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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
W, X, Y and Z are four different positive integers.
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Updated on: 12 Mar 2019, 07:54
Updated on: 12 Mar 2019, 07:54
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Question Stats:
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82% (02:45) wrong based on 76 sessions
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W, X, Y and Z are four different positive integers. When X is divided by Y, the quotient is Z and the remainder is W.
If W = X – 4, which of the following is a possible value of X?
Indicate ALL possible values of X.
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
(F) 9
(G) 10
(H) 11
If W = X – 4, which of the following is a possible value of X?
Indicate ALL possible values of X.
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
(F) 9
(G) 10
(H) 11
Originally posted by GreenlightTestPrep on 08 Mar 2019, 10:11.
Last edited by GreenlightTestPrep on 12 Mar 2019, 07:54, edited 3 times in total.
Last edited by GreenlightTestPrep on 12 Mar 2019, 07:54, edited 3 times in total.
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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
W, X, Y and Z are four different positive integers.
[#permalink]
12 Mar 2019, 07:58
12 Mar 2019, 07:58
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GreenlightTestPrep wrote:
W, X, Y and Z are four different positive integers. When X is divided by Y, the quotient is Z and the remainder is W.
If W = X – 4, which of the following is a possible value of X?
Indicate ALL possible values of X.
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
(F) 9
(G) 10
(H) 11
If W = X – 4, which of the following is a possible value of X?
Indicate ALL possible values of X.
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
(F) 9
(G) 10
(H) 11
Given: When X is divided by Y, the quotient is Z and the remainder is W.
------ASIDE-------
There's a nice rule that says, "If N divided by D equals Q with remainder R, then N = DQ + R"
For example, since 17 divided by 5 equals 3 with remainder 2, then we can write 17 = (5)(3) + 2
Likewise, since 53 divided by 10 equals 5 with remainder 3, then we can write 53 = (10)(5) + 3
---------------------------------------
So, from the given information, we can write: X = YZ + W
Also given: W = X – 4
Take X = YZ + W and replace X with X - 4 to get: X = YZ + X - 4
Subtract X from both sides: 0 = YZ - 4
Rewrite as: 4 = YZ
We're told that Y and Z are DIFFERENT positive INTEGERS. So, there are only 2 possible cases:
case i: Y = 1 and Z = 4
case ii: Y = 4 and Z = 1
case i yields a CONTRADICTION.
If Y = 1, then we are dividing X by 1, and if we divide by 1, the remainder will always be ZERO.
In other words, if Y = 1, then W = 0
However, we are told that W is a POSITIVE integer.
So, we can definitely rule out case i, which means it MUST be the case that Y = 4 and Z = 1 (case ii)
So, we have: When X is divided by 4, the quotient is 1 and the remainder is W
This tells us that 4 divides into X 1 time
So, the possible values of X are: 4, 5, 6, and 7 (since 4 divides into each value 1 time).
Let's check each case.
If X = 4, then the remainder (W) is 0. Since we're told W is POSITIVE, we can ELIMINATE this case.
If X = 5, then the remainder (W) is 1. Since Y already equals 1, and since we're told the numbers are DIFFERENT, we can ELIMINATE this case.
If X = 6, then the remainder (W) is 2. So, when X (6) is divided by 4 (Y), the quotient (Z) is 1, and the remainder (W) is 2. PERFECT!
If X = 7, then the remainder (W) is 3. So, when X (7) is divided by 4 (Y), the quotient (Z) is 1, and the remainder (W) is 3. PERFECT!
Answer: C, D
Cheers,
Brent
RELATED VIDEO
General Discussion
Re: W, X, Y and Z are four different positive integers.
[#permalink]
29 Jun 2020, 19:16
29 Jun 2020, 19:16
Why in the second case answer can't be 5? Because previously you outlined that the case Y=1 and X=4 is incorrect since when Y is divided by 1 remainder always 0, and now you say that 5 is inviavle, cause Y is already 1? However we should only consider the second case where Y is predetermined as 4? Could you please explain?
