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16 Sep 2014, 01:31
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What is the 101st digit that appears after the decimal point in the decimal representation of the sum \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}\)?
A. \(0\)
B. \(1\)
C. \(5\)
D. \(7\)
E. \(8\)
A. \(0\)
B. \(1\)
C. \(5\)
D. \(7\)
E. \(8\)
16 Sep 2014, 01:31
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Official Solution:
What is the 101st digit that appears after the decimal point in the decimal representation of the sum \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}\)?
A. \(0\)
B. \(1\)
C. \(5\)
D. \(7\)
E. \(8\)
If you have a fraction where the denominator is 9, 99, 999, etc., the decimal equivalent will have a repeating pattern in the decimal part. The repeating part is just the numerator of the fraction, with enough leading zeroes added to match the number of 9s in the denominator.
Examples:
• \(\frac{2}{9} = 0.2222...\)
• \(\frac{3}{99} = 0.030303...\)
• \(\frac{45}{999} = 0.045045045...\)
Thus, for any fraction \(\frac{n}{999...}\), the decimal representation will be \(0.nnn...\), with the number \(n\) (padded with leading zeroes if necessary) recurring as many times as there are 9s in the denominator.
Notice that each term in the provided sum can be expressed as a fraction with a certain number of 9s in the denominator:
\(\frac{1}{3} = \frac{3}{9} =0.333...\)
\(\frac{1}{9} = 0.111...\)
\(\frac{1}{27} =\frac{37}{999} =0.037...\)
\(\frac{1}{37} =\frac{27}{999} =0.027...\)
Hence, \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}=0.333... + 0.111... + 0.037... + 0.027... = 0.508...\)
Alternatively, since the denominators of each fraction are factors of 999, we can express all fractions with 999 in the denominator:
\(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}=\)
\(=\frac{333}{999} + \frac{111}{999} + \frac{37}{999} + \frac{27}{999}=\)
\(=\frac{508}{999}=\)
\(=0.508508...\).
In this decimal representation, the repeating part is 508. Therefore, every third digit is 8 (such as the 3rd, 6th, 9th, ... digits). The 102nd digit will be 8, because 102 is a multiple of 3. Hence, the digit preceding it, which is the 101st digit, must be 0.
Answer: A
What is the 101st digit that appears after the decimal point in the decimal representation of the sum \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}\)?
A. \(0\)
B. \(1\)
C. \(5\)
D. \(7\)
E. \(8\)
If you have a fraction where the denominator is 9, 99, 999, etc., the decimal equivalent will have a repeating pattern in the decimal part. The repeating part is just the numerator of the fraction, with enough leading zeroes added to match the number of 9s in the denominator.
Examples:
• \(\frac{2}{9} = 0.2222...\)
• \(\frac{3}{99} = 0.030303...\)
• \(\frac{45}{999} = 0.045045045...\)
Thus, for any fraction \(\frac{n}{999...}\), the decimal representation will be \(0.nnn...\), with the number \(n\) (padded with leading zeroes if necessary) recurring as many times as there are 9s in the denominator.
Notice that each term in the provided sum can be expressed as a fraction with a certain number of 9s in the denominator:
\(\frac{1}{3} = \frac{3}{9} =0.333...\)
\(\frac{1}{9} = 0.111...\)
\(\frac{1}{27} =\frac{37}{999} =0.037...\)
\(\frac{1}{37} =\frac{27}{999} =0.027...\)
Hence, \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}=0.333... + 0.111... + 0.037... + 0.027... = 0.508...\)
Alternatively, since the denominators of each fraction are factors of 999, we can express all fractions with 999 in the denominator:
\(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}=\)
\(=\frac{333}{999} + \frac{111}{999} + \frac{37}{999} + \frac{27}{999}=\)
\(=\frac{508}{999}=\)
\(=0.508508...\).
In this decimal representation, the repeating part is 508. Therefore, every third digit is 8 (such as the 3rd, 6th, 9th, ... digits). The 102nd digit will be 8, because 102 is a multiple of 3. Hence, the digit preceding it, which is the 101st digit, must be 0.
Answer: A
General Discussion
24 Oct 2014, 03:00
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Bunuel
I was trying though a different approach,
1/3 - 0.333333....Thus 101 digit in this sequence will be 3
1/9 - 0.111111....Thus 101 digit in this sequence will be 1
1/27-0.037037....Thus 101 digit in this sequence will be 3
1/37-0.027027....Thus 101 digit in this sequence will be 2
Their sum must be 9...
What's wrong with this approach ?
Regards
LS
