When positive integer x is divided by 5, the remainder is 3;

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

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
x-y=5(a-c) or 7(b-d)
take LCM of 5 and 7 = 35 (it's easy here bacause both are prime)

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.

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..

x-y is divisible by 35..

hence, answer is E..

Acc to me,in these kind of questions, plugging numbers is the best approach.

x = 5a + 3 = 7b + 4
y = 5c + 3 = 7d + 4
x-y = 5(a-c) = 7(b-d)

Since, 5 and 7 are prime number, hence 5*7 must be a factor of (x-y).

 


In addition to the links shared by Bunuel, you can also check out this video on grouping in division.

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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 x-y 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.

Trial and error method is lengthy than the equation method provided by bigtreezl.

Hi Bunuel , could u explain this concept in terms of grouping objects. Thanks a lot for all the posts as always.

I'm not entirely certain what you're referring to, but you can find additional information on remainders and similar questions in the links provided below:

Remainders

• Theory
  REMAINDERS
  Remainders: Tips and hints
  Divisibility and Remainders on the GMAT
  Remainders on the GMAT
  A Remainders Shortcut for the GMAT
  All About Negative Remainders on the GMAT
  A Tricky Question on Negative Remainders
  Compilation of tips and tricks to deal with remainders
  Cyclicity and the remainders on the GMAT
  GMAT Question Patterns: Remainders
  
  • Questions
    Units digits, exponents, remainders problems
    DS Questions
    PS Questions

For other subjects:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

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Here is a one minute solution to this problem.

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