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W, X, Y and Z are four different positive integers. When X
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08 Mar 2019, 12:27
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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 – 7, what is the sum of all possible values of W? A) 19 B) 20 C) 21 D) 22 E) 23
Re: W, X, Y and Z are four different positive integers. When X
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08 Mar 2019, 18:36
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GMATPrepNow 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 – 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
Re: W, X, Y and Z are four different positive integers. When X
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09 Mar 2019, 10:30
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GMATPrepNow 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 – 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
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15% (02:36) correct 
