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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 Integer

First 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.
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Thank you...it's clear now...
The problem was hard for me because of the wordiness, but after your paraphrasing it's a lot easier to understand the problem.
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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.

The questions difficulty level is set to low, But I think it is not an easy question.
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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.


Suppose a=4 and b=5 then
X = 4 x 4 x 5 x 5 x 5 ------> 3 X 4 = 12 Factor.

I am facing Challenge understanding how Y is 11? why not Y be the multiple of 11 or less than 11?
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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.


Suppose a=4 and b=5 then
X = 4 x 4 x 5 x 5 x 5 ------> 3 X 4 = 12 Factor.

I am facing Challenge understanding how Y is 11? why not Y be the multiple of 11 or less than 11?

y is not 11. y can take 11 values: all factors of x except 1.

Does this make sense?
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honchos
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.


Suppose a=4 and b=5 then
X = 4 x 4 x 5 x 5 x 5 ------> 3 X 4 = 12 Factor.

I am facing Challenge understanding how Y is 11? why not Y be the multiple of 11 or less than 11?

y is not 11. y can take 11 values: all factors of x except 1.

Does this make sense?

That means X/Y<X This is the Condition.
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honchos


Suppose a=4 and b=5 then
X = 4 x 4 x 5 x 5 x 5 ------> 3 X 4 = 12 Factor.

I am facing Challenge understanding how Y is 11? why not Y be the multiple of 11 or less than 11?

y is not 11. y can take 11 values: all factors of x except 1.

Does this make sense?

That means X/Y<X This is the Condition.

We need to find the number of values of y, so that x/y=integer. From (1) we have that the number of factors of x is 12. So, y can take 12-1=11 values, so y can be ANY factor of x but 1.
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honchos
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.


Suppose a=4 and b=5 then
X = 4 x 4 x 5 x 5 x 5 ------> 3 X 4 = 12 Factor.

I am facing Challenge understanding how Y is 11? why not Y be the multiple of 11 or less than 11?



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