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.
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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

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.
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awesome explanation!

Thanks for the explanation Bunuel
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This is definitely a 800+ question
Good explanation and thnx for the theory, very useful!

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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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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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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I understand now Thank you so much!
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