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At a particular moment, a restaurant has \(x\) biscuits and \(y\) patron(s), with \(x \ge 2\) and \(y \ge 1\). 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) \(x = a^2b^3\), where \(a\) and \(b\) are different prime numbers

(2) \(b = a + 1\)
[Reveal] Spoiler: OA

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If we want to distribute \(x\) biscuits among \(y\) patrons equally and with no split or left-over biscuits, then \(x\) must be divisible by \(y\). Note that since both \(x\) and \(y\) count physical objects, both variables must be positive integers. The value of \(x\) is also constrained to be at least 2.

Since \(x\) must be divisible by \(y\), we can also say that \(y\) must be a factor of \(x\). Asking how many values of \(y\) satisfy the conditions is equivalent to asking how many factors \(x\) has.

(1) SUFFICIENT. If we can write the prime factorization of \(x\) as \(a^2b^3\), where \(a\) and \(b\) are different prime numbers, then we can in fact count the factors of \(x\) - even though we do not know the values of \(x\), \(a\), or \(b\). The reason is that we can construct every factor of \(x\) uniquely out of powers of \(a\) and powers of \(b\). No factor of \(x\) can contain any primes other than \(a\) and \(b\). Moreover, in any factor of \(x\), the power of \(a\) cannot be larger than 2 (since \(x = a^2b^3\), and if the factor had a higher power of \(a\), then when we divide \(x\) by the factor, we would be left with uncanceled \(a\)'s in the denominator). By the same reasoning, the power of \(b\) in the factor cannot be larger than 3. Finally, both powers must be non-negative integers (0 or positive integers. Thus, we can construct a table to see all the possibilities. Simply multiply together the row and column labels to get each entry:
\(a^0 = 1\) \(a^1 = a\) \(a^2\) \(b^0 = 1\) \(1\) \(a\) \(a^2\) \(b^1 = b\) \(b\) \(ab\) \(a^2b\) \(b^2\) \(b^2\) \(ab^2\) \(a^2b^2\) \(b^3\) \(b^3\) \(ab^3\) \(a^2b^3\)
Thus, there are 12 unique factors of \(x\). In fact, we do not have to enumerate the factors. A shortcut is to add 1 to each prime's power in the factorization (to account for the possibility of \(a^0\) or \(b^0\)) and then multiply the results together. In this case, since \(x = a^2b^3\), we write \((2 + 1)(3 + 1) = (3)(4) = 12\).

(2) INSUFFICIENT. By itself, the statement does not refer to \(x\) or \(y\), so it cannot be sufficient to answer the given question.

Incidentally, one trap in this problem is that if you put the statements together, you can actually figure out the values of \(a\) and \(b\), and therefore the value of \(x\). Since \(b = a + 1\), we can conclude that \(a = 2\) and \(b = 3\). The only primes that differ by exactly 1 are 2 and 3. Since all primes greater than 2 are odd, the minimum difference between all other pairs of primes is 2.


Answer: A
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New post 01 Sep 2016, 06:15
I didn't understand the explanation of first statement

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New post 01 Sep 2016, 09:16
gthrb22 wrote:
I didn't understand the explanation of first statement


Please find below alternative solution:

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.

Hope it helps.
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New post 25 Oct 2017, 11:37
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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S99-13   [#permalink] 25 Oct 2017, 11:37
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