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Re: Every day a certain bank calculates its average daily [#permalink]

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01 Aug 2014, 08:22

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Re: Every day a certain bank calculates its average daily [#permalink]

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08 Oct 2014, 11:21

Every day a certain bank calculates its average daily deposit for that calendar month up to and including that day. If on a randomly chosen day in June the sum of all deposits up to and including that day is a prime integer greater than 100, what is the probability that the average daily deposit up to and including that day contains fewer than 5 decimal places?

(A) 1/10 (B) 2/15 (C) 4/15 (D) 3/10 (E) 11/30

Any prime number (Except 2 & 5) or its multiple when in denominator, gives recurring decimal number. Any multiple of 2, except a multiple with prime number, will result in non-recurring decimals.

Till 30, the following will result in non-recurring decimals less than 5 - 1, 2, 4, 5, 8, 10, 16, 20, 25 = 9 numbers

Re: Every day a certain bank calculates its average daily [#permalink]

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24 Oct 2015, 21:18

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Every day a certain bank calculates its average daily [#permalink]

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01 Jul 2016, 07:30

Bunuel wrote:

mariyea wrote:

Bunuel wrote:

Nominator is a prime number>100 (on a randomly chosen day in June the sum of all deposits up to and including that day is a prime integer greater than 100) and denominator is that day: {the average daily deposit}={sum of deposits up to and including that day}/{# of days}=p/d.

Oh so since we are limited to having five decimal places we were able to control that by choosing a using a denominator that gives us a terminating decimal and we multiply that by 10000 to make that number greater than 100. And the denominator is the 30 days out of which we can choose the 9 days.

But if the Nominator is supposed to be a prime number then we are either supposed to choose from 2 and 5 and nothingelse. Please have patience with me. I just need to understand.

I'm not sure understood your question.

Anyway: \(average=\frac{prime}{day}\). In order this value (p/d) to be terminating decimal \(d\) must be of a type \(2^n5^m\). Because if the chosen day, \(d\), is NOT of a type \(2^n5^m\) then \(average=\frac{p}{d}\) will not be a terminating decimal and thus will have more than 5 decimal places.

There are 9 such days: 1, 2, 4, 5, 8, 10, 16, 20, 25. For example \(average=\frac{prime}{1}\) or \(average=\frac{prime}{2}\) or \(average=\frac{prime}{4}\), ..., \(average=\frac{prime}{25}\) all will be terminating decimals (and for ALL other values of d: 3, 6, 7, ... p/d will be recurring decimal thus will have infinite number of decimal places so more than 5). Also for all these 9 values of d average=p/d not only be terminating decimal but also will have fewer than 5 decimal places (\(average=\frac{prime}{16}\) will have max # of decimal places which is 4).

Hope it's clear.

Hi Bunuel How do we know that these numbers would terminate within 5 digits? That's what I'm not able to understand from your explanation. Thanks

Re: Every day a certain bank calculates its average daily [#permalink]

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08 Nov 2016, 09:48

Bunuel wrote:

mariyea wrote:

Bunuel wrote:

Nominator is a prime number>100 (on a randomly chosen day in June the sum of all deposits up to and including that day is a prime integer greater than 100) and denominator is that day: {the average daily deposit}={sum of deposits up to and including that day}/{# of days}=p/d.

Oh so since we are limited to having five decimal places we were able to control that by choosing a using a denominator that gives us a terminating decimal and we multiply that by 10000 to make that number greater than 100. And the denominator is the 30 days out of which we can choose the 9 days.

But if the Nominator is supposed to be a prime number then we are either supposed to choose from 2 and 5 and nothingelse. Please have patience with me. I just need to understand.

I'm not sure understood your question.

Anyway: \(average=\frac{prime}{day}\). In order this value (p/d) to be terminating decimal \(d\) must be of a type \(2^n5^m\). Because if the chosen day, \(d\), is NOT of a type \(2^n5^m\) then \(average=\frac{p}{d}\) will not be a terminating decimal and thus will have more than 5 decimal places.

There are 9 such days: 1, 2, 4, 5, 8, 10, 16, 20, 25. For example \(average=\frac{prime}{1}\) or \(average=\frac{prime}{2}\) or \(average=\frac{prime}{4}\), ..., \(average=\frac{prime}{25}\) all will be terminating decimals (and for ALL other values of d: 3, 6, 7, ... p/d will be recurring decimal thus will have infinite number of decimal places so more than 5). Also for all these 9 values of d average=p/d not only be terminating decimal but also will have fewer than 5 decimal places (\(average=\frac{prime}{16}\) will have max # of decimal places which is 4).

Hope it's clear.

Bunuel -

How do we know that Prime / 16 = 4 decimal places will be our maximum amount of decimal places? Is there a trick to this?

