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If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and wxyz = 9
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Updated on: 02 Jul 2017, 01:32
Question Stats:
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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.. https://gmatclub.com/forum/ifwxyandzareintegerssuchthat1wxyzandwx243581.html
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Originally posted by chetan2u on 01 Jul 2017, 20:18.
Last edited by Bunuel on 02 Jul 2017, 01:32, edited 1 time in total.
Renamed the topic.



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Re: If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and wxyz = 9
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05 Jul 2017, 12:17
[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?



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Re: If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and wxyz = 9
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05 Jul 2017, 13:26
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
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Re: If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and wxyz = 9
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05 Jul 2017, 13:36
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 wxyz 22377 221121 23722 231114 26711 261114 34711(Option C)
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Re: If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and wxyz = 9
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09 Jul 2017, 11:27
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



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Re: If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and wxyz = 9
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14 Jul 2017, 03:02
pushpitkc wrote: 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 wxyz 22377 221121 23722 231114 26711 261114 34711(Option 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?
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If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and wxyz = 9
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15 Jul 2017, 21:39
chetan2u wrote: 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
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
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If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and wxyz = 9
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15 Jul 2017, 21:39






