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If w, x, y and z are integers

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If w, x, y and z are integers

by jack0997 » Mon Jun 12, 2017 4:45 am
If w, x, y, and z are integers such that 1 < w ≤ x ≤ y≤ z and w*x*y*z =ƒ 924, then how many possible values exist for z?

(A) Three
(B) Four
(C) Five
(D) Six
(E) Seven

OA D
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by Jay@ManhattanReview » Mon Jun 12, 2017 5:21 am
jack0997 wrote:If w, x, y, and z are integers such that 1 < w ≤ x ≤ y≤ z and w*x*y*z =ƒ 924, then how many possible values exist for z?

(A) Three
(B) Four
(C) Five
(D) Six
(E) Seven

OA D
924 ƒ = 2^2 x 3 x 7 x 11

We see that there is an extra '2' since the exponent of 2 is 2.
This extra '2' can be combined with any of the other factors to generate different values of z.
Also, keeping the two 2's separate, the other factors may be combined to generate different values of z.

Thus, possible values of w;x;y and z such that 1 < w ≤ x ≤ y≤ z are:

1. w = 3; x = 2*2=4; y = 7; z = 11
2. w = 2; x = 2*3=6; y = 7; z = 11
3. w = 2; x = 3; y = 7; z = 2*11=22
4. w = 2; x = 3; y = 11; z = 2*7=14
5. w = 2; x = 2; y = 7; z = 3*11=33
6. w = 2; x = 2; y = 11; z = 3*7=21
7. w = 2; x = 2; y = 3; z = 7*11=77

Thus, there are six possible values of z, which are 11, 14, 21, 22, 33 and 77.

The correct answer: D

Hope this helps!

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by ceilidh.erickson » Sun Jun 18, 2017 11:21 am
There is no realistic alternative to solving this problem other than the strategy Jay outlined above.*

Questions that ask you "how many solutions are there?" can be very time-consuming, because you're never quite sure if you came up with every possibility. You might have thought of 5 of these, but missed the 6th.

My recommendation: if you see a question like this, look at the clock. If you're ahead of time, go ahead and list as many possibilities as you can. If you're behind, just guess and move on. It's easy to make a mistake on these, and there's no way to easily check your work, so it might not be worth your time.


*You might realize the combinatoric pattern that if we're choosing 4 values [w, x, y, z] out of 5 primes, we'll have 1 with 2 primes in each set, so 5C2. Since there are two 2's, several of these will be double-counted, and some of these pairings will *not* be the greatest value in the set. So it's still best to list out all the possibilities.
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by Matt@VeritasPrep » Thu Jun 22, 2017 5:32 pm
I think this is a similar, but shorter and easier way:

924 = 2² * 3 * 7 * 11

Since there are exactly four factors greater than 1, one and only one of w, x, y, and z must contain two prime factors. (We're assigning five prime factors to four variables, with no empty variables (e.g. = 1) allowed.)

Since z must be ≥ 11, we can take every pair of primes greater than 11:

2*7
3*7
2*11
3*11
7*11

and also 11 itself, for a total of 6.

EDIT: Reading above, this is Ceilidh's idea. Nothing new under the sun!
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