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Positive integer n leaves a remainder of 4 after division by

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Positive integer n leaves a remainder of 4 after division by [#permalink] New post   Updated on: 20 Apr 2012, 03:05
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Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5. If n is greater than 30, what is the remainder that n leaves after division by 30?

A. 3
B. 12
C. 18
D. 22
E. 28
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E

Originally posted by bchekuri on 05 May 2010, 23:49.
Last edited by Bunuel on 20 Apr 2012, 03:05, edited 1 time in total.
Edited the question and added the OA
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Re: Manhattan Remainder Problem [#permalink] New post   06 May 2010, 04:18
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To elaborate more.

Suppose we are told that:
Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 2 after division by 8. What is the remainder that n leaves after division by 12?

The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: \(n=6p+4\). Thus according to this particular statement \(n\) could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ...

The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: \(n=8q+2\). Thus according to this particular statement \(n\) could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ...

The above two statements are both true, which means that the only valid values of \(n\) are the values which are common in both patterns. For example \(n\) can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer \(q\).

So we should derive general formula (based on both statements) that will give us only valid values of \(n\).

How can these two statement be expressed in one formula of a type \(n=kx+r\)? Where \(x\) is divisor and \(r\) is a remainder.

Divisor \(x\) would be the least common multiple of above two divisors 6 and 8, hence \(x=24\).

Remainder \(r\) would be the first common integer in above two patterns, hence \(r=10\).

Therefore general formula based on both statements is \(n=24k+10\). Thus according to this general formula valid values of \(n\) are: 10, 34, 58, ...

Now, \(n\) divided by 12 will give us the reminder of 10 (as 24k is divisible by 12).

Hope it helps.
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Re: Manhattan Remainder Problem [#permalink] New post   29 Nov 2010, 04:40
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Friends,

IT IS VERY COMMON IN GMAT to solve this kind of qtns using "NEGATIVE REMAINDER" theory.

The theory says:

if a # x is devided by y and leave the positive # r as the remainder then it can also leave negative # (r-y) as the remainder.

e.g:

9 when devided by 5 leves the remainder 4 : 9=5*1+4
it can also leave the remainder 4-5 = -1 : 9=5*2 -1


back to the original qtn:
n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5
==> n leaves a remainder of -2 (i.e. 4-6) after division by 6 and a remainder of -2 (i.e. 3-5) after division by 5
==> n when devided by 5 and 6 leaves the same remainder -2.
what is n?
LCM (5,6)-2 = 30-2 = 28
CHECK: 28 when devided by 6 leaves the remainder 4 and when devided by 5 leaves the remainder 3

However, the qtn says n > 30

so what is the nex #, > 28, that can give the said remainders when devided by 6 and 5
nothing but 28 + (some multiple of 6 and 5) as this "some multiple of 6 and 5" will not give any remainder when devided by 5 or 6 but 28 will give the required remainders.

hence n could be anything that is in the form 28 + (some multiple of 6 and 5)
observe that "some multiple of 6 and 5" is always a multiple of 30 as LCM (5,6) = 30.

hence when n (i.e. 28 + some multiple of 6 and 5) is devided by 30 gives the remainder 28.

ANSWER "E"

Regards,
Murali.

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Re: Manhattan Remainder Problem [#permalink] New post   06 May 2010, 00:22
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bchekuri wrote:
Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5. If n is greater than 30, what is the remainder that n leaves after division by 30?
(A) 3
(B) 12
(C) 18
(D) 22
(E) 28

How to approach this Problem?


Positive integer n leaves a remainder of 4 after division by 6 --> \(n=6p+4\) --> 4, 10, 16, 22, 28, ...
Positive integer n leaves a remainder of 3 after division by 5 --> \(n=5q+3\) --> 3, 8, 13, 18, 23, 28, ...

\(n=30k+28\) - we have 30 as lcm of 5 and 6 is 30 and we have 28 as the first common integer in the above patterns is 28.

Hence remainder when positive integer n is divided by 30 is 28.

Answer: E.

P.S. n>30 is a redundant information.
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Re: Manhattan Remainder Problem [#permalink] New post   06 May 2010, 12:09
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bchekuri wrote:
Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5. If n is greater than 30, what is the remainder that n leaves after division by 30?
(A) 3
(B) 12
(C) 18
(D) 22
(E) 28

How to approach this Problem?


Once we get the concept, We can also use the options to answer faster.

