2015 AMC 12A Problems/Problem 15

Problem

What is the minimum number of digits to the right of the decimal point needed to express the fraction $\frac{123456789}{2^{26}\cdot 5^4}$ as a decimal?

$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 104$

Solution 1

We can rewrite the fraction as $\frac{123456789}{2^{22} \cdot 10^4} = \frac{12345.6789}{2^{22}}$ . Since the last digit of the numerator is odd, a $5$ is added to the right if the numerator is divided by $2$ , and this will continuously happen because $5$ , itself, is odd. Indeed, this happens twenty-two times since we divide by $2$ twenty-two times, so we will need $22$ more digits. Hence, the answer is $4 + 22 = \boxed{\textbf{(C)}\ 26}$

Solution 2

Multiply the numerator and denominator of the fraction by $5^{22}$ (which is the same as multiplying by 1) to give $\frac{5^{22} \cdot 123456789}{10^{26}}$ . Now, instead of thinking about this as a fraction, think of it as the division calculation $(5^{22} \cdot 123456789) \div 10^{26}$ . The dividend is a huge number, but we know it doesn't have any digits to the right of the decimal point. Also, the dividend is not a multiple of 10 (it's not a multiple of 2), so these 26 divisions by 10 will each shift the entire dividend one digit to the right of the decimal point. Thus, $\boxed{\textbf{(C)}\ 26}$ is the minimum number of digits to the right of the decimal point needed.

Solution 3

The denominator is $10^4 \cdot 2^{22}$ . Each $10$ adds one digit to the right of the decimal, and each additional $2$ adds another digit. The answer is $4 + 22 = \boxed{\textbf{(C)}\ 26}$ .

2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2015 AMC 12A Problems/Problem 15

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also