Any help on these questions would be tremendously appreciated. Here is a "700 to 800" level question from Manhattan GMAT.
Q1: 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 =
a2
b3, where
a and
b are different prime numbers. (2)
b =
a + 1
My thoughts: I don't get how the answer is A. I understand that X has 12 factors, but what does that tell us about y?
A1:
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
a2
b3, 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 =
a2
b3, 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:
a0 = 1
a1 = a
a2
b0 = 1
1
a
a2
b1 = b
b
ab
a2b
b2
b2
ab2
a2b2
b3
b3
ab3
a2b3
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
a0 or
b0) and then multiply the results together. In this case, since
x =
a2
b3, 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.
The correct answer is (A).