Last visit was: 19 Jul 2025, 21:21 It is currently 19 Jul 2025, 21:21
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
805+ Level|   Remainders|      
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 13 May 2024
Posts: 6,755
Own Kudos:
34,133
 [51]
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,755
Kudos: 34,133
 [51]
W, X, Y and Z are four different positive integers. When X 08 Mar 2019, 12:27
12
Kudos
Add Kudos
39
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

30% (02:36) correct 70% (02:38) wrong based on 342 sessions

History

Date
Time
Result
Not Attempted Yet
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 – 7, what is the sum of all possible values of W?
A) 19
B) 20
C) 21
D) 22
E) 23
ShowHide Answer
Official Answer
B
Most Helpful Reply
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 19 Jul 2025
Posts: 11,294
Own Kudos:
41,843
 [21]
Given Kudos: 333
Status:Math and DI Expert
Products:
My Reviews
Expert
Expert reply
Posts: 11,294
Kudos: 41,843
 [21]
W, X, Y and Z are four different positive integers. When X 08 Mar 2019, 18:36
13
Kudos
Add Kudos
8
Bookmarks
Bookmark this Post
GMATPrepNow
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 – 7, what is the sum of all possible values of W?
A) 19
B) 20
C) 21
D) 22
E) 23


Let us convert the information into an equation..
X=YZ+W..
Substitute W=X-7, so X=YZ+X-7, or YZ=7=1*7..
Since 7 is prime only product of YZ will be 1*7, and Y cannot be 1, as X and Z will be same then 7=1*7+0
So Y=7, now the remainder W has to be less than Y, so less than 7, but it cannot be 1 as Z is 1 and all numbers are different.
Thus, the possible values of W are 2, 3, 4, 5 and 6 and sum of these values is 2+3+4+5+6=20

B
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 13 May 2024
Posts: 6,755
Own Kudos:
34,133
 [10]
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,755
Kudos: 34,133
 [10]
W, X, Y and Z are four different positive integers. When X Updated on: 17 May 2021, 08:35
Originally posted by BrentGMATPrepNow on 09 Mar 2019, 10:30.
Last edited by BrentGMATPrepNow on 17 May 2021, 08:35, edited 1 time in total.
7
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
GMATPrepNow
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 – 7, what is the sum of all possible values of W?
A) 19
B) 20
C) 21
D) 22
E) 23

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 – 7
Take X = YZ + W and replace X with X - 7 to get: X = YZ + X - 7
Subtract X from both sides: 0 = YZ - 7
Rewrite as: 7 = YZ

We're told that Y and Z are positive INTEGERS. So, there are only 2 possible cases:
case i: Y = 1 and Z = 7
case ii: Y = 7 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 = 7 and Z = 1 (case ii)

So, we have: When X is divided by 7, the quotient is 1 and the remainder is W
This tells us that 7 divides into X 1 time
So, the possible values of X are: 7, 8, 9, 10, 11, 12 and 13 (since 7 divides into each value 1 time.
Let's check each case.

If X = 7, then the remainder (W) is 0. Since we're told W is POSITIVE, we can ELIMINATE this case.
If X = 8, 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 = 9, then the remainder (W) is 2. So, when X (9) is divided by 7 (Y), the quotient (Z) is 1, and the remainder (W) is 2. PERFECT!
If X = 10, then the remainder (W) is 3. So, when X (10) is divided by 7 (Y), the quotient (Z) is 1, and the remainder (W) is 3. PERFECT!
If X = 11, then the remainder (W) is 4. So, when X (11) is divided by 7 (Y), the quotient (Z) is 1, and the remainder (W) is 4. PERFECT!
If X = 12, then the remainder (W) is 5. So, when X (12) is divided by 7 (Y), the quotient (Z) is 1, and the remainder (W) is 5. PERFECT!
If X = 13, then the remainder (W) is 6. So, when X (13) is divided by 7 (Y), the quotient (Z) is 1, and the remainder (W) is 6. PERFECT!

What is the sum of all possible values of W?
Sum = 2 + 3 + 4 + 5 + 6
= 20

Answer: B

Cheers,
Brent

RELATED VIDEO
General Discussion
User avatar
Fdambro294
User avatar
Joined: 10 Jul 2019
Last visit: 06 Apr 2025
Posts: 1,353
Own Kudos:
707
Given Kudos: 1,658
Posts: 1,353
Kudos: 707
W, X, Y and Z are four different positive integers. When X 30 May 2021, 00:35
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Start with setting up the actual division or the division algorithms for integers:


X / Y = (Z) + (W/Y)

X = YZ + W


Constraint: the Integer Remainder = W = X - 7 ——-> substitute for W

X = YZ + X - 7

Y*Z = 7

Since Y is the divisor and each of the Variables is a positive integer, we can’t make Y = 1 because then there would be no remainder of W

