M28-36

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:

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:

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:

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
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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

You'd have a carry over 1, from the sum of 102nd digits.
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Thanks ...another thing i learn today !!!

Regards
LS

Hi bunuel

how do you come up with 333/999+111/999+27/999+37/999

can you explain please..

jimmy

Jimmy02 -

999 is the common demoninator for 3, 9, 27 and 37. 3, 9 and 27 are all divisible by 3, so 27 would be a common denominator for them. 37 is prime, so the lowest common multiple between 37 and 27 is 37 x 27, or 999. As for the fractions, 333/999 = 1/3; 111/999 = 1/9; 37/999 = 1/27; and 27/999 = 1/37.
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Hi

yes 999 is common denominator.. I just want to know what if i get different values, how would i deduce them to same expression.

thanks for your reply.

jimmy

jimmy02 -

I am not sure what you mean. Can you give a little more information?
If 999 is the common denominator, we need to set every fraction up with 999 as the denominator.
Thus 1/3 becomes 333/999 (think of it as a proportion; if I multiply the denominator by 333, I need to multiply the numerator by 333)
1/9 becomes 111/999
1/27 becomes 37/999
and, 1/37 becomes 27/999

Is this what you were asking? Let me know.
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I think this is a high-quality question and I agree with explanation. good question

Converting Decimals to Fractions

• To convert a terminating decimal to fraction:
1. Calculate the total numbers after decimal point
2. Remove the decimal point from the number
3. Put 1 under the denominator and annex it with "0" as many as the total in step 1
4. Reduce the fraction to its lowest terms

Example: Convert \(0.56\) to a fraction.
1: Total number after decimal point is 2.
2 and 3: \(\frac{56}{100}\).
4: Reducing it to lowest terms: \(\frac{56}{100}=\frac{14}{25}\)

• To convert a recurring decimal to fraction:
1. Separate the recurring number from the decimal fraction
2. Annex denominator with "9" as many times as the length of the recurring number
3. Reduce the fraction to its lowest terms

Example #1: Convert \(0.393939...\) to a fraction.
1: The recurring number is \(39\).
2: \(\frac{39}{99}\), the number \(39\) is of length \(2\) so we have added two nines.
3: Reducing it to lowest terms: \(\frac{39}{99}=\frac{13}{33}\).

• To convert a mixed-recurring decimal to fraction:
1. Write down the number consisting with non-repeating digits and repeating digits.
2. Subtract non-repeating number from above.
3. Divide 1-2 by the number with 9's and 0's: for every repeating digit write down a 9, and for every non-repeating digit write down a zero after 9's.

Example #2: Convert \(0.2512(12)\) to a fraction.
1. The number consisting with non-repeating digits and repeating digits is 2512;
2. Subtract 25 (non-repeating number) from above: 2512-25=2487;
3. Divide 2487 by 9900 (two 9's as there are two digits in 12 and 2 zeros as there are two digits in 25): 2487/9900=829/3300.
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[quote="Bunuel"]Official Solution:

What is the 101st digit after the decimal point in the decimal representation of \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}\)?

A. \(0\)
B. \(1\)
C. \(5\)
D. \(7\)
E. \(8\)


\(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{37}=\frac{333}{999} + \frac{111}{999} + \frac{37}{999} + \frac{27}{999}=\frac{508}{999}=0.508508...\).

102nd digit will be 8, thus 101st digit will be 0.


Hi Bunuel

Can you please explain how did you figure out that 102nd digit will be 8?

thanks

Every 3rd digit is 8 (3rd, 6th, 9th, ...). 102nd digit (because it's a multiple of 3) will be 8, so the previous digit must be 0.
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Hi Bunuel, thanks for such a comprehensive explanation, could you help solve a question where we had 1/7+1/49+1/343, should we apply the same logic of converting everything into 9s?

No. How can you convert these denominators into 9's?
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