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When positive integer x is divided by 5, the remainder is 3;
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12 Mar 2008, 13:16
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When positive integer x is divided by 5, the remainder is 3; and when x is divided by 7, the remainder is 4. When positive integer y is divided by 5, the remainder is 3; and when y is divided by 7, the remainder is 4. If x > y, which of the following must be a factor of x  y? (A) 12 (B) 15 (C) 20 (D) 28 (E) 35
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Re: When positive integer x is divided by 5, the remainder is 3;
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13 Jan 2012, 10:31
Baten80 wrote: Trial and error method is lengthy than the equation method provided by bigtreezl. The way to derive general formula is described in the solution below:When positive integer x is divided by 5, the remainder is 3; and when x is divided by 7, the remainder is 4. When positive integer y is divided by 5, the remainder is 3; and when y is divided by 7, the remainder is 4. If x > y, which of the following must be a factor of x  y? (A) 12 (B) 15 (C) 20 (D) 28 (E) 35 When the positive integer x is divided by 5 and 7, the remainder is 3 and 4, respectively: \(x=5q+3\) (x could be 3, 8, 13, 18, 23, ...) and \(x=7p+4\) (x could be 4, 11, 18, 25, ...). There is a way to derive general formula based on above two statements:Divisor will be the least common multiple of above two divisors 5 and 7, hence \(35\). Remainder will be the first common integer in above two patterns, hence \(18\) > so, to satisfy both this conditions x must be of a type \(x=35m+18\) (18, 53, 88, ...); The same for y (as the same info is given about y): \(y=35n+18\); \(xy=(35m+18)(35n+18)=35(mn)\) > thus xy must be a multiple of 35. Answer: E. More about this concept: manhattanremainderproblem93752.html?hilit=derive#p721341goodproblem90442.html?hilit=derive#p722552Hope it helps.
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Re: gprep remainders
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14 Mar 2008, 20:18
35
Interesting thing to note here is it doesn't matter what the reminders are, the only thing that matters is if the reminders are same for x and y when devided by the same number.
x can be written as 5a+3 or 7b+4 y can be written as 5c+3 or 7d+4 xy=5(ac) or 7(bd) take LCM of 5 and 7 = 35 (it's easy here bacause both are prime)



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Re: factor
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11 May 2008, 10:19
Hi , I tried two ways of reaching the answer and I get E 35. Not sure if this is right though ! For x we have two equations: x=3+5m x=4+7n Similarly for y: y=3+5l 4= 4 +7j xy =3+5m  (3+5l) =5(ml) xy = 4+7n (4+7j) = 7(nj) 7 and 5 are factors of xy, therefore 7*5 =35 must be a factor of xy. Another way I tried to solve this is by using numbers, If x=3+5m, then c must be a number with units digit 8 or 3 since x=7 +5n , the values of x that satisfy these two equations are 18 and 53, since y has similar rules and x>y; x=53 and y =18, xy= 5318 =35, therefore 35 must be a factor of xy.



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Re: factor
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12 May 2008, 14:11
fresinha12 wrote: x=5N+3 => 8,13,18,23,28,33,39 x=7N+4 => 11,18,25,32,39..
so if i pick x=23 y=11 then xy=12..
A can be a factor of xy..
if you pick x=39 y=11 then 28 can also be a factor..
so D can also be an answer..
not sure where this question is from? Fresinha, the highlighted should be 38, hence ur error. One way to look at this question is that the same reminders will be yielded every 5x7 times (common denominator offset by the same reminders).



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Re: factor
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12 May 2008, 17:49
Agree with E
Remainder 3: x = 3 8 13 18 23 28 33 38 .... 53 58 Remainder 4: x = 4 11 18 25 ... 53
Remainder 3: y = 3 8 13 18 23 28 33 38 .... 53 58 Remainder 4: y = 4 11 18 25 ... 53
values of 18 and 53 are valid for x and y
x > y so x = 53, y = 18
x  y = 35



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Re: GMAT Set  Q8
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17 Oct 2008, 16:07
I got E
5n+3=x 7m+4=x
5n+3=y 7m+4=y
for x 5n+3=7m+4 5n=7m+1 5*10 = 7*7+1 (find numbers that fit) x=50
for y 5n+3=7m+4 5n=7m+1 5*3=7*2+1 y=15
xy=5015=35, 35 is a factor of 35



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Re: GMAT Set  Q8
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07 Nov 2008, 04:59
x = 5a + 3 = 7b + 4 y = 5c + 3 = 7d + 4 xy = 5(ac) = 7(bd)
Since, 5 and 7 are prime number, hence 5*7 must be a factor of (xy).