Every day a certain bank calculates its average daily [#permalink]

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27 Jan 2017, 16:43

Bunuel wrote:

bmillan01 wrote:

Every day a certain bank calculates its average daily deposit for that calendar month up to and including that day. If on a randomly chosen day in June the sum of all deposits up to and including that day is a prime integer greater than 100, what is the probability that the average daily deposit up to and including that day contains fewer than 5 decimal places?

(A) 1/10 (B) 2/15 (C) 4/15 (D) 3/10 (E) 11/30

Theory: Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be terminating decimal.

(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.)

BACK TOT THE ORIGINAL QUESTION:

Question: does \(average=\frac{p}{d}\) has less than 5 decimal places? Where \(p=prime>100\) and \(d\) is the chosen day.

If the chosen day, \(d\), is NOT of a type \(2^n5^m\) (where \(n\) and \(m\) are nonnegative integers) then \(average=\frac{p}{d}\) will not be a terminating decimal and thus will have more than 5 decimal places.

How many such days are there of a type \(2^n5^m\): 1, 2, 4, 5, 8, 10, 16, 20, 25 (\(1=2^0*5^0\), \(2=2^2\), \(4=2^2\), \(5\), \(8=2^3\), \(10=2*5\), \(16=2^4\), \(20=2^2*5\), \(25=5^2\)), total of 9 such days (1st of June, 4th of June, ...).

Now, does \(p\) divided by any of these \(d's\) have fewer than 5 decimal places? Yes, as \(\frac{p}{d}*10,000=integer\) for any such \(d\) (10,000 is divisible by all these numbers: 1, 2, 4, 5, 8, 10, 16, 20, 25).

So, there are 9 such days out of 30 in June: \(P=\frac{9}{30}=\frac{3}{10}\) .

Answer: D.

Hope it's clear.

"Now, does \(p\) divided by any of these \(d's\) have fewer than 5 decimal places? Yes, as \(\frac{p}{d}*10,000=integer\) for any such \(d\) (10,000 is divisible by all these numbers: 1, 2, 4, 5, 8, 10, 16, 20, 25)."

p/d*100,000=integer as well, for any of the d's. So how can you tell it does not have 5 decimal places? I don't understand the rationale.

Re: Every day a certain bank calculates its average daily [#permalink]

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27 Jan 2017, 18:42

1

This post received KUDOS

plalud wrote:

"Now, does \(p\) divided by any of these \(d's\) have fewer than 5 decimal places? Yes, as \(\frac{p}{d}*10,000=integer\) for any such \(d\) (10,000 is divisible by all these numbers: 1, 2, 4, 5, 8, 10, 16, 20, 25)."

p/d*100,000=integer as well, for any of the d's. So how can you tell it does not have 5 decimal places? I don't understand the rationale.

A number that has less than 5 decimal places can be represented as \(\frac{k}{10,000}\), where k is an integer. Essentially, \(\frac{p}{d}\)=\(\frac{k}{10,000}\), if \(\frac{p}{d}\) has less than 5 decimal places.

This is the rationale behind multiplying \(\frac{p}{d}\) with 10,000 to check if the product is an integer. Since, 10,000 is divisible by all possible denominators, \(\frac{p}{d}*10,000\) is an integer.

Of course, \(\frac{p}{d}*100,000\) is also an integer as you pointed out, but as we have already established \(\frac{p}{d}*10,000\) is an integer, \(\frac{p}{d}*100,000\) will be a multiple of 10 (The 5th decimal place is 0) Hope that helps!

Every day a certain bank calculates its average daily [#permalink]

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27 Jan 2017, 20:11

CholericQuill wrote:

plalud wrote:

"Now, does \(p\) divided by any of these \(d's\) have fewer than 5 decimal places? Yes, as \(\frac{p}{d}*10,000=integer\) for any such \(d\) (10,000 is divisible by all these numbers: 1, 2, 4, 5, 8, 10, 16, 20, 25)."

p/d*100,000=integer as well, for any of the d's. So how can you tell it does not have 5 decimal places? I don't understand the rationale.

A number that has less than 5 decimal places can be represented as \(\frac{k}{10,000}\), where k is an integer. Essentially, \(\frac{p}{d}\)=\(\frac{k}{10,000}\), if \(\frac{p}{d}\) has less than 5 decimal places.

This is the rationale behind multiplying \(\frac{p}{d}\) with 10,000 to check if the product is an integer. Since, 10,000 is divisible by all possible denominators, \(\frac{p}{d}*10,000\) is an integer.

Of course, \(\frac{p}{d}*100,000\) is also an integer as you pointed out, but as we have already established \(\frac{p}{d}*10,000\) is an integer, \(\frac{p}{d}*100,000\) will be a multiple of 10 (The 5th decimal place is 0) Hope that helps!

