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 $4 + 22 = \boxed{\textbf{(C)}\ 26}$ .

Solution 4 (for those like myself who couldn't think of something smarter to do)

First, we can write the denominator as $2^{22}\cdot 10^4$ and forget about the $10^4$ (however, we will need to add $4$ back to our final answer). Noticing that $2^{22}=\left(2^{11} \right)^2=2048^2,$ we divide 123456789 by 2048 using long division. We get 60281.63525390625 as the result (though the process seems intimidating, it actually doesn't take that long, just a couple of minutes). From there, we notice that there are 11 places after the decimal point in the quotient, which means there will be another 11 after we divide this quotient by 2048 again. So, there will be $22$ places after the decimal in the final quotient. We add back $4$ for a total of $26.$

2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AMC 12 Problems and Solutions

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