BTGmoderatorDC wrote: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
Source: Manhattan Prep
So, in a nutshell, we have to find out the possible values of x/y such that x/y is an integer, x > 2 and y > 1.
Let's take each statement one by one.
(1) x = a^2*b^3, where a and b are different prime numbers
Given that x = a^2*b^3, the number of factors of x are (2 + 1)*(3 + 1) = 12. Since 12 factors of x also contains 1, the qualified values of y would be 12 - 1 = 11 as y > 1.
Let's take an example:
x = 2^2*3^3 = 108
Thus, 2^2*3^3 = 108 has (2 + 1)*(3 + 1) = 12. The factors are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, and 108. for x/y to be an integer, y can take and of the 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, and 108 values, i.e. 11 values. Sufficient.
Don't get confused with 2 as a factor of x, given that x > 2. Given x = a^2*b^3, where a and b are different prime numbers, the minimum value of x = 108 > 2.
Even if you take any other example, you will get the same answer. You may try with x = 5^2*7^3. The number of factors of x would be 12.
(2) b = a + 1
Certainly insufficient.
The correct answer:
A
Hope this helps!
-Jay
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