Great answer, kudos. To build on that:

since 1,000 is divisible by all possible denominators, \(\frac{p}{d}*1,000\) is an integer. Therefore, the number has less than 4 decimal places.

Moreover, we can find what is the maximum amount of decimal places the number could have: 3, since 100 is the minimum factor of 10 that makes 25 a potential divisor.

Re: Every day a certain bank calculates its average daily [#permalink]

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23 Mar 2017, 20:39

Thank you for the explanation. It's really helpful. Just one concern - "Does the nominator need to be prime greater than 100?". I seems to be a redundant feature

Re: Every day a certain bank calculates its average daily [#permalink]

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29 Jul 2017, 23:01

Bunuel wrote:

gmat1011 wrote:

Actually wanted to seek 1 clarification to better understand this:

p/d *10,000=integer

p is a prime integer greater than 100

d can be one of the 9 numbers

To test the tendency to leave a certain desired number of decimal places, upon division of p by d why is it ok to multiply p by a common multiple (10k here) of the 9 numbers in the denominator?

very crude general example (which I am hoping is an analogy): 37 divided by 7 leaves R of 2 and certain decimal places; 37 * 14 divided by 7 leaves no remainder ---> how can the later scenario be used to test whether a certain desired number of decimal places are left by the first scenario...

Your example is not good as \(\frac{37}{7}\) will be recurring decimal (will have infinite number of decimal places).

How many decimal places will terminating decimal \(\frac{p}{2^n*5^m}\) have? (p is prime number)

Consider following examples: \(0.2\) has 1 decimal place --> 1.2*10=12=integer (multiplying by 10 with 1 zero); \(0.25\) has 2 decimal places --> 1.25*10^2=125=integer (multiplying by 100 with 2 zeros); \(0.257\) has 3 decimal places --> 1.257*10^3=1257=integer (multiplying by 100 with 3 zeros); \(0.2571\) has 4 decimal places --> 1.2571*10^4=12571=integer (multiplying by 100 with 4 zeros); ...

So, terminating decimal, \(\frac{p}{2^n*5^m}\) (where p is prime number), will have \(k\) decimal places, where \(k\) is the least value in \(10^k\) for which \(\frac{p}{2^n*5^m}*10^k=integer\).

In our original question least value of \(k\) for which \(\frac{p}{d}*10^k=integer\), for all 9 d's, is 4 or when \(10^k=10,000\) (k=4 is needed when d=16).

Hope it's clear.

Hey Bunuel, I tried several prime numbers eg 101,109 and all of them give integers when multiplied by 10,000. However, I didn't understand HOW to get this k=4 ? Is it by trial and error ? Once again, I understood your method to find if p/q is terminating or not. Only confused about the less than 5 decimal places part. Thanks !!

Every day a certain bank calculates its average daily [#permalink]

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31 Aug 2017, 04:45

Bunuel wrote:

bmillan01 wrote:

Every day a certain bank calculates its average daily deposit for that calendar month up to and including that day. If on a randomly chosen day in June the sum of all deposits up to and including that day is a prime integer greater than 100, what is the probability that the average daily deposit up to and including that day contains fewer than 5 decimal places?

(A) 1/10 (B) 2/15 (C) 4/15 (D) 3/10 (E) 11/30

Theory: Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be terminating decimal.

(We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.)

BACK TOT THE ORIGINAL QUESTION:

Question: does \(average=\frac{p}{d}\) has less than 5 decimal places? Where \(p=prime>100\) and \(d\) is the chosen day.

If the chosen day, \(d\), is NOT of a type \(2^n5^m\) (where \(n\) and \(m\) are nonnegative integers) then \(average=\frac{p}{d}\) will not be a terminating decimal and thus will have more than 5 decimal places.

How many such days are there of a type \(2^n5^m\): 1, 2, 4, 5, 8, 10, 16, 20, 25 (\(1=2^0*5^0\), \(2=2^2\), \(4=2^2\), \(5\), \(8=2^3\), \(10=2*5\), \(16=2^4\), \(20=2^2*5\), \(25=5^2\)), total of 9 such days (1st of June, 4th of June, ...).

Now, does \(p\) divided by any of these \(d's\) have fewer than 5 decimal places? Yes, as \(\frac{p}{d}*10,000=integer\) for any such \(d\) (10,000 is divisible by all these numbers: 1, 2, 4, 5, 8, 10, 16, 20, 25).

So, there are 9 such days out of 30 in June: \(P=\frac{9}{30}=\frac{3}{10}\) .

Answer: D.

Hope it's clear.

hi man

any of the 9 days (1, 2, 4, 5, 8, 10, 16, 20, 25) when set as denominator, the ultimate fraction will culminate in a number less than 5 decimal places, sufficient. So why is that, "10,000" is necessary here to solve this question, bro?