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Biscuits & Patrons

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Biscuits & Patrons

by das.ashmita » Mon Aug 27, 2012 11:42 am
At a particular moment, a restaurant has x biscuits and y patron(s), with x≥2 and y≥1. How many values of y are there, such that all the biscuits can be distributed among the patrons, with each patron receiving an equal number of whole biscuits and with no biscuits left over?

(1) x=a^2*b^3, where a and b are different prime numbers

(2) b=a+1

OA: A
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Source: — Data Sufficiency |
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by vk_vinayak » Mon Aug 27, 2012 11:56 am
Good question. I initially thought that answer is C, but I can see why A is correct.
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by das.ashmita » Tue Aug 28, 2012 12:22 am
Hi Vinayak

Can u please explain how you got A?
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by vk_vinayak » Tue Aug 28, 2012 12:37 am
x = a^2 * b^3

a and b are different prime numbers. Number of factors (including 1 and x itself) of x are: (2+1)*(3+1) = 12

Since we are looking for factors of x, and for all the possible values of x, if x can be expressed as above, it will have 12 factors.

Therefore, since y>1, y can have (12-1)=11 values.

A is sufficient.
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by das.ashmita » Tue Aug 28, 2012 3:37 am
thanks a lot Vinayak. :)
I read the question again and your explanation. it makes sense now. I missed out on the factors part.
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