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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

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

*New project from GMAT Club!!! Check HERE

Theory on remainders problems: remainders-144665.html

All DS remainders problems to practice: search.php?search_id=tag&tag_id=198
All PS remainders problems to practice: search.php?search_id=tag&tag_id=199
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shopaholic
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): \(y=35n+18\);

\(x-y=(35m+18)-(35n+18)=35(m-n)\) --> thus x-y must be a multiple of 35.

Answer: E.

More about this concept:
manhattan-remainder-problem-93752.html?hilit=derive#p721341
good-problem-90442.html?hilit=derive#p722552

Hope 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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Bunuel
shopaholic
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): \(y=35n+18\);

\(x-y=(35m+18)-(35n+18)=35(m-n)\) --> thus x-y must be a multiple of 35.

Answer: E.

More about this concept:
manhattan-remainder-problem-93752.html?hilit=derive#p721341
good-problem-90442.html?hilit=derive#p722552

Hope 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 53-18 is a multiple of only 35 from answer choices.
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on remainders problems: remainders-144665.html

All DS remainders problems to practice: search.php?search_id=tag&tag_id=198
All PS remainders problems to practice: search.php?search_id=tag&tag_id=199

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

*New project from GMAT Club!!! Check HERE

Theory on remainders problems: remainders-144665.html

All DS remainders problems to practice: search.php?search_id=tag&tag_id=198
All PS remainders problems to practice: search.php?search_id=tag&tag_id=199

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.
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AccipiterQ

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)
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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.
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shopaholic
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

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
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Classic remainders deal:

­
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Bunuel
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): \(y=35n+18\);

\(x-y=(35m+18)-(35n+18)=35(m-n)\) --> thus 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.
Hi Bunuel,
why we can know x=35m+18 according x=5q+3 & x=7p+4?
[ltr]Is the conclusion drawn by exhaustive method?[/ltr]
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fayey
Bunuel
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): \(y=35n+18\)
;

\(x-y=(35m+18)-(35n+18)=35(m-n)\) --> thus 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.
Hi Bunuel,
why we can know x=35m+18 according x=5q+3 & x=7p+4?
[ltr]Is the conclusion drawn by exhaustive method?[/ltr]

Please check the highlighted parts.

Hope it helps.
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Bunuel
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): \(y=35n+18\)
;

\(x-y=(35m+18)-(35n+18)=35(m-n)\) --> thus 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.
Hi Bunuel,
why we can know x=35m+18 according x=5q+3 & x=7p+4?
[ltr]Is the conclusion drawn by exhaustive method?[/ltr]

Please check the highlighted parts.

Hope it helps.
Thank you!
I just confused about the common remainder. So anyway we have to list them separately and find the first common integer in the above two patterns to be the new remainder.
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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


Here's my solution:

Concept: Whenever you are subtracting two numbers, you are completely subtracting the remainders..... let's call this (I)
Let me explain: Consider X = 21 & Y = 11 -------- (X mod 5) = 1 || (Y mod 5) = 1 - hence the remainder is 1 for both

Now, from (I),
Subtracting both X and Y should remove the remainder.

X-Y = 21 - 11 = 10 ----> (10 mod 5) = 0, hence (I) stands.

Let's project this concept into our numerical,

When positive integer x is divided by 5, the remainder is 3; and when x is divided by 7, the remainder is 4 -
What if there were no remainders, X would have been a multiple of 5 and 7, hence -----> a multiple of 5 *7 = 35

When positive integer y is divided by 5, the remainder is 3; and when y is divided by 7, the remainder is 4.
What if there were no remainders, Y would have been a multiple of 5 and 7, hence -----> a multiple of 5 *7 = 35

If x > y, which of the following must be a factor of x - y?
Now since the question itself is asking you to subtract both the numbers (X-Y) remainders will Nil out, therefore the X-Y has to be nothing but a multiple of 35.

Hence the Answer is (E)
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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
There are two methods to solving this question but the simplest one is here

Let, Positive Integer = N

When positive integer x is divided by 5, the remainder is 3
i.e. x = 5a+3
i.e. x may be {3, 8, 13, 18, 23, 28, 33, 38, 43, 48, 53, 58,...}


when x is divided by 7, the remainder is 4
i.e. x = 7b+4
i.e. x may be {4, 11, 18, 25, 32, 39, 46, 53, 60, 67,...}

Looking at common numbers we see that
x may be {18, 53, 88...}
[Notice: It's an arithmetic progression with a common difference of 35 which is LCM of 5 and 7 (the divisors)]


Positive integer y is divided by 5, the remainder is 3;
i.e. y = 5c+3
i.e. y may be {3, 8, 13, 18, 23, 28, 33, 38, 43, 48, 53, 58,...}

and when y is divided by 7, the remainder is 4
i.e. x = 7b+4
i.e. x may be {4, 11, 18, 25, 32, 39, 46, 53, 60, 67,...}

Looking at common numbers we see that
y may be {18, 53, 88...}
[Notice: It's an arithmetic progression with a common difference of 35 which is LCM of 5 and 7 (the divisors)]

we know that x > y

so, we can choose x to be 53 when y is 18
so x-y = 53-18 = 35
Answer: Option E

To check more videos on Number properties check the kink Below (200 Videos)
Number Properties Videos
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x = 5a+3, 3,8,13,18,23,28,33, ....
x = 7b+4, 4,11,18,25,32,....

Thinking of generic formula, x = 35k+18

y follows the same generic formula, y = 35p+18

x - y = 35(k-p)

Answer: 35
shopaholic
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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Let me help you see the elegant pattern behind this remainder question.

When I first encounter students struggling with this type of problem, they often try to find the exact values of x and y using the Chinese Remainder Theorem. That's unnecessarily complex for the GMAT. Let me show you the streamlined approach.

Key Insight: Both x and y have identical remainder patterns:
- Remainder 3 when divided by 5
- Remainder 4 when divided by 7

Step 1: Express the conditions algebraically
For any number with these remainder properties, we can write:
\(x = 5k + 3\) for some integer k
\(x = 7m + 4\) for some integer m

Similarly for y:
\(y = 5j + 3\) for some integer j
\(y = 7n + 4\) for some integer n

Step 2: Apply the difference property
Here's the crucial pattern - when two numbers have identical remainders, their difference eliminates those remainders:

\(x - y = (5k + 3) - (5j + 3) = 5(k - j)\)
\(x - y = (7m + 4) - (7n + 4) = 7(m - n)\)

This tells us that \(x - y\) is divisible by both 5 and 7.

Step 3: Find what MUST divide x - y
Since 5 and 7 are coprime (share no common factors), and \(x - y\) is divisible by both, \(x - y\) must be divisible by their product: \(5*7 = 35\)

Want to master this remainder pattern that appears in 3-4 GMAT questions yearly? The complete solution walks through the coprime divisor principle and shows you how to spot this pattern instantly.

The framework there also covers what happens when divisors aren't coprime - a common trap the GMAT loves to test.
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