kashishh
Bunuel
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?
Basically the questions asks: how many values of y are there, such that x/y=integer, which means that we are asked to find the # of factors of x.
(1) x=a^2*b^3, where a and b are different prime numbers --> the # of factors of x (including 1 and x itself) is (2+1)(3+1)=12, so y can take 12-1=11 values (y can take all the values except 1 since given that y>1)
(2) b=a+1. Not sufficient.
Answer: A.
Dear Bunuel,
Could you pls elaborate on "the # of factors of x (including 1 and x itself) is (2+1)(3+1)=12, so y can take 12-1=11 values (y can take all the values except 1 since given that y>1)"
How you have zeroed on 2 and 3, of all prime numbers?
Finding the Number of Factors of an IntegerFirst make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.
The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\).
NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)
Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
So, the # of factors of x=a^2*b^3, where a and b are different prime numbers is (2+1)(3+1)=12.
Hope it's clear.