Therefore: since we found that —- Y * Z = 7

Z = Quotient = 1

And

Y = Divisor = 7


Now plugging these values back into the original division we set up in the beginning:

X/7 = 1 + W/7

Rule: the integer remainder must always be less than < divisor

W < 7

And we know that: W = X - 7

So: X - 7 < 7

X < 14

Also, since we need a remainder W that is a positive integer, X must exceed Y (which we have as 7)

7 < X < 14


Now we can create the possible values of X with our values for Y = 7 and Z = 1 and then find which values of W satisfy

Case 1: X = 8

8/7 = 1 + 1/7

W = X - 7 ———> 1 = 8 - 7 —- works

However, in this case Z = W, so we must throw this case out as we are told that the variables are different positive integers in the question stem

Case 2: X = 9

9/7 = 1 + 2/7

W = X - 7 ——-> 2 = 9 - 7

Works: W = 2 works


.....can keep doing cases, creating consecutive integer values for W up until

Case: X = 13

13/7 = 1 + 6/7

W = X - 7 ———> 6 = 13 - 7 —— satisfies


valid cases: W can equal = 2 , 3, 4, 5, 6


If we try to make X = 14 or any higher integer:

15/7 = 2 + 1/7

W = X - 7 ———-> 1 does not equal 15 - 7 —- does NOT satisfy


Summing up the possible values for remainder W:

2 + 3 + 4 + 5 + 6 =

20

Posted from my mobile device
User avatar
BrushMyQuant
User avatar
Joined: 05 Apr 2011
Last visit: 19 Jul 2025
Posts: 2,253
Own Kudos:
2,468
Given Kudos: 100
Status:Tutor - BrushMyQuant
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE:Information Technology (Computer Software)
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
Posts: 2,253
Kudos: 2,468
W, X, Y and Z are four different positive integers. When X 23 Sep 2023, 04:55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
↧↧↧ Detailed Video Solution to the Problem ↧↧↧




Given that 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 – 7, what is the sum of all possible values of W?

X is divided by Y, the quotient is Z and the remainder is W
=> X = Y*Z + W = YZ + W

W = X – 7
=> X = 7 + W = YZ + W
=> YZ = 7 = 1*7 = 7*1

=> We will have two cases



Divisor (Y) = 1 or 7 and Quotient(Z) = 7 or 1
-Case 1: Y = 1 , Z = 7

Divisor(Y) is 1
And we know that any positive integer divided by 1 will always give 0 as remainder

=> W = 0
=> X = 1*7+0 = 7
And Z = 7

=> X and Z become same, but they are different integers

Which is NOT POSSIBLE		-Case 2: Y = 7 , Z = 1

Divisor(Y) is 7 => Reminders will be between 0 to 6
=> Following cases are possible
=> W = 0, same as Case 1, so we can ignore this => NOT POSSIBLE
=> W = 1, Both W and Z = 1 => NOT POSSIBLE
X, Y, Z, W all are different in below cases
=> W = 2, X = 7*1+2 = 9 => POSSIBLE
=> W = 3, X = 7*1+3 = 10 => POSSIBLE
=> W = 4, X = 7*1+4 = 11 => POSSIBLE
=> W = 5, X = 7*1+5 = 12 => POSSIBLE
=> W = 6, X = 7*1+6 = 13 => POSSIBLE

=> Sum of all possible values of W = 2 + 3 + 4 + 5 + 6 = 20

So, Answer will be B
Hope it helps!

Watch the following video to learn the Basics of Remainders

User avatar
MDSSourov
User avatar
Joined: 28 Dec 2024
Last visit: 24 Feb 2025
Posts: 2
Given Kudos: 3
Posts: 2
Kudos: 0
W, X, Y and Z are four different positive integers. When X 17 Jan 2025, 13:26
Kudos
Add Kudos
Bookmarks
Bookmark this Post
BrentGMATPrepNow
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 – 7, what is the sum of all possible values of W?
A) 19
B) 20
C) 21
D) 22
E) 23
X=YZ+W..
So X=YZ+X-7 (w=x-7),
YZ=7=1*7..
Since 7 is prime only product of YZ will be 1*7, and Y cannot be 1, but z can.
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 y=7 and z=1
So Y=7, now the remainder W has to be less than Y, so less than 7, but it cannot be 1 as Z is 1 because all numbers are different. If w =7 it will again divide by 7, that’s why less than 7
Thus, the possible values of W are 2, 3, 4, 5 and 6 and sum of these values is 2+3+4+5+6=20
B
Moderators:
Bunuel
Math Expert
102627 posts
Krunaal
PS Forum Moderator
698 posts