If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and wxyz = 9

[quote="chetan2u"]If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and wxyz = 924, then how many possible combinations exist for value of w,x,y,z?

(A) Three
(B) Five
(C) seven
(D) Eight
(E) Nine


Source: Modification of Bunuel's Q..

w.x.y.z = 924
Factors of 924 = 1 x 2 x 2 x 3 x 7 x 11
as given 1 < w ≤ x ≤ y ≤ z, 1 can not be considered.
remaining 2, 2, 3, 7, 11
Now we have to arrange these five numbers in 4 positions - w,x,y,z such that, 1 < w ≤ x ≤ y ≤ z
W = 1 Way (there are two 2s so only 1 way)
X = 2 Ways (This can be filled in by 2 or 3)
Y = 2 Ways (This can be filled in by 3 or 7)
Z = 2 Ways (This can be filled in by 7 or 11)

so, total combinations -1 x 2 x 2 x 2 = 8 ANS (D)

Is this correct?

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

factors of 924 = 2 * 2 * 3 * 7 * 11


2 * 6 * 7 * 11
3 * 4 * 7 * 11
2 * 3 * 11 * 14
2 * 3 * 7 * 22
2 * 2 * 11 * 21
2 * 2 * 7 * 33
2 * 2 * 3 * 77

I am getting only 7

924, when prime factorized will given us \(2^2 * 3 * 7 * 11\)
From the question stem, we are supplied with the following information,
w > 1, w ≤ x, x ≤ y, y ≤ z
The only possibility for w is 2.

Now coming for values of x,y,z we can have
There are 7 ways of writing the combinations
w---x---y-----z
2---2---3----77
2---2---11---21
2---3---7----22
2---3---11--14
2---6---7----11
2---6---11--14
3---4----7---11(Option C)

Factors of 924= 1*2*2*3*7*11.
since, value for w,x,y and z must me greater than 1 we need not consider 1 here.

So, the values we can have for w,x,y and z must come from {2,2,3,7 & 11}
2 can be multiplied with the other digits in the set giving 4 different values which are 4,6,14 & 22
3 can be multiplied with the two digits 7 & 11 which give 2 different values which are 21 & 33
7 can be multiplied with 11 which gives 77 as value

All the above digits can be used as values for w,x,y and z. Therefore, 4+2+1=7

Answer: C

I don't want to count manually, and I am kind of sure that there must be another way. I just can't figure it out. I tried using factors and multiples concept to split 924 into 4, but that isn't working.

Could you maybe suggest something, please?

Hi rekhabishop,

Ofcourse we start with finding factors..
924=2*2*3*7*11....
so w,x,y,z can take values of any of 5 factors so 4 out of available 5 factors. Means only one number of w,x,y,z will be multiple of two factors But w cannot because then its SMALLEST value will be 2*2 and thus will become MORE than x.

Z is the largest of all number if the numbers are different..
Let's find ways..
1) when 11 is a factor
11....3*4*7*11 or 2*6*7*11
22...2*3*7*22
33...2*2*7*33
77...2*2*3*77
TOTAL 2+1+1+1=5

2) when 11 is not a factor..
Here the product of the two factors should be>11
2*2*11*21.....21
2*3*11*14...14
No other two numbers can have product more than 11

Total 5+2=7
C
_________________

We have 5 Prime Bases when we break down 924 into its prime factorization:

(2) (2) (3) (7) (11)

Since none of the variables can equal 1 AND it is possible for the variables to be EQUAL AND we are told that:

(W) (X) (Y) (Z) = 924

We have to take those 5 prime bases and combine them in such a way so that 4 positive integers multiply to equal = 924

Each distinct case will involve one combination of 2 of the prime bases together, combining to make ONE factor

And

the other 3 prime bases filling in for the remaining variables.

We can answer the question by counting how many unique ways we can take 2 of the 5 and combine them to make a unique factor——- with the remaining 3 Prime Bases filling in for the 3 other Variables.

Set: (2 , 2, 3, 7 , 11)

How many ways can we make unique groups of 2?

Case 1: 2-2
Case 2: 2-3
Case 3: 2-7
Case 4: 2-11
Case 5: 3-7
Case 6: 3-11
Case 7: 7-11

The answer is 7 different ways we can have the variables W, X, Y, Z multiply to equal = 924 given the constraints

(C) 7

Posted from my mobile device