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Math Expert V
Joined: 02 Sep 2009
Posts: 60647
If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and w × x ×  [#permalink]

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15 00:00

Difficulty:   95% (hard)

Question Stats: 27% (02:58) correct 73% (02:52) wrong based on 179 sessions

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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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Math Expert V
Joined: 02 Aug 2009
Posts: 8337
If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and w × x ×  [#permalink]

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1
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,

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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Target Test Prep Representative V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 9142
Location: United States (CA)
Re: If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and w × x ×  [#permalink]

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2
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.

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Manager  P
Joined: 01 Aug 2017
Posts: 220
Location: India
GMAT 1: 500 Q47 V15
GPA: 3.4
WE: Information Technology (Computer Software)
Re: If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and w × x ×  [#permalink]

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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
Intern  B
Joined: 21 Feb 2019
Posts: 37
Re: If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and w × x ×  [#permalink]

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

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Manager  G
Joined: 23 Jan 2018
Posts: 220
Location: India
WE: Information Technology (Computer Software)
Re: If w, x, y and z are integers such that 1 < w ≤ x ≤ y ≤ z and w × x ×  [#permalink]

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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 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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