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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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If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and w × x ×  [#permalink]

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New post 28 Jun 2017, 01:45
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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 values exist for z?

(A) Three
(B) Four
(C) Five
(D) Six
(E) Seven
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If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and w × x ×  [#permalink]

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New post 28 Jun 2017, 04:28
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1
Bunuel wrote:
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


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
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Re: If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and w × x ×  [#permalink]

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New post 30 Jun 2017, 09:23
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1
Bunuel wrote:
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



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
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Re: If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and w × x ×  [#permalink]

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New post 23 Nov 2018, 00:15
Bunuel wrote:
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


\(924 = 2 * 2 *3 * 7 * 11\)

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
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Re: If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and w × x ×  [#permalink]

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New post 02 Apr 2019, 08:49
Why didnt we say 4×7 28. which makes the total possibility 7?

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Re: If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and w × x ×  [#permalink]

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New post 04 Apr 2019, 10:51
Aderonke01 wrote:
Why didnt we say 4×7 28. which makes the total possibility 7?

Posted from my mobile device


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
GMAT Club Bot
Re: If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and w × x ×   [#permalink] 04 Apr 2019, 10:51
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