Since the number leaves the remainder 4 after division by 6 we know it is an even number. Only even number that leaves a remainder of 3 after being divisible by 5 has to end with 8.

So narrows the options to 18 and 28.

since 30+18 = 48 is divisible by 6 the answer is 30+28 = 54
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Positive integer n leaves a remainder of 4 after division by [#permalink] New post   Updated on: 08 Oct 2022, 22:02
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kuttingchai wrote:

Hello Murali,

Can you please explain how will you solve following example using "NEGATIVE REMAINDER" theory discussed above?
"Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 2 after division by 8. What is the remainder that n leaves after division by 12?"


Question: Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 2 after division by 8. What is the remainder that n leaves after division by 12?

Solution:
n = 6a + 4 (or remainder = -2)
n = 8b + 2 (or remainder = -6)

The negative remainder is not the same in the two cases. So don't use negative remainders here.

Check out posts on this concept on the blog link given in my signature below.

Originally posted by KarishmaB on 11 Jun 2012, 21:15.
Last edited by KarishmaB on 08 Oct 2022, 22:02, edited 1 time in total.
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Re: Manhattan Remainder Problem [#permalink] New post   28 Nov 2010, 23:29
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Possible values of n are:

n = 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, ...

If R = 3 after division of 5, then possible values of n are

n = 3, 8, 13, 18, 23, 28, 33, 38, 43, 48, 53, 58, ...

Therefore according to give conditions, some possible values of n are

n = 28, 58, ...

n > 30

Let's say n = 58

58/30 gives us a remainder of : 28.
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Re: Manhattan Remainder Problem [#permalink] New post   10 Jun 2012, 12:21
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muralimba wrote:
Friends,

IT IS VERY COMMON IN GMAT to solve this kind of qtns using "NEGATIVE REMAINDER" theory.

The theory says:

if a # x is devided by y and leave the positive # r as the remainder then it can also leave negative # (r-y) as the remainder.

e.g:

9 when devided by 5 leves the remainder 4 : 9=5*1+4
it can also leave the remainder 4-5 = -1 : 9=5*2 -1


back to the original qtn:
n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5
==> n leaves a remainder of -2 (i.e. 4-6) after division by 6 and a remainder of -2 (i.e. 3-5) after division by 5
==> n when devided by 5 and 6 leaves the same remainder -2.
what is n?
LCM (5,6)-2 = 30-2 = 28
CHECK: 28 when devided by 6 leaves the remainder 4 and when devided by 5 leaves the remainder 3

However, the qtn says n > 30

so what is the nex #, > 28, that can give the said remainders when devided by 6 and 5
nothing but 28 + (some multiple of 6 and 5) as this "some multiple of 6 and 5" will not give any remainder when devided by 5 or 6 but 28 will give the required remainders.

hence n could be anything that is in the form 28 + (some multiple of 6 and 5)
observe that "some multiple of 6 and 5" is always a multiple of 30 as LCM (5,6) = 30.

hence when n (i.e. 28 + some multiple of 6 and 5) is devided by 30 gives the remainder 28.

ANSWER "E"

Regards,
Murali.

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Hello Murali,

Can you please explain how will you solve following example using "NEGATIVE REMAINDER" theory discussed above?
"Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 2 after division by 8. What is the remainder that n leaves after division by 12?"
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Re: Manhattan Remainder Problem [#permalink] New post   03 Nov 2012, 01:46
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Bunuel wrote:
To elaborate more.

Suppose we are told that:
Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 2 after division by 8. What is the remainder that n leaves after division by 12?

The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: \(n=6p+4\). Thus according to this particular statement \(n\) could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ...

The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: \(n=8q+2\). Thus according to this particular statement \(n\) could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ...

The above two statements are both true, which means that the only valid values of \(n\) are the values which are common in both patterns. For example \(n\) can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer \(q\).

So we should derive general formula (based on both statements) that will give us only valid values of \(n\).

How can these two statement be expressed in one formula of a type \(n=kx+r\)? Where \(x\) is divisor and \(r\) is a remainder.

Divisor \(x\) would be the least common multiple of above two divisors 6 and 8, hence \(x=24\).

Remainder \(r\) would be the first common integer in above two patterns, hence \(r=10\).

Therefore general formula based on both statements is \(n=24k+10\). Thus according to this general formula valid values of \(n\) are: 10, 34, 58, ...

Now, \(n\) divided by 12 will give us the reminder of 10 (as 24k is divisible by 12).

Hope it helps.



