MGMAT Challenge Problems

[jsloan01]

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 = a2b3, 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 a2b3, 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 = a2b3, 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 = a2b3, 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).

[GMATQuantCoach]

How many values of y are there

The question is asking how many possible values of y are there, not the value of y.

1) Once you've understood x has 12 unique factors, each factor is a possible value of y.

Hope this helps!

[audia8]

I go with A. The question I have is I get the 15 unique factors for X from 1)

x=a2b3=>4C1+4C2+4C3+4C4=15. Correct me, if I'm wrong. Thanks.

[india_chintan]

Bingo Quantcoach. It asks for different possible values of y which is 12. u ot the answer but misundertood the question.

[audia8]

I thought the question stated x = a*2*b*3. Now I realize that x = a*a*b*b*b. In that case, I agree that y could have 12 different values.

[finsisher]

this is a quality question. , that covers all the concepts of number theory and the word problem...

since we know that x = ky so to know how many possible pairs of (x,y) exist we just need to know number of factors of x now stmt1 says x= a^2*b^3.. then just by using number theory formula to find factors of composite number which can be expressed as multiples of prime , number of factors of x = 3*4 =12 hence A is itself sufficient.. I loved this question great job jsloan

[Daan]

I thought I'd explain the logic behind it a bit more, so people are reassured that (extremely) clear thinking is all you need, even if you know little number theory.

The question provides us with the following information:

x = y.k | x ≥ 2, y ≥ 1

The question is: how many values may y assume?
i.e. how many factors does x have?

f.e.

Suppose x = 20
y could be 1 or 20, since 20 = 1. 20
y could be 2 or 10, since 20 = 2. 10
y could be 4 or 5 since 20 = 4. 5

for a total of 6 possible factors.
This does seem like a bit of work though, especially for higher values of x.



Suppose x = 36
How do we determine all distinct factors?

We simply break up 36 into factors and then those factors into factors untill they cant be broken up anymore.

36 = 6 . 6 = (3 . 2) . ( 3 . 2) = 2^2 . 3^2
Both 2 and 3 are primenumbers and primenumbers have the intrinsic property that they cannot be broken up into factors anymore (beside 1 and the prime itself).

So 36 = 2 . 2 . 3 . 3
But from this we know all the factors of 36, because those primes can build any factor of 36.
36 = 2 . 2. 3 . 3 = 2 . (2.2.3) = (2.2) . (3.3) = (2.3).(2.3) etc.

So to know the different factors of a number x, you should break down x into its primefactors. All different possible combinations of those primefactors are the building blocks for all the factors of x.

For 36, the question is:
How many different numbers are 'hidden' in 2.2.3.3?

2.2.3.3 = 2^2 . 3^2
2.2.3 = 2^2 . 3^1
2.2 = 2^2 . 3 ^0

2.3.3 = 2^1 . 3 ^2
2.3 = 2^1 . 3^1
2 = 2^1. 3^0

3.3 = 2^0 . 3^2
3 = 2^0 . 3^1
1 =2^0 . 3^0

So 3*3 = 9 different factors because you can use both primefactors 2 and 3, in three different ways: multiply by 2 or 3 either once, twice or not at all.
3 possible ways to use 2 * 3 possible ways to use 3 = 9 possible ways total.

Note that the actual values of the primefactors are irrelevant in determining how many factors there are in total. You only need to know the power to which those primefactors are raised.

So in statement 1) for a^2 * b^3, there are:
3 possible ways of using 2 (0 times, 1 times, 2 times)
4 possible ways of using 3 (0, 1, 2 and 3 times)

and the total amount of factors is 12.

-- --

Now that's a lot of explaining, but the logical reasoning is actually quite easy to grasp:
*We (should) know primenumbers can build any number
**So if we know the primefactors of x, we can construct all factors of x simply by multiplying those primefactors in all possible ways

To know all possible ways (but not all values), statement 1) suffices

[aimster7888]

Can anybody show how to get 12 using combinations?

[sonikamadala]

Hi jsloan01,
U would have posted the FS1 as x = (a^2)*(b^3) rather x = a2b3.

because it is confusing many people in this chain.

Just suggestion