If w, x, y and z are integers such that 1

Question

If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and wxyz = 924, then how many possible values exist for z?

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

chetan2u · Answer

Hi,

Ofcourse we start with finding factors..
924=2*2*3*7*11

Z is the largest of all number if the numbers are different.. 
Let's find ways..
 1) when 11 is a factor of Z 
11....3*4*7*11
22...2*3*7*22
33...2*2*7*33
77...2*2*3*77
 2) when 11 is not a factor of Z.. 
Here the product of the two numbers should be>11
2*2*11*21.....21
2*3*11*14...14
No other two numbers can have product more than 11

Total 4+2=6
D

ScottTargetTestPrep · Answer

We can break 924 into prime factors:
 
924 = 3 x 308 = 3 x 4 x 77 = 3 x 2 x 2 x 11 x 7
 
We see that 924 is a product of 5 (not necessarily distinct) prime factors. If we want to express 924 as a product of 4 factors (each of which is greater than 1), one of the factors must be the product of two of 5 prime factors. Since 1 < w ≤ x ≤ y ≤ z, we can express 924 as:
 
xwyz = 3 x (2 x 2) x 7 x 11 = 3 x 4 x 7 x 11
 
We can see that in the case above, z = 11. Instead of listing other ways we can write the product xwyz, let’s just focus on the number of ways we can make z > 11. We can see that z must now be the product of exactly two of the 5 prime factors, so z could be:
 
2 x 7 = 14, 2 x 11 = 22, 3 x 7 = 21, 3 x 11 = 33, or 7 x 11 = 77
 
Thus, z could be any one the following 6 numbers: 11, 14, 21, 22, 33, or 77.
 
Answer: D

SSM700plus · Answer

924 = 2 * 2 *3 * 7 * 11

There are two possibilities. A) Largest number z is multiple of 11 B) z is not multiple of 11

A) Largest number z is multiple of 11

w<=x<=y<=z 
i)  3<4<7<11 
ii)  2<3<7<22 
iii)  2<=2<7<33 
iv)  2<=2<3<77

B) z is not multiple of 11
i)  2<3<11<14 
ii)  2<=2<11<21

Total possible values - 4 + 2 = 6.

Ans D

Aderonke01 · Answer

Why didnt we say 4×7 28. which makes the total possibility 7?

Posted from my mobile device

ArupRS · Answer

would we have 4 distinct factors then? 924=wxyz
if we consider 28 then 924=3*11*28 --- the fourth factor is missing.

Regards,
Arup

beeunoia · Answer

1. Let's find all factors of 924
924 =  2  * 462
462 =  2  * 231
231 =  3  * 77
77 =  11  *  7

2. Set of factors = {2, 2, 3, 7, 11}

3. Let's build combinations taking into consideration the condition we've  1 < w ≤ x ≤ y ≤ z   and that  z  should be the biggest number 
 z  can be:
3.1 All combinations with 11:  22, 33, 44, 77  
3.2 All combinations with 7:  28, 21

4. It means that the answer is 6.

Jaya6 · Answer

Seems like 11*4=44 is missing here, which makes the possibilty 7.

chetan2u · Answer

That will not be correct. 
If one of them is 44, the others would be 3 and 7 as 924=44*3*7. The only possibility for the 4th number, thereafter, is 1. But all are greater than 1.

santosh93 · Answer

924 can be prime factorized to 2*2*3*7*11

So below are the possible values of (w,x,y,z) that satisfy 1 <w   <  x  <  y  <  z and w*x*y*z=924

2,3,7,22
2,2,7,33
2,2,3,77
2,3,11,14
2,2,11,21
3,4,7,11

So total possible values for z = 11,14,21,22,33,77

bumpbot · Answer

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If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and w × x ×

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