W, X, Y and Z are four different positive integers. When X

Question

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

chetan2u · Accepted Answer

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

BrentGMATPrepNow · Answer

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
 https://www.youtube.com/watch?v=XCxgIR042tE

Fdambro294 · Answer

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

BrushMyQuant · Answer

↧↧↧ Detailed Video Solution to the Problem ↧↧↧

https://www.youtube.com/watch?v=EnWAZrhIwOI

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

https://www.youtube.com/watch?v=P0S6j2uTaj4

MDSSourov · Answer

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

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W, X, Y and Z are four different positive integers. When X

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