Every day a certain bank calculates its average daily deposit for that

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.