Sorry , i didn't understand the last statement written marked in red... instead of 12 if we have other values like 14 or 20 then how remainder would vary.. just lil but curious to understand the gravity
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Re: Manhattan Remainder Problem [#permalink] New post   03 Nov 2012, 01:50
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breakit wrote:
Bunuel wrote:
To elaborate more.

Suppose we are told that:
Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 2 after division by 8. What is the remainder that n leaves after division by 12?

The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: \(n=6p+4\). Thus according to this particular statement \(n\) could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ...

The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: \(n=8q+2\). Thus according to this particular statement \(n\) could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ...

The above two statements are both true, which means that the only valid values of \(n\) are the values which are common in both patterns. For example \(n\) can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer \(q\).

So we should derive general formula (based on both statements) that will give us only valid values of \(n\).

How can these two statement be expressed in one formula of a type \(n=kx+r\)? Where \(x\) is divisor and \(r\) is a remainder.

Divisor \(x\) would be the least common multiple of above two divisors 6 and 8, hence \(x=24\).

Remainder \(r\) would be the first common integer in above two patterns, hence \(r=10\).

Therefore general formula based on both statements is \(n=24k+10\). Thus according to this general formula valid values of \(n\) are: 10, 34, 58, ...

Now, \(n\) divided by 12 will give us the reminder of 10 (as 24k is divisible by 12).

Hope it helps.



Sorry , i didn't understand the last statement written marked in red... instead of 12 if we have other values like 14 or 20 then how remainder would vary.. just lil but curious to understand the gravity


\(n=24k+10=12(2k)+10\) --> \(n\) can be: 10, 34, 58, ... \(n\) divided by 12 will give us the reminder of 10.

As, you can see, n divided by 14 can give different remainders. If n=10, then n divided by 14 yields the remainder of 10 but if n=34, then n divided by 14 yields the remainder of 6.
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink] New post   13 Apr 2014, 22:37
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Maybe I did something wrong, or I just got lucky, so please inform me. I also had 28 as the solution, by just plugging in the values.

a. 3 / 6 = 0 r 6 INCORRECT
b. 12 / 6 = 2 r 0 INCORRECT
c. 18 / 6 = 3 r 0 INCORRECT
d. 22 / 6 = 3 r 4 & 22 / 5 = 4 r 2 INCORRECT
e. 28 / 6 = 4 r 4 & 28 / 5 = 4 r 3 CORRECT

Super-simple math, so there must be something wrong here, haha. :-D
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink] New post   14 Apr 2014, 00:08
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Saabs wrote:
Maybe I did something wrong, or I just got lucky, so please inform me. I also had 28 as the solution, by just plugging in the values.

a. 3 / 6 = 0 r 6 INCORRECT
b. 12 / 6 = 2 r 0 INCORRECT
c. 18 / 6 = 3 r 0 INCORRECT
d. 22 / 6 = 3 r 4 & 22 / 5 = 4 r 2 INCORRECT
e. 28 / 6 = 4 r 4 & 28 / 5 = 4 r 3 CORRECT

Super-simple math, so there must be something wrong here, haha. :-D


What you did is correct. The reason you did it will tell you whether you got lucky or used good reasoning.

The question tells you that "n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5"

The options are not the values of n; they are the values of remainder that is leftover after you divide n by 30. You divided the options by 6 and 5 and got the remainder as 4 and 3 respectively. Was that a mistake you made? If yes, then you got lucky.

Though, if you had used logic and said, "Ok, so when n is divided by 30, groups of 30 are made. What is leftover is given in the options. 30 is completely divisible by 6 and by 5 hence the groups of 30 can be evenly divided into groups of 6 as well as groups of 5. So whatever is leftover after division by 30, we need to split that into further groups of 6 and 5. When we split it into groups of 6, we must have remainder as 4 since n leaves remainder 4. When we split it into groups of 5, we must have remainder as 3 since n leaves remainder 3. And, if that is the reason you divided the options by 6 and 5, checked their remainders and got your answer, then well done! You got the logic!"
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Re: Manhattan Remainder Problem [#permalink] New post   18 May 2014, 12:55
Bunuel wrote:
To elaborate more.

Suppose we are told that:
Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 2 after division by 8. What is the remainder that n leaves after division by 12?

The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: \(n=6p+4\). Thus according to this particular statement \(n\) could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ...

The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: \(n=8q+2\). Thus according to this particular statement \(n\) could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ...

The above two statements are both true, which means that the only valid values of \(n\) are the values which are common in both patterns. For example \