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# M15-25

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Math Expert
Joined: 02 Sep 2009
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M15-25  [#permalink]

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16 Sep 2014, 00:56
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Difficulty:

45% (medium)

Question Stats:

67% (01:07) correct 33% (01:08) wrong based on 117 sessions

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In the decimal notation of number $$(\frac{2}{23})^3$$, what is the third digit to the right of the decimal point?

A. 0
B. 1
C. 2
D. 4
E. 8

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Re M15-25  [#permalink]

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16 Sep 2014, 00:56
Official Solution:

In the decimal notation of number $$(\frac{2}{23})^3$$, what is the third digit to the right of the decimal point?

A. 0
B. 1
C. 2
D. 4
E. 8

Because $$(\frac{2}{23})^3 \lt (\frac{2}{20})^3 = (\frac{1}{10})^3 = 0.001$$, the third digit to the right of the decimal point in the decimal notation of $$(\frac{2}{23})^3$$ is 0.

Answer: A
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Re: M15-25  [#permalink]

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21 Jul 2015, 02:32
Bunuel , the way i did it was divide 2 by 23 which would give something like (.0xyz) so when this is cubed,it would be something like (.0000000x) so the digit in question would be zero.Is this logic correct?
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Re M15-25  [#permalink]

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05 Aug 2015, 21:54
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. i think we can not use upper limit in here. we should divide the fraction and apply number properties to find the answer
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Re: M15-25  [#permalink]

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17 Aug 2015, 03:11
amirzohrevand wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. i think we can not use upper limit in here. we should divide the fraction and apply number properties to find the answer

Hope alternative solution HERE help to understand better.
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Re: M15-25  [#permalink]

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04 Nov 2016, 09:32
Bunuel wrote:
amirzohrevand wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. i think we can not use upper limit in here. we should divide the fraction and apply number properties to find the answer

I guess you can assume 11-rule --> 2/23 =(roughly) 2/22 = 1/11 = 0,09090909. Third position is 0
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Re: M15-25  [#permalink]

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26 Jan 2017, 11:03
I thought the digit 1 is the third from the right of the decimal point. So, if 1 is not the third, in which position is it? Are not the decimal points supposed to be the ones after the comma, in which case 1 would be the third and the 2 zeros would be first and second. I don't understand this concept at all.... Can someone pplease help?!
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Re: M15-25  [#permalink]

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27 Jan 2017, 07:12
danipetu wrote:
I thought the digit 1 is the third from the right of the decimal point. So, if 1 is not the third, in which position is it? Are not the decimal points supposed to be the ones after the comma, in which case 1 would be the third and the 2 zeros would be first and second. I don't understand this concept at all.... Can someone pplease help?!

(2/23)^3 = 0.000657516232431988164707816224213035259307964165365332456... 0 is the third digit to the right of the decimal point.

1 is the third digit to the right of the decimal point of $$(\frac{2}{20})^3= 0.001$$. Because $$(\frac{2}{23})^3 \lt (\frac{2}{20})^3$$, then the third digit to the right of the decimal point of (2/23)^3 must be 0.
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Re: M15-25  [#permalink]

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27 Jan 2017, 11:51
Bunuel wrote:
danipetu wrote:
I thought the digit 1 is the third from the right of the decimal point. So, if 1 is not the third, in which position is it? Are not the decimal points supposed to be the ones after the comma, in which case 1 would be the third and the 2 zeros would be first and second. I don't understand this concept at all.... Can someone pplease help?!

(2/23)^3 = 0.000657516232431988164707816224213035259307964165365332456... 0 is the third digit to the right of the decimal point.

1 is the third digit to the right of the decimal point of $$(\frac{2}{20})^3= 0.001$$. Because $$(\frac{2}{23})^3 \lt (\frac{2}{20})^3$$, then the third digit to the right of the decimal point of (2/23)^3 must be 0.

Thank you a lot for repplying. It is very clear now!! Thanks for the help!!
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M15-25  [#permalink]

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13 Feb 2018, 05:26
Hi Bunuel,

I used the method below...leading me to a correct answer...wanted to clarify if my method will work all the time even though it's an estimate. The other example used an estimate too. I did the following:

(2/23)^3 > (2/25)^3

multiplied the inner fraction (2/25) by 4/4 to get the percentage equivalent (8/100)

cubed the respective numerator and denominator to get (512/10^6)

the answer being..0.000512

Tosin
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Re: M15-25  [#permalink]

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13 Feb 2018, 05:35
ttaiwo wrote:
Hi Bunuel,

I used the method below...leading me to a correct answer...wanted to clarify if my method will work all the time even though it's an estimate. The other example used an estimate too. I did the following:

(2/23)^3 > (2/25)^3

multiplied the inner fraction (2/25) by 4/4 to get the percentage equivalent (8/100)

cubed the respective numerator and denominator to get (512/10^6)

the answer being..0.000512

Tosin

Since it's only estimation then it won't work for all cases. For example, if the question asked about the third digit to the right of the decimal point of (2/20)^3, then your approach would give you a wrong answer.
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Re: M15-25  [#permalink]

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13 Feb 2018, 06:14
Bunuel wrote:
ttaiwo wrote:
Hi Bunuel,

I used the method below...leading me to a correct answer...wanted to clarify if my method will work all the time even though it's an estimate. The other example used an estimate too. I did the following:

(2/23)^3 > (2/25)^3

multiplied the inner fraction (2/25) by 4/4 to get the percentage equivalent (8/100)

cubed the respective numerator and denominator to get (512/10^6)

the answer being..0.000512

Tosin

Since it's only estimation then it won't work for all cases. For example, if the question asked about the third digit to the right of the decimal point of (2/20)^3, then your approach would give you a wrong answer.

Ah understood.

I thought my method was similar to yours as it was an estimate but the other way round...rather than choosing an estimate that greater than (2/23)^3 like you did...I went for an estimate that was smaller than (2/23)^3...

I see how I could come unstuck here...is the takeaway from this that I should try and get an estimate that is smaller than the amount if such a question were asked? I guess this in itself is a very specific question...but I kinda find it hard to think that one would have to go through long division for such.
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Re: M15-25  [#permalink]

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18 Feb 2018, 12:20
1
Simple and quickest approach:

(2/23) - > (0.0x..) no need to calculate after x

(0.0x..)^3 -> (0.000z....) (one 0 will multiply 3 times when cubed)

Therefore the third digit after the decimal point is 0.
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Re: M15-25  [#permalink]

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22 Jul 2018, 02:07
dileeprk, what if your 'z' value in your solution or the fourth digit comes >5, wont the third digit will be rounded off to 1?
In this situation, only Bunuel's estimation approach is what is giving correct ans. considering rounding off.

Take a look again.
Re: M15-25 &nbs [#permalink] 22 Jul 2018, 02:07
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# M15-25

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