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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 \(2*5^2\). 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.


Great explanation!!! I learn a lot from this
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bunuel... this is great. thanks...

the way you piece it together is sometimes scary... just curious - what was your gmat score :wink:
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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...
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Great - many thanks.
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awesome explanation!
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Brilliant explanation. Never knew that dividing by (2^m)(5^n) gives a terminating decimal :P .. Thanks for the info
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Thanks for the explanation Bunuel
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Thanks for the explanation....
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This is definitely a 800+ question :o
Good explanation and thnx for the theory, very useful!
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One question about this problem. The problem doesn't ask if the decimal will be terminating, but rather if the decimal will have less than 5 places. Your solution checks for termination, but how do you check for the number of decimal places? Couldn't some of the possibilities result in termination with more than 5 decimal places?
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TehJay
One question about this problem. The problem doesn't ask if the decimal will be terminating, but rather if the decimal will have less than 5 places. Your solution checks for termination, but how do you check for the number of decimal places? Couldn't some of the possibilities result in termination with more than 5 decimal places?

You should read the last part:
"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."

This issue is also discussed in the posts following the one with solution.
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Bunuel
bmillan01
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 \(2*5^2\). 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.

So you're saying that by controlling the terminating decimal using d=2^m*5^n and, and you can make it an integer by multiplying it by 10000 (if the number must be five decimal places to the left.
Therefore, you can choose of days 1, 2, 4, 5, 8, 10, 16, 20, or 25. (making that 9 days). - But they're not prime??? What am I missing here... I'm the slow one of the lot please help me out :)
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mariyea
Bunuel
bmillan01
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 \(2*5^2\). 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.

So you're saying that by controlling the terminating decimal using d=2^m*5^n and, and you can make it an integer by multiplying it by 10000 (if the number must be five decimal places to the left.
Therefore, you can choose of days 1, 2, 4, 5, 8, 10, 16, 20, or 25. (making that 9 days). - But they're not prime??? What am I missing here... I'm the slow one of the lot please help me out :)

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 (so denominator d must not be a prime it should be of a type d=2^m*5^n. It's nominator p which is given to be a prime>100).
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Bunuel
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. :)
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mariyea
Bunuel
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.
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Bunuel
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Bunuel
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.

I understand now Thank you so much!
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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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