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M28-36

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M28-36  [#permalink]

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New post 16 Sep 2014, 00:31
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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. \(8\)

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Re M28-36  [#permalink]

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New post 16 Sep 2014, 00:31
2
3
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.


Answer: A
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Re: M28-36  [#permalink]

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New post 24 Oct 2014, 02:00
1
Bunuel wrote:
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. \(8\)


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

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


Answer: A



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
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Re: M28-36  [#permalink]

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New post 24 Oct 2014, 02:03
2
lastshot wrote:
Bunuel wrote:
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. \(8\)


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

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


Answer: A



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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Re: M28-36  [#permalink]

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New post 24 Oct 2014, 02:05
Bunuel wrote:
lastshot wrote:
Bunuel wrote:
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. \(8\)


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

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


Answer: A



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.



Thanks ...another thing i learn today !!!

Regards
LS
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Re: M28-36  [#permalink]

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New post 03 Aug 2015, 09:51
Hi bunuel

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

can you explain please..

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Re: M28-36  [#permalink]

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New post 03 Aug 2015, 10:03
1
1
jimmy02 wrote:
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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Re: M28-36  [#permalink]

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New post 03 Aug 2015, 21:21
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

VeritasPrepDennis wrote:
jimmy02 wrote:
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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Re: M28-36  [#permalink]

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New post 06 Aug 2015, 06:35
1
jimmy02 wrote:
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

VeritasPrepDennis wrote:
jimmy02 wrote:
Hi bunuel

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

can you explain please..

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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Re: M28-36  [#permalink]

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New post 18 Oct 2015, 06:40
3
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Three things:
1. This question is the same as M05-04
2. In the explanation, should read 333/999 + 111/999 + 37/999 + 27/999 = 508/999 instead of 333/999 + 111/999 + 27/999 + 37/999 = 508/999 because it implies that 27/999 corresponds to 1/27 and 37/999 corresponds to 1/37 and it is the other way around. Not written incorrectly, as order does not matter, just confusing if you are trying to figure it out.
3. Would be helpful to include in explanation the following: We are dealing with a repeating decimal in this question. It's helpful to know that there's a way to write these kinds of decimals as a fraction. For example, the repeating decimal 0.444444444(4) may be written as 4/9. So, 5/9, 7/9 and 8/9 will all be repeating decimals. You might check it in your calculator. In order to make two decimal points repeat, you have to divide the two digit number by 99. For example, 23/99=0.232323232323(23). In order to make 3 decimal points repeat, you have to divide the three digit number by 999. For example, 508/999=0.508508508508(508)
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Re: M28-36  [#permalink]

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New post 19 Jan 2016, 09:30
danjbon wrote:
Three things:
1. This question is the same as M05-04
2. In the explanation, should read 333/999 + 111/999 + 37/999 + 27/999 = 508/999 instead of 333/999 + 111/999 + 27/999 + 37/999 = 508/999 because it implies that 27/999 corresponds to 1/27 and 37/999 corresponds to 1/37 and it is the other way around. Not written incorrectly, as order does not matter, just confusing if you are trying to figure it out.

3. Would be helpful to include in explanation the following: We are dealing with a repeating decimal in this question. It's helpful to know that there's a way to write these kinds of decimals as a fraction. For example, the repeating decimal 0.444444444(4) may be written as 4/9. So, 5/9, 7/9 and 8/9 will all be repeating decimals. You might check it in your calculator. In order to make two decimal points repeat, you have to divide the two digit number by 99. For example, 23/99=0.232323232323(23). In order to make 3 decimal points repeat, you have to divide the three digit number by 999. For example, 508/999=0.508508508508(508)


Updated the question.
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Re M28-36  [#permalink]

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New post 18 Aug 2016, 17:11
I think this is a high-quality question and I agree with explanation. good question
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Re: M28-36  [#permalink]

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New post 28 Nov 2018, 04:20
Bunuel How did you figure out that 1/37 is the same as 27/999 ?
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Re: M28-36  [#permalink]

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New post 03 Dec 2018, 05:59
(1/3) + (1/9) + (1/27) + (1/37) =(13716/26973)=0.508508.....
Since after the decimal point the number in the third position happens to be 8 and 99th position is a multiple of the 3rd position, it follows that - according to the rule of cyclicity - 5 will take the 100th position, 0 will take 101th position. So A is the answer.
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Re: M28-36  [#permalink]

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New post 04 Dec 2018, 23:40
Bunuel wrote:
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.

I'm a bit confused as to how you converted 508/999 to .508508... etc. Can you just assume that the 999 is close enough to 1,000?


Answer: A
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Re: M28-36  [#permalink]

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New post 05 Dec 2018, 04:25
thinkpad18 wrote:
Bunuel wrote:
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.

I'm a bit confused as to how you converted 508/999 to .508508... etc. Can you just assume that the 999 is close enough to 1,000?


Answer: A


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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Re: M28-36  [#permalink]

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New post 05 Dec 2018, 07:53
Bunuel wrote:
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.

@


Answer: A


Bunuel :
really appreciate your solution to the problem , but can you please debrief on how you decided to go by taking 999 as the common denominator for all , usually in fractions we take the LCM of the terms or common multiplier term... i agree that here its 999 but how did you come up with .. any input would be great..
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Re: M28-36 &nbs [#permalink] 05 Dec 2018, 07:53
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