Difference between revisions of "2015 AMC 12A Problems/Problem 15"

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<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 104</math>
 
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 104</math>
  
==Solution==
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==Solution 1==
 
We can rewrite the fraction as <math>\frac{123456789}{2^{22} \cdot 10^4} = \frac{12345.6789}{2^{22}}</math>. Since the last digit of the numerator is odd, a <math>5</math> is added to the right if the numerator is divided by <math>2</math>, and this will continuously happen because <math>5</math>, itself, is odd. Indeed, this happens twenty-two times since we divide by <math>2</math> twenty-two times, so we will need <math>22</math> more digits. Hence, the answer is <math>4 + 22 = 26 \textbf{ (C)}</math>.
 
We can rewrite the fraction as <math>\frac{123456789}{2^{22} \cdot 10^4} = \frac{12345.6789}{2^{22}}</math>. Since the last digit of the numerator is odd, a <math>5</math> is added to the right if the numerator is divided by <math>2</math>, and this will continuously happen because <math>5</math>, itself, is odd. Indeed, this happens twenty-two times since we divide by <math>2</math> twenty-two times, so we will need <math>22</math> more digits. Hence, the answer is <math>4 + 22 = 26 \textbf{ (C)}</math>.
  
 
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==Solution 2==
==Alternate Solution==
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Multiply the numerator and denominator of the fraction by <math>5^{22}</math> (which is the same as multiplying by 1) to give <math>\frac{5^{22} \cdot 123456789}{10^{26}}</math>. Now, instead of thinking about this as a fraction, think of it as the division calculation <math>(5^{22} \cdot 123456789) \div 10^{26}</math> . 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 further to the right of the decimal point. Thus,
Note that <math>123456789</math> is not a multiple of <math>2</math> or <math>5</math>, and therefore shares no factors with the original denominator. Multiply the numerator and denominator of the fraction by <math>5^{22}</math> to give <math>\frac{5^{22} \cdot 123456789}{10^{26}}</math>. This fraction will require <math>26</math> divisions by ten to write as a decimal, and since the original fraction is less than <math>1</math> all of the digits will be to the right of the decimal point. Answer: <math>\textbf{ (C)}</math>
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<math>\boxed{\textbf{(C)}\ 26}</math> is the minimum number of digits to the right of the decimal point needed.
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2015|ab=A|num-b=14|num-a=16}}
 
{{AMC12 box|year=2015|ab=A|num-b=14|num-a=16}}

Revision as of 14:45, 18 August 2017

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

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 further 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.

See Also

2015 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
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