1971 AHSME Problems/Problem 13

Problem

If $(1.0025)^{10}$ is evaluated correct to $5$ decimal places, then the digit in the fifth decimal place is

$\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }5\qquad \textbf{(E) }8$

Solution

We can rewrite $(1.0025)^{10}$ as $(1+\tfrac{25}{10^4})^{10}$ . By the Binomial Theorem, we know that this expression equals $1+ \tbinom{10}1\cdot\tfrac{25}{10^4}+\tbinom{10}2\cdot\tfrac{25^2}{10^8}+\tbinom{10}3\cdot\tfrac{25^3}{10^{12}}+\cdots+(\tfrac{25}{10^4})^{10}$ . We are looking for the $\tfrac1{10^5}$ term (the fifth decimal digit), so we can disregard the first two terms, because they will not affect our final result. Thus, we are left with $\tbinom{10}2\cdot\tfrac{25^2}{10^8}+\tbinom{10}3\cdot\tfrac{25^3}{10^{12}}+\cdots = \tfrac{45\cdot25^2}{10^8}+\tfrac{120\cdot25^3}{10^{12}}+\cdots = \tfrac{28,125}{10^8}+\tfrac{1,875,000}{10^{12}}+\cdots$ . The first term here has a $\tfrac1{10^5}$ term of $8$ , and the second term does not contribute to the fifth decimal place, because it does not have enough digits in the numerator. So, we can disregard the subsequent terms, because they will not have enough digits either. Thus, the fifth digit of the expansion is $\boxed{\textbf{(E) }8}$ .

1971 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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1971 AHSME Problems/Problem 13

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