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# S99-13

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Math Expert
Joined: 02 Sep 2009
Posts: 94342
Own Kudos [?]: 640708 [13]
Given Kudos: 85011
Math Expert
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Given Kudos: 85011
Intern
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Math Expert
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gthrb22 wrote:
I didn't understand the explanation of first statement

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.

Hope it helps.
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No.of biscuits= x
No.of customers= y

For equal and whole number distribution of biscuits among customers, x/y must be an Interger.

So, we need to find all different values of y, i.e all factors of x.

St.1: x= (prime no.1)^2 x (prime no.2)^3

Therefore, Total Factors of x inclusive of 1 and x=
(2+1)x(3+1)= 3x4=12.
Hence, y can be 12 different factors of x.
Sufficient.

St.2:prime no.1 and prime no.2 are consecutive numbers.
In other words, P.no.1= 2 and P.no.2= 3.
This statement alone is of no use. Insufficient.

Hence, Ans. A

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Statement (1): Sufficient! x=a2b3, where a,b are different prime numbers --> we can determine the number of factors of x by basing on a,a,b,b,b.

Statement (2): Insufficient because we dont know value of x and y
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IMO
For different values of a & b there will be different values of y. So there are infinite possible values y.
So, the answer must be C.
Please explain why m i wrong?
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Your confusion is understandable. I guess the question should be rephrased to ask - how many ways are there such that each patron gets equal parts. Nos are infinite, but there are only 12 ways in which equal distribution is possible.
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Bunuel wrote:
gthrb22 wrote:
I didn't understand the explanation of first statement

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.

Hope it helps.

Hi Bunuel,

You say that "Y can take all values except 1 since given that y>1". But it is not. the question states that it is AT LEAST 1.
So shouldn't the answer be E? Since we do not know if Y is 1 or higher.

Thanks.