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Re: PS: When a number x is divided by 5 it leaves a reminder of
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27 Aug 2009, 07:55
The difference must be the multiple of 35, which is LCM of 5 and 7. 1) In order for x and y to leave the same remainder when divided by 5, the gap between two numbers should be a multiple of 5. 2)In order for x and y to leave the same remainder when divided by 7, the gap between two numbers should be a multiple of 7. But x and y leave the same remainders when divided by both 5 and 7...so the gap between x and y should be a multiple of 5 AND a multiple of 7 or simply it should be a multiple of 35, which is LCM (5,7).
The only number that is a multiple of 35 is E, hence E is an answer.



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Re: PS: When a number x is divided by 5 it leaves a reminder of
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27 Aug 2009, 09:39
Lets find out some numbers which leaves a reminder of 3 when divided by 5.
3,8,13,18,23,28,33,38,43,48,53,58,63,68,73,78,83,88
out of these numbers,
numbers which leaves a reminder of 4 when divided by 7 are
18,53,88....
as x is greater than y..
hence, if x = 18 , then y = 53 and if x = 53 , then y = 88 or if x = 18 , then y = 88
in all cases..
xy is divisible by 35..
hence, answer is E..
Acc to me,in these kind of questions, plugging numbers is the best approach.



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Re: PS: When a number x is divided by 5 it leaves a reminder of
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27 Aug 2009, 09:51
gmate2010 wrote: Lets find out some numbers which leaves a reminder of 3 when divided by 5.
3,8,13,18,23,28,33,38,43,48,53,58,63,68,73,78,83,88
out of these numbers,
numbers which leaves a reminder of 4 when divided by 7 are
18,53,88....
as x is greater than y..
hence, if x = 18 , then y = 53 and if x = 53 , then y = 88 or if x = 18 , then y = 88
in all cases..
xy is divisible by 35..
hence, answer is E..
Acc to me,in these kind of questions, plugging numbers is the best approach. If you realize that the difference between the numbers is a multiple of 35, it takes about 30 sec to solve this one...plugging numbers you lose your time....but of course, people have different ways of solving...whatever works better for you



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Re: NS Factor Combo!!
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06 Sep 2011, 20:08
Here we are dealing with two numbers which give the same remainder by 5 and by 7. It's useful to know that if you were to list all such numbers, you would get an equally spaced list, where the numbers are separated by the LCM of 5 and 7, so by 35. So xy must be divisible by 35 here. Of course, you could come up with sample numbers if you weren't familiar with the underlying theory. We need two numbers which give a remainder of 3 when divided by 5, and a remainder of 4 when divided by 7. We can start by listing small numbers which give a remainder of 3 when divided by 5. This list is equally spaced, by 5, so it's straightforward to generate a long list quickly: 3, 8, 13, 18, 23, 28, 33, 38, 43, 48, 53, 58, 63, 68, ... Now if you scan this list looking for numbers which give a remainder of 4 when divided by 7, you'll see that 18 and 53 both work. So it might be that x=53 and y=18, and their difference is 35, from which we also get answer E.
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Re: When positive integer x is divided by 5, the remainder is 3;
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13 Jan 2012, 09:32



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Re: When positive integer x is divided by 5, the remainder is 3;
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01 Nov 2014, 23:09
1) x=5Q+3 also x=7q+4
2) y=5R+3 also y=7r+4
Taking difference of above equations separately :
xy=5(QR) also xy=7(qr) we don't get any remainders in the equations that follows above. so it implies (xy) is both divisible by 5 & 7.
Looking at the answer choices one should prefer the option which is both divisible by 5 & 7 that is 35.
Correct Answer is choice E



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When positive integer x is divided by 5, the remainder is 3;
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08 Jan 2017, 16:45
marcodonzelli wrote: When positive integer x is divided by 5, the remainder is 3; and when x is divided by 7, the remainder is 4. When positive integer y is divided by 5, the remainder is 3; and when y is divided by 7, the remainder is 4. If x > y, which of the following must be a factor of x  y?
(A) 12 (B) 15 (C) 20 (D) 28 (E) 35 x will always equal y plus a multiple of the product of the two divisors, 5 and 7 e.g., least values would be x=53, y=18 if x=y+35, then xy=35 35



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Re: When positive integer x is divided by 5, the remainder is 3;
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07 Feb 2018, 20:10
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Re: When positive integer x is divided by 5, the remainder is 3; &nbs
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