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When positive integer x is divided by 5, the remainder is 3
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Updated on: 06 Jun 2013, 06:31
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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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Originally posted by shopaholic on 02 Mar 2012, 10:39.
Last edited by Bunuel on 06 Jun 2013, 06:31, edited 2 times in total.
Edited the question, added the answer choices and OA




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Re: How to solve this problem
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02 Mar 2012, 11:16
shopaholic 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?
thanks in advance Welcome to GMAT Club. Below is a solution to your question. 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. P.S. Please post answer choices for PS questions.
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Re: When positive integer x is divided by 5, the remainder is 3
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06 Jun 2013, 06:32



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Re: When positive integer x is divided by 5, the remainder is 3
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06 Jun 2013, 09:50
I do the "for dummies" way on this, because it is the first thing that popped in my mind, and the first thing I would have tried. It actually doesn't take that long on paper. Easily under 2 minutes.
5x+3: 5+3=8 10+3=13 15+3=18 20+3=23 25+3=28 30+3=33 35+3=38 40+3=43 45+3=48 50+3=53
7x+4: 7+4=11 14+4=18 21+4=25 28+4=32 35+4=39 42+4=46 49+4=53 56+4=60 63+4=67 70+4=74
The thing about PS problems is that there can only be one answer. So as long as you can find it one time, it will be the same for all other times. The question is asking for the larger number minus the smaller number so, 5318 = 35
Answer is E



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Re: How to solve this problem
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23 Oct 2013, 19:45
Bunuel wrote: shopaholic 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?
thanks in advance Welcome to GMAT Club. Below is a solution to your question. 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. P.S. Please post answer choices for PS questions.there must be some other way to solve this; I have no idea how anyone that reads that question could sit there and think of what you wrote out, in under two minutes. It's a great solution, but I think there must be some other way to crack this nut.



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Re: How to solve this problem
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24 Oct 2013, 00:29
AccipiterQ wrote: Bunuel wrote: shopaholic 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?
thanks in advance Welcome to GMAT Club. Below is a solution to your question. 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. P.S. Please post answer choices for PS questions.there must be some other way to solve this; I have no idea how anyone that reads that question could sit there and think of what you wrote out, in under two minutes. It's a great solution, but I think there must be some other way to crack this nut. If you know the trick to derive general formula you can solve the question just under 2 minutes. Else, you can find common numbers (18 and 53) and see that 5318 is a multiple of only 35 from answer choices.
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Re: When positive integer x is divided by 5, the remainder is 3
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24 Oct 2013, 09:01
Bunuel wrote: Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HEREActually, on this problem, if 5 is a factor of it, as is 7, couldn't you just look at the stem, and multiply 5 by 7 without doing any extra work, to get the answer?



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Re: When positive integer x is divided by 5, the remainder is 3
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24 Oct 2013, 09:04



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Re: When positive integer x is divided by 5, the remainder is 3
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25 Nov 2014, 21:30
AccipiterQ wrote: there must be some other way to solve this; I have no idea how anyone that reads that question could sit there and think of what you wrote out, in under two minutes. It's a great solution, but I think there must be some other way to crack this nut. Yes, there is. You just read the question and the answer will be there by the time you are done (doing everything Bunuel did above). Consider the question one line at a time: "When positive integer x is divided by 5, the remainder is 3; and when x is divided by 7, the remainder is 4."Think  You will get x by finding the first such value which when divided by 5 leaves remainder 3 and when divided by 7 leaves remainder 4. x will be 35N + First such value "When positive integer y is divided by 5, the remainder is 3; and when y is divided by 7, the remainder is 4."Go back and compare the divisors and remainders. You see that they are exactly the same. This means y also belongs to the same series. y = 35M + First such value So you know now that x will be some multiple of 35 plus a number and y will be another multiple of 35 plus the same number. So x and y will differ by some multiple of 35. "If x > y, which of the following must be a factor of x  y?"So 35 must be a factor of xy. Answer (E)
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Re: When positive integer x is divided by 5, the remainder is 3
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20 Jun 2016, 21:10
There is one other way to solve this problem. I earlier went with enumerating choices and it was taking time. So here is another way x>y is given x=5a1+3 and y=5a2+3.. also x=7b1+4 and y = 7b2+4 xy = 5a15a2+3+3=> 5(a1a2) also xy => 7b17b2+44=> 7(b1b2). so xy is a multiple of both 5 and 7. 35 fits the bill it is a factor of 7 and 5.



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Re: When positive integer x is divided by 5, the remainder is 3
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06 Oct 2017, 10:54
shopaholic 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 Let’s list some integers that have a remainder of 3 when divided by 5: 3, 8, 13, 18, 23, … Let’s list some integers that have a remainder of 4 when divided by 7: 4, 11, 18, 25, … We see that 18 is common in both lists and is the smallest positive integer that satisfies both conditions. Thus, y could be 18. To get a value for x, we can simply add the LCM of 5 and 7, i.e, 35, to 18. So, x could be 53 (notice that 53 also satisfies both conditions). We see that in this case, x  y = 35, and of all the choices, only 35 is factor of x  y. Alternate Solution: Since x produces a remainder of 3 when divided by 5, we can write x = 5p + 3 for some integer p. Since x produces a remainder of 4 when divided by 7, we can write x = 7q + 4 for some integer q. Since y produces a remainder of 3 when divided by 5, we can write y = 5s + 3 for some integer s. Since y produces a remainder of 4 when divided by 7, we can write y = 7r + 4 for some integer r. If we subtract the third equality from the first, we obtain: x  y = 5p  5s = 5(p  s). Thus, x  y is a multiple of 5. If we subtract the fourth equality from the second, we obtain: x  y = 7q  7r = 7(q  r). Thus, x  y is a multiple of 7. Since x  y is a multiple of both 5 and 7, it is a multiple of the LCM of 5 and 7 as well, namely 35. Answer: E
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