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Re: x and y are positive integers. When 16x is divided by y,
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27 Feb 2017, 14:21

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GMATPrepNow wrote:

x and y are positive integers. When 16x is divided by y, the quotient is x, and the remainder is 4. What is the sum of all possible y-values?

A) 7 B) 12 C) 19 D) 26 E) 41

There's a nice rule that say, "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

------NOW ONTO THE QUESTION------------------------------

When 16x is divided by y, the quotient is x, and the remainder is 4. Applying the above rule, we can write: 16x = (y)(x) + 4 Subtract xy from both sides: 16x - xy = 4 Factor: x(16 - y) = 4

Since x and (16 - y) are both positive integers, there are 3 possible solutions: #1: x = 1 and (16 - y) = 4, in which case y = 12 #2: x = 2 and (16 - y) = 2, in which case y = 14 #3: x = 4 and (16 - y) = 1, in which case y = 15

What is the sum of all possible y-values? SUM = 12 + 14 + 15 = 41

Re: x and y are positive integers. When 16x is divided by y,
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27 Feb 2017, 15:19

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Great question! I solved it a very similar way to you all. (16x)/y = x + 4 --> 16x = xy + 4 --> y = (16x - 4)/x ---> y = 16 - 4/x.

We know from the prompt that Y is an integer, therefore, the only values of X that will make Y and integer is 0, 1, 2, 4. However, the prompt says that X & Y are positive integers. Therefore we can't use 0. So plugging in the values of 1, 2, and 4 into the equation, you're left with 15 + 14 + 12 = 41.