 12 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 1989 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

(-1)^{5^2}+1^{2^5}=

-7

-2

0

1

57

Evaluate the exponents from the top down

The parity of the exponent on

-1

determines its sign

Because

5^2=25

is odd,

(-1)^{5^2}=-1.

Also

1^{2^5}=1,

so the sum is

0.

Thus the correct answer is C.

Problem 1 in other years