1971 AMC 12 Problem 13

Attempt Problem 13 of the 1971 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 1971 AMC 12 solutions, or check the answer key.

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13.

(1.0025)^{10}

is evaluated correct to

5

decimal places, then the digit in the fifth decimal place is:

0

1

2

5

8

Write

1.0025=1+0.0025

In the binomial expansion, retain terms large enough to affect the fifth decimal place

The binomial expansion gives

\begin{gathered} (1+0.0025)^{10}\\ =1+10(0.0025)\\ \quad+45(0.0025)^2\\ \quad+120(0.0025)^3+\cdots\\ =1.025283125\ldots \end{gathered}

which rounds to

1.02528.

The fifth decimal digit is

8.

Therefore, the correct answer is E.

Problem 13 in other years