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

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

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Bunuel · Accepted Answer

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 a positive integer x is divided by 5 and 7, the remainders are 3 and 4, respectively: x = 5q + 3 (where x could be 3, 8, 13, 18 , 23, ...) and x = 7p + 4 (where x could be 4, 11, 18 , 25, ...). A general formula can be derived from the above two statements: The divisor will be the least common multiple of the two divisors 5 and 7, hence 35. The remainder will be the first common integer in the two patterns, hence 18. Therefore, to satisfy both conditions, x must be of the form x = 35m + 18 (where x could be 18, 53, 88, ...). The same applies to y (as the same information is given about y): y = 35n + 18 . Henxe, x-y=(35m+18)-(35n+18)=35(m-n) . Therefore, x - y must be a multiple of 35. Answer: E. More about this concept: https://gmatclub.com/forum/manhattan-re ... ve#p721341 https://gmatclub.com/forum/good-problem ... ve#p722552 Hope it helps.

ventivish · Answer

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 x-y =3+5m - (3+5l) =5(m-l) x-y = 4+7n -(4+7j) = 7(n-j) 7 and 5 are factors of x-y, therefore 7*5 =35 must be a factor of x-y. 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, x-y= 53-18 =35, therefore 35 must be a factor of x-y.

sreehari · Answer

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)

gmatnub · Answer

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

scthakur · Answer

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).

LenaA · Answer

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.

gmate2010 · 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.

IanStewart · Answer

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.

MHIKER · Answer

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

Yes2GMAT · Answer

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

Bunuel · Answer

KarishmaB · Answer

Here is a  one minute solution to this problem.

KarishmaB · Answer

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

bumpbot · Answer

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

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

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