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== | + | ==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>. | ||
| − | + | ==Solution 2== | |
| − | == | + | 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, |
| − | + | <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 13:45, 18 August 2017
Contents
Problem
What is the minimum number of digits to the right of the decimal point needed to express the fraction 
as a decimal?
Solution 1
We can rewrite the fraction as 
. Since the last digit of the numerator is odd, a 
is added to the right if the numerator is divided by 
, and this will continuously happen because , itself, is odd. Indeed, this happens twenty-two times since we divide by twenty-two times, so we will need 
more digits. Hence, the answer is 
.
Solution 2
Multiply the numerator and denominator of the fraction by 
(which is the same as multiplying by 1) to give 
. Now, instead of thinking about this as a fraction, think of it as the division calculation 
. 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,
is the minimum number of digits to the right of the decimal point needed.
See Also
| 2015 AMC 12A (Problems • Answer Key • Resources) | |
| 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 | |
