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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

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↧↧↧ 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

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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
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