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16 Nov 2020, 04:00
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B
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Difficulty:
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59% (01:20) correct
41%
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wrong
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A bar over a sequence of digits in a decimal indicates that the sequence repeats indefinitely. If \(\frac{m}{n} = 0.\overline{36}\), where \(m\) and \(n\) are positive integers, what is the least possible value of \(m\) ?
A. 3
B. 4
C. 7
D. 13
E. 22
D01-08
A. 3
B. 4
C. 7
D. 13
E. 22
D01-08
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05 May 2025, 00:22
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Official Solution:
A bar over a sequence of digits in a decimal indicates that the sequence repeats indefinitely. If \(\frac{m}{n} = 0.\overline{36}\), where \(m\) and \(n\) are positive integers, what is the least possible value of \(m\) ?
A. 3
B. 4
C. 7
D. 13
E. 22
Let's begin by converting the recurring decimal \(0.\overline{36}\) to a fraction. There are several methods to do this.
APPROACH 1. Put the recurring number in the numerator and put as many nines in the denominator as there are digits in the recurring number. Hence, for \(0.\overline{36}\), we put 36 in the numerator and 99 (two 9's) in the denominator, since there are two recurring digits. Thus, \(\frac{m}{n} = 0.\overline{36}=\frac{36}{99}\). To find the least positive integer value of \(m\), we simplify the fraction to its lowest terms, which is \(\frac{m}{n} =\frac{36}{99}=\frac{4}{11}\). Therefore, the least possible value of \(m\) is 4.
P.S. You won't need this for the GMAT, but if you're still interested, here is how to convert mixed recurring decimals such as \(0.2\overline{18}\), into fractions. First, write a number, consisting of the non-repeating digits and the repeating digits together, which in this case is 218. Next, subtract the non-repeating digits (2) from this number to get 216, and put it in the numerator. To form the denominator, put as many nines as there are repeating digits, followed by as many zeros as there are non-repeating digits. For \(0.2\overline{18}\), there are two repeating digits and one non-repeating digit, so the denominator would be 990. We get \(0.2\overline{18}=\frac{218-2}{990}=\frac{216}{990}=\frac{12}{55}\).
APPROACH 2. Equate the recurring decimal to \(x\). We get \(x = 0.\overline{36}\). Multiply \(x\) by a power of 10, such that only the recurring part appears after the decimal point. For \(0.\overline{36}\), it would mean multiplying it by 100 to get \(100x=36.\overline{36}\). Subtract \(x\) from this equation to eliminate the recurring part: \(100x - x = 99x=36.\overline{36}-0.\overline{36}=36\). Therefore, we get that \(x=\frac{m}{n} = 0.\overline{36}=\frac{36}{99}\). From here we proceed the same way as in the first approach to get \(m=4\).
P.S. You won't need this for the GMAT, but if you're still interested, here is how to convert mixed recurring decimals such as \(0.2\overline{18}\), into fractions with this approach. Equate the recurring decimal to \(x\). We get \(x = 0.2\overline{18}\). Next, multiply the recurring decimal by a power of 10 so that only the recurring numbers appear after the decimal point. For \(0.2\overline{18}\), this means multiplying by 10, giving \(10x=2.\overline{18}\). Then, multiply the recurring decimal by a power of 10 so that the recurring numbers also appear before the decimal point. For \(0.2\overline{18}\), this means multiplying by 1000, giving \(1000x=218.\overline{18}\). Subtracting one from the other yields \(1000x-10x=218.\overline{18}-2.\overline{18}\), which simplifies to \(990x=216\). Therefore, we can write \(x=\frac{216}{990}=\frac{12}{55}\).
Answer: B
A bar over a sequence of digits in a decimal indicates that the sequence repeats indefinitely. If \(\frac{m}{n} = 0.\overline{36}\), where \(m\) and \(n\) are positive integers, what is the least possible value of \(m\) ?
A. 3
B. 4
C. 7
D. 13
E. 22
Let's begin by converting the recurring decimal \(0.\overline{36}\) to a fraction. There are several methods to do this.
APPROACH 1. Put the recurring number in the numerator and put as many nines in the denominator as there are digits in the recurring number. Hence, for \(0.\overline{36}\), we put 36 in the numerator and 99 (two 9's) in the denominator, since there are two recurring digits. Thus, \(\frac{m}{n} = 0.\overline{36}=\frac{36}{99}\). To find the least positive integer value of \(m\), we simplify the fraction to its lowest terms, which is \(\frac{m}{n} =\frac{36}{99}=\frac{4}{11}\). Therefore, the least possible value of \(m\) is 4.
P.S. You won't need this for the GMAT, but if you're still interested, here is how to convert mixed recurring decimals such as \(0.2\overline{18}\), into fractions. First, write a number, consisting of the non-repeating digits and the repeating digits together, which in this case is 218. Next, subtract the non-repeating digits (2) from this number to get 216, and put it in the numerator. To form the denominator, put as many nines as there are repeating digits, followed by as many zeros as there are non-repeating digits. For \(0.2\overline{18}\), there are two repeating digits and one non-repeating digit, so the denominator would be 990. We get \(0.2\overline{18}=\frac{218-2}{990}=\frac{216}{990}=\frac{12}{55}\).
APPROACH 2. Equate the recurring decimal to \(x\). We get \(x = 0.\overline{36}\). Multiply \(x\) by a power of 10, such that only the recurring part appears after the decimal point. For \(0.\overline{36}\), it would mean multiplying it by 100 to get \(100x=36.\overline{36}\). Subtract \(x\) from this equation to eliminate the recurring part: \(100x - x = 99x=36.\overline{36}-0.\overline{36}=36\). Therefore, we get that \(x=\frac{m}{n} = 0.\overline{36}=\frac{36}{99}\). From here we proceed the same way as in the first approach to get \(m=4\).
P.S. You won't need this for the GMAT, but if you're still interested, here is how to convert mixed recurring decimals such as \(0.2\overline{18}\), into fractions with this approach. Equate the recurring decimal to \(x\). We get \(x = 0.2\overline{18}\). Next, multiply the recurring decimal by a power of 10 so that only the recurring numbers appear after the decimal point. For \(0.2\overline{18}\), this means multiplying by 10, giving \(10x=2.\overline{18}\). Then, multiply the recurring decimal by a power of 10 so that the recurring numbers also appear before the decimal point. For \(0.2\overline{18}\), this means multiplying by 1000, giving \(1000x=218.\overline{18}\). Subtracting one from the other yields \(1000x-10x=218.\overline{18}-2.\overline{18}\), which simplifies to \(990x=216\). Therefore, we can write \(x=\frac{216}{990}=\frac{12}{55}\).
Answer: B
General Discussion
16 Nov 2020, 04:34
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Bunuel
Notice the pattern:
0.777... --> \(\frac{7}{9}\)
0.717171... --> \(\frac{71}{99}\)
0.712712712... --> \(\frac{712}{999}\)
Thus:
0.363636... --> \(\frac{36}{99} = \frac{4}{11}\) --> smallest m=4, smallest n=11
Proof:
x = 0.363636...
100x = 36.363636...
If we subtract the first equation from the second, the blue portions disappear, as follows:
\(100x - x = 36 - 0\)
\(99x = 36\)
\(x = \frac{36}{99} = \frac{4}{11}\)
More practice:
https://gmatclub.com/forum/express-the- ... 14446.html
