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

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
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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

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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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