u1983 wrote:At a particular moment, a restaurant has x biscuits and y patrons, 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
The question is posted correctly. I edited it.
Given: a restaurant has x biscuits and y patrons, with x > 2 and y > 1.
Question: 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?
Let's understand the question.
It means that each patron will get x/y biscuits such that x/y is an integer. We have to find out the number of possible values of y. Note that we are not asked to find out the possible values of y.
Let's take each statement one by one.
(1) x = a^2*b^3, where a and b are different prime numbers.
=> a^2*b^3 / y is an integer. Since a and b are prime numbers, we can count the number of factors of x = a^2*b^3. The number of factors of x = a^2*b^3 equals to (2 + 1)*(3 + 1) = 12. This means that there are 12 values of x, including 1, so a total of 12 values can divide x. Since it is given that y > 1, we have 12 - 1 = 11 possible values of y. Sufficient.
(2) b = a + 1
Certainly insufficient.
The correct answer:
A
Hope this helps!
-Jay
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