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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
(A) 1/10
(B) 2/15
(C) 4/15
(D) 3/10
(E) 11/30
18 Jul 2010, 11:42
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bmillan01
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.
21 Jul 2010, 13:13
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gmat1011
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.
.. Thanks for the info 
