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

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

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
x-y = 5a1-5a2+3+3=> 5(a1-a2)
also x-y => 7b1-7b2+4-4=> 7(b1-b2).
so x-y is a multiple of both 5 and 7. 35 fits the bill it is a factor of 7 and 5.

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 x-y.

Answer (E)

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

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 53-18 is a multiple of only 35 from answer choices.

Actually, 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?

No. Neither 5 nor 7 is a factor of either x or y.

because the divisors and remainders are the same for both dividends,
all values of x-y will equal a multiple of the product of the two divisors, 5 and 7
5*7=35
E

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.