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D01-08

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D01-08 [#permalink]

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New post 16 Sep 2014, 00:11
1
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A
B
C
D
E

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  45% (medium)

Question Stats:

63% (00:50) correct 37% (01:20) wrong based on 183 sessions

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Re D01-08 [#permalink]

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New post 16 Sep 2014, 00:11
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Official Solution:

\(m\) and \(n\) are positive integers. What is the smallest possible value of integer \(m\) if \(\frac{m}{n}\) = 0.3636363636...?

A. 3
B. 4
C. 7
D. 13
E. 22


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 \(\frac{4}{9}\). So, \(\frac{5}{9}\), \(\frac{7}{9}\) and \(\frac{8}{9}\) 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, \(\frac{23}{99} = 0.232323232323(23)\). Similarly, \(\frac{36}{99} = \frac{4}{11} = 0.36363636(36)\). Now it's clear that the minimum value of \(m = 4\).

Alternate Solution:

In case you did not know the formula for the repeating decimal (most probably did not), there is another approach to solving this question - backsolving. This is not a typical backsolving question, however, since both of the variables are unknown and we have to make some assumptions to get to the solution.

1. Looking at the repeating decimal - 0.36.... - the ratio between m and n has to be slightly less than 1:3.

2. Let's run through the answer choices:

A. 3 - the number that's slightly less than 3*3 is 8. \(\frac{3}{8} = 0.375\). Does not work.

B. 4 - the number that's slightly less than 3*4 is 11. \(\frac{4}{11} = 0.3636\). Works!

C. 7

D. 13

E. 22

We could continue going through answer choices C, D, and E, but the question asks us for the smallest possible value of m, and 4 is the smallest of the ones that work (even if multiple do) so there is no value to check others.


Answer: B
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Re: D01-08 [#permalink]

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New post 01 Dec 2014, 02:36
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When you look at the question , it does strike that 0.3636.. is 4 * 0.0909.. . I think from there on it becomes very simple.
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D01-08 [#permalink]

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New post 01 Dec 2014, 03:39
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1
m and n are positive integers. What is the smallest possible value of integer m if \frac{m}{n} = 0.3636363636...?

A. 3
B. 4
C. 7
D. 13
E. 22
------------------------------------------------------------------------------------------------------------------------------------------------

it is easy to memorize that
n will be 11 because if we choose any other value smaller than 11 it will not give 2 repeating non terminating values in return.

for eg. 1/3 = .3333...(1 repeating decimal)
1/7 = .142857 142857. ... (6 repeating decimals)
1/9 = .11111... (1 repeating decimals)
1/11= .090909 (2 repeating decimals) --> this is wat we willl pick

Now look at the ans choices
a. 3 upon dividing with 11 will not yield .363636... out
b. 4 upon dividing with 11 will yield .363636 ... correct

Hence B ans!

Regards
SG
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Re: D01-08 [#permalink]

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New post 01 Dec 2014, 07:40
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Bunuel wrote:
Official Solution:

\(m\) and \(n\) are positive integers. What is the smallest possible value of integer \(m\) if \(\frac{m}{n}\) = 0.3636363636...?

A. 3
B. 4
C. 7
D. 13
E. 22


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 \(\frac{4}{9}\). So, \(\frac{5}{9}\), \(\frac{7}{9}\) and \(\frac{8}{9}\) 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, \(\frac{23}{99} = 0.232323232323(23)\). Similarly, \(\frac{36}{99} = \frac{4}{11} = 0.36363636(36)\). Now it's clear that the minimum value of \(m = 4\).

Answer: B

Alternatively,
\(\frac{m}{n}\) = 0.363636... - Eq 1
100\(\frac{m}{n}\) = 36.363636... - Eq 2
(Multiplied by 100 as we have 2 repeating decimals, in case of 3 repeating decimals we'd multiply by 1000 and so on)

Subtract equation1 from 2
99\(\frac{m}{n}\) = 36
\(\frac{m}{n}\) = \(\frac{4}{11}\)
Smallest possible value of m is 4.

It can be used in the similar way for any number of repeating decimals.
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Re: D01-08 [#permalink]

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New post 25 Nov 2015, 04:24
It is easy to spot the answer once we know that 1/11 yields 0,0909
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Re D01-08 [#permalink]

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New post 18 Dec 2015, 07:43
I think this is a high-quality question and I agree with explanation.
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Re: D01-08 [#permalink]

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New post 18 Dec 2015, 07:47
james2329 wrote:
Bunuel wrote:
Official Solution:

\(m\) and \(n\) are positive integers. What is the smallest possible value of integer \(m\) if \(\frac{m}{n}\) = 0.3636363636...?

A. 3
B. 4
C. 7
D. 13
E. 22


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 \(\frac{4}{9}\). So, \(\frac{5}{9}\), \(\frac{7}{9}\) and \(\frac{8}{9}\) 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, \(\frac{23}{99} = 0.232323232323(23)\). Similarly, \(\frac{36}{99} = \frac{4}{11} = 0.36363636(36)\). Now it's clear that the minimum value of \(m = 4\).

Answer: B

Alternatively,
\(\frac{m}{n}\) = 0.363636... - Eq 1
100\(\frac{m}{n}\) = 36.363636... - Eq 2
(Multiplied by 100 as we have 2 repeating decimals, in case of 3 repeating decimals we'd multiply by 1000 and so on)

Subtract equation1 from 2
99\(\frac{m}{n}\) = 36
\(\frac{m}{n}\) = \(\frac{4}{11}\)
Smallest possible value of m is 4.

It can be used in the similar way for any number of repeating decimals.


This is exactly how Bunuel's formula is derived.
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Re: D01-08 [#permalink]

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New post 14 Jul 2016, 02:16
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1. Looking at the repeating decimal - 0.36.... - the ratio between m and n has to be slightly less than 1:3.

Could you please explain how did we arrive at this when we know 0.36> 0.33 how is it that the ratio is slightly less than 1:3.

Could you please clarify this point.

Thanks :)
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Re D01-08 [#permalink]

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New post 20 Aug 2016, 04:55
I think this is a high-quality question and I agree with explanation.
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Re: D01-08 [#permalink]

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New post 02 Jan 2018, 09:07
all 'repeating decimals" can be written as a fraction with a denominator that is made up of only 9s i.e. 9, 99, 999. below is an explanation how.
but before that, an interesting fact to know is that when the repeating digits are 0 and 1, then the nominator is ALWAYS 1 and the denominator is ALWAYS made up of 9s.
i.e. 0.11111.... = 1/9
0.010101.... = 1/99
0.001001001..... = 1/999 i.e
note that for every additional 9 in the denominator in addition to the first 9, a 0 is added before and after the 1s. except the first 9. (see above)
now, knowing this simple fact, you can tackle all such questions.

0.363636.... = 0.36 + 0.0036 + 0.000036 +......
= 36 (0.01 + 0.0001 + 0.000001 + .....)
= 36 (0.010101......) = 36 (1/99) = 36/99 = 4/11
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Re D01-08 [#permalink]

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New post 07 May 2018, 03:32
I think this is a high-quality question and I agree with explanation.
Re D01-08   [#permalink] 07 May 2018, 03:32
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