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x and y are positive integers. When 16x is divided by y,

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x and y are positive integers. When 16x is divided by y, [#permalink]

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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

*Kudos for all correct solutions

[Reveal] Spoiler:
EDIT: Changed answer choice E
[Reveal] Spoiler: OA

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Last edited by GMATPrepNow on 27 Feb 2017, 09:21, edited 1 time in total.
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Re: x and y are positive integers. When 16x is divided by y, [#permalink]

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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) 39

*Kudos for all correct solutions



16x = yx+4
y=(16x-4)/x

if x=1 then y =12
if x= 2 then y= 14
if x=3 y is no longer integer
if x=4 then y = 15

upto here y = 12+14+15 = 41...
but i m not finding any such option.....

can any one explain!!!!!

thanks
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Re: x and y are positive integers. When 16x is divided by y, [#permalink]

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Yeesh, I added those values several times and got 39 each time.
I've changed answer choice E to the correct value.

Cheers and thanks,
Brent
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Re: x and y are positive integers. When 16x is divided by y, [#permalink]

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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) 39

*Kudos for all correct solutions


Hi

It seems I've missed something.

16x = yx + 4

16x - yx = 4

x(16 - y) = 4 (1*4), (2*2) and (4*1). I'm discarding negativ values because our x an y are >0.

x=1, 16 - y = 4 ---> y=12, check: 16*1 = 12*1 + 4

x = 2, 16 - y = 2 -----> y = 14, check: 32 = 14*2 + 4

x = 4, 16 - y = 1 -------> y=15, chek: 64 = 15*4 + 4

4<y<16

12+14+15 = 41

I'm not getting 39. Please correct me.
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Re: x and y are positive integers. When 16x is divided by y, [#permalink]

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New post 27 Feb 2017, 09:37
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You're correct, vitaliyGMAT

It seems I have not yet mastered addition. :oops:

Cheers,
Brent
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Re: x and y are positive integers. When 16x is divided by y, [#permalink]

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New post 27 Feb 2017, 09:56
GMATPrepNow wrote:
You're correct, vitaliyGMAT

It seems I have not yet mastered addition. :oops:

Cheers,
Brent


Thanks

Your questions are really good! Please keep posting, just don't hurry next time :-D
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Re: x and y are positive integers. When 16x is divided by y, [#permalink]

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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

Answer: E

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Re: x and y are positive integers. When 16x is divided by y, [#permalink]

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