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Given: x > 2 and y > 1 and x is completely divisible by y or in other words y is a factor of x.wilson4mba 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
But x = a²b³, does it not mean that x has 12 factors, and thus y can have 11 values (y has to be different than 1)?Statement 1: x = a²b³, a and b are different prime numbers.
There are infinite numbers of x of the given form.
]Rahul@gurome wrote:Given: x > 2 and y > 1 and x is completely divisible by y or in other words y is a factor of x.wilson4mba 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
Thus, we have to find the number of factors of x.
Statement 1: x = a²b³, a and b are different prime numbers.
There are infinite numbers of x of the given form.
Not sufficient.
Statement 2: b = (a + 1)
a and b has no relation with question.
Not sufficient.
1 & 2 Together: x = a²b³, a and b are different prime numbers and b = (a + 1)
Only possible values of a and b are 2 and 3 respectively.
Thus, x = 2²3³ = 4*27 = 108
Now we can easily find the number of factors of x!
Sufficient.
The correct answer is C.
Even i thought that answer should be A,Curguar wrote:But x = a²b³, does it not mean that x has 12 factors, and thus y can have 11 values (y has to be different than 1)?Statement 1: x = a²b³, a and b are different prime numbers.
There are infinite numbers of x of the given form.
In which case, OA would be A?
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