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Every day a certain bank calculates its average daily

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

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

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New post 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

Probability = 9 / 30 = 3/10 (D)
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Re: Every day a certain bank calculates its average daily [#permalink]

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

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New post 14 May 2016, 04:16
Hi debbiem,
refer your PM...

the SUM could be anything, so do not get distracted by PRIME integer >100...
Average = SUM/ DAYS...

Decimals depend depend upon the denominator and FINITE decimal will always have ONLY 2s and 5s in the denominator...

NOW what can be the denominator..
it has to be one of the numbers from 1 to 30..
out of these how many are multiple of ONLY 2 and 5..


\(2 :----- 2, 2^2,2^3, 2^4..............
5 :--------- 5, 5^2.......................
2 & 5 :--------- 10, 20.....................
and.. 1...\)
so total = 4+2+2+1 = 9..
so prob = \(\frac{9}{30} = \frac{3}{10}\)
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Re: Every day a certain bank calculates its average daily [#permalink]

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New post 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

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

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New post 01 Jul 2016, 07:32
chetan2u wrote:
Hi debbiem,
refer your PM...

the SUM could be anything, so do not get distracted by PRIME integer >100...
Average = SUM/ DAYS...

Decimals depend depend upon the denominator and FINITE decimal will always have ONLY 2s and 5s in the denominator...

NOW what can be the denominator..
it has to be one of the numbers from 1 to 30..
out of these how many are multiple of ONLY 2 and 5..


\(2 :----- 2, 2^2,2^3, 2^4..............
5 :--------- 5, 5^2.......................
2 & 5 :--------- 10, 20.....................
and.. 1...\)
so total = 4+2+2+1 = 9..
so prob = \(\frac{9}{30} = \frac{3}{10}\)


How are we sure that it would terminate within 5 digits? It could terminate with 5 digits, right? Or am I missing out something

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

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New post 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?

Thanks for your help as always!

Ryan

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

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New post 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.

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

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New post 27 Jan 2017, 18:42
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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!

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New post 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.

Is that correct?

Best of luck

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

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New post 27 Jan 2017, 20:20
plalud wrote:

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.

Is that correct?

Best of luck


Thanks
Partially correct - All except 16. All the other denominators will result in less than 4 decimal places. 16 will result in 4.
Cheers!

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

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New post 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

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

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New post 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 !!

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

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New post 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?

thanks in advance...

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