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when a positive integer x is divided by 3, the remainder is [#permalink]
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12 Jun 2008, 23:00
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This topic is locked. If you want to discuss this question please repost it in the respective forum. when a positive integer x is divided by 3, the remainder is 1, when positive integer y is divided by 5 , the remainder is 1. If x+y is divisible by 5 what is the least possible value of y 1 6 7 11 13 this is how i solve such questions (it takes me 11.5 mins to solve this) my point is that is this approach an optimal one (if no then please suggest an optimal approach) X =3N+1 Y=5N+1 X+Y=15N+16 16/15 remainder is 1 therefore the least value that y could have is 1
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Re: when a positive integer x is divided by [#permalink]
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13 Jun 2008, 00:51
vdhawan1 wrote: when a positive integer x is divided by 3, the remainder is 1, when positive integer y is divided by 5 , the remainder is 1. If x+y is divisible by 5 what is the least possible value of y
1 6 7 11 13
this is how i solve such questions (it takes me 11.5 mins to solve this) my point is that is this approach an optimal one (if no then please suggest an optimal approach)
X =3N+1 Y=5N+1 X+Y=15N+16 16/15 remainder is 1 therefore the least value that y could have is 1 I don't get why you assume that the 'N' in X=3N+1 and Y=5N+1 is the same (there is no reason for that). And I don't understand what you mean after that either (why do you divide 16 by 15 ? How does it translate into Y least value is 1 ?) Correct answer is still 1 : just try X=4 and Y=1 and you can verify it works.



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Re: when a positive integer x is divided by [#permalink]
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13 Jun 2008, 00:57
how can y = 1, when question says y = 5N+1, N = 1,2,3 .... N can not be zero.
The answer should be 11,
x = 4, y = 11 x+y = 15, divisible by 5.



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Re: when a positive integer x is divided by [#permalink]
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13 Jun 2008, 01:06
durgesh79 wrote: how can y = 1, when question says y = 5N+1, N = 1,2,3 .... N can not be zero.
The answer should be 11,
x = 4, y = 11 x+y = 15, divisible by 5. Why do you assume N cannot be 0 ? I don't get it. 1 is a positive integer whose remainder is 1 when divided by 5 (since 1=5*0+1) Edit : and even if you assume (for no reason) that N should be positive, then the answer would be 6 and not 11. Try x=4 and y=6, you'll see
Last edited by Oski on 13 Jun 2008, 02:17, edited 1 time in total.



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Re: when a positive integer x is divided by [#permalink]
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13 Jun 2008, 01:23
Oski wrote: Edit : and even if you assume (for no reason) that N should be positive, then the answer would be 6 and not 11. Try x=4 and y=6, you'll see oops... yes the answer should be 6. Regarding the point why N can not be zero, I thought when we talk of y divided by x and remainder r, it is assumed that y>x, so that the integer quotient is greater than or equal to 1. Please correct me if I'm wrong.



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Re: when a positive integer x is divided by [#permalink]
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13 Jun 2008, 01:32
Oh k the OA for this one is A
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Re: when a positive integer x is divided by [#permalink]
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13 Jun 2008, 02:20
durgesh79 wrote: Regarding the point why N can not be zero, I thought when we talk of y divided by x and remainder r, it is assumed that y>x, so that the integer quotient is greater than or equal to 1. Please correct me if I'm wrong. I see no reason for assuming that. For me 72 is divisible by 9 even if 72 is not greater than 9. Plus 0 is divisible by any integer (and that would not be the case if we make your assumption)



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Re: when a positive integer x is divided by [#permalink]
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13 Jun 2008, 05:55
Oski wrote: durgesh79 wrote: Regarding the point why N can not be zero, I thought when we talk of y divided by x and remainder r, it is assumed that y>x, so that the integer quotient is greater than or equal to 1. Please correct me if I'm wrong. I see no reason for assuming that. For me 72 is divisible by 9 even if 72 is not greater than 9. Plus 0 is divisible by any integer (and that would not be the case if we make your assumption) My assumption was only in the case of +ve remainders, but i guess the assumption is wrong. thanks for pointing it out. Remainder rule from today : if y is divided by x and the remainder is r, then y = Nx + r, where N = 0,1,2,3........



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Re: when a positive integer x is divided by [#permalink]
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13 Jun 2008, 05:58
durgesh79 wrote: Remainder rule from today : if y is divided by x and the remainder is r, then y = Nx + r, where N = 0,1,2,3........ Not in this particular question, but in general N can even be negative. General rule would be : if y is divided by x and the remainder is r, then y = Nx + r, where N is integer



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Re: when a positive integer x is divided by [#permalink]
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13 Jun 2008, 05:59
vdhawan1 wrote: when a positive integer x is divided by 3, the remainder is 1, when positive integer y is divided by 5 , the remainder is 1. If x+y is divisible by 5 what is the least possible value of y
1 6 7 11 13
this is how i solve such questions (it takes me 11.5 mins to solve this) my point is that is this approach an optimal one (if no then please suggest an optimal approach)
X =3N+1 Y=5N+1 X+Y=15N+16 16/15 remainder is 1 therefore the least value that y could have is 1 That was pretty simple. 1/5 > remainder is 1. A



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Re: when a positive integer x is divided by [#permalink]
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13 Jun 2008, 06:01
GMATBLACKBELT wrote: That was pretty simple.
1/5 > remainder is 1.
A This is not a proof since it misses every properties of x (which has some).



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Re: when a positive integer x is divided by [#permalink]
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13 Jun 2008, 06:02
This is a good question. I wasn't even thinking of 0 as an option. I immediately started out with x=4, y = 6, and when 4 + 6 = 10/5 = 2, then I thought y must = 6, but x = 4, divided by 3, is 1 w/ remainder 1. y= 1, 1/5 = 0 w/remainder 1, and 1 + 4 =5, obviously divisible by 5. i guess the point of this question is we must think about 0 as a number and not just go for integers unless the question stem says so. vdhawan1 wrote: Oh k
the OA for this one is A
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Re: when a positive integer x is divided by [#permalink]
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13 Jun 2008, 06:08
jallenmorris wrote: huh? I'm just saying this is not a proof that the answer is (1). You also have to show that it works for at least one value of x that verifies the specified properties.



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Re: when a positive integer x is divided by [#permalink]
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17 Sep 2008, 17:18
durgesh79 wrote: Oski wrote: Edit : and even if you assume (for no reason) that N should be positive, then the answer would be 6 and not 11. Try x=4 and y=6, you'll see oops... yes the answer should be 6. Regarding the point why N can not be zero, I thought when we talk of y divided by x and remainder r, it is assumed that y>x, so that the integer quotient is greater than or equal to 1. Please correct me if I'm wrong. I don't agree. For example, when you divide 3 by 5, the remainder is 3. On the other hand, when you divide 8 by 5, the remainder is also 3. So the numerator doesn't always have to be bigger than the denominator.



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Re: when a positive integer x is divided by [#permalink]
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17 Sep 2008, 20:45
tarek99 wrote: GMATBLACKBELT wrote: vdhawan1 wrote: when a positive integer x is divided by 3, the remainder is 1, when positive integer y is divided by 5 , the remainder is 1. If x+y is divisible by 5 what is the least possible value of y
1 6 7 11 13
this is how i solve such questions (it takes me 11.5 mins to solve this) my point is that is this approach an optimal one (if no then please suggest an optimal approach)
X =3N+1 Y=5N+1 X+Y=15N+16
am i missing something here? shouldn't 3N+5N equal to 8N? and 1+1=2? where from the 16??
16/15 remainder is 1 therefore the least value that y could have is 1 That was pretty simple. 1/5 > remainder is 1. A Thats not a valid reasoning. The same quotient, N, may not be true for both x and y when divided by 3 and 5 respectively. x = 3n + 1 y = 5p + 1 since (x+y) is divisible by 5, x+y can be as small as 5 but not 0 or smaller than 0 because x and y each is a +ve integer. so our concern is the smallest value for y. if that is the case, what is the possible smallest value for y? first it is 1, then 2 and so on. lets see whether 1 is possible value for y. y's value depends on x's value. since x + y is divisible by 5, x should be 5k1 to have y's value 1. the possible value for x = 5k1 are 4, 9, 19, 29, 39, 49 etc. but 9, 29, 39 cannot be the values of x cuz they do not yeild 1 as reminder if they each are divided by 3. so the possible values are 4 or 19, 49 etc. in each case y is 1. so this is the smallest possible value for y. //A//
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Re: when a positive integer x is divided by [#permalink]
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17 Sep 2008, 22:23
This is how I approaced.
x+y = 5k for k = 0,1,2,3,.....
or y = 5k x or y = 5k (3m+1) for m = 0,1,2,3,......
Now, trying substitution, k = 0, m = 0, y is ve....not possible. k=1, m=1, y = 1.
Hence, the smallest positive value of y will be 1.



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Re: when a positive integer x is divided by [#permalink]
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18 Sep 2008, 04:43
GMAT TIGER, how did you get x to be 5k1? Also, i'm trying to understand this problem abstractly because I feel that this can be solved without doing any math. First of all, I remember I read from the Manhattan GMAT book "Number Properties" that when x and y are divisible by, for example, 5, then both x+y and xy will also be divisible by 5. However, in this problem, y is definitely not divisible by 5 otherwise the remainder wouldn't be 1. And we don't know much about x either. So how can we go about this abstractly? if you guys know what I mean? After all, this is suppose to be a GMAT question, so there is always an easy and sneaky way to reach to an answer with almost minimal effort.



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Re: when a positive integer x is divided by [#permalink]
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18 Sep 2008, 05:56
I found a quick approach!!! Like I said before, if a is divisible by 5 and if a= x+y, then it means that both x and y separately should be divisible by 5. However, sometimes a can be divisible by 5, but x and y may not necessarily be divisible by 5. In that case, the sum of the remainders of x and y should be a multiple of 5., hence divisible by 5. We already know that when y is divided by 5, the remainder is 1. So when x is divided by 5, x's remainder should add up to y's remainder of 1 to give us 5, hence, x should give us a remainder of 4. With this approach, the value of y automatically becomes 1, 6, 11, or etc....the least value is obviously 1! This is how you're pretty much expected to approach the math section of the GMAT. If you follow this approach, you can answer this type of question literally in less than 10 seconds! Hope this can make your life MUCH EASIER!



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Re: when a positive integer x is divided by [#permalink]
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07 Nov 2008, 13:12
Can someone explain how is the answer 1 or A The question says that Y divided with 5 has a remainder 1 that means Y should be at least 6 to produce remainder 1
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Re: when a positive integer x is divided by [#permalink]
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07 Nov 2008, 17:23
spiridon wrote: Can someone explain how is the answer 1 or A
The question says that Y divided with 5 has a remainder 1 that means Y should be at least 6 to produce remainder 1 1 / 5 = 0, remander 1 Keep this in your mind and don't assume (Y should be at least 6 to produce remainder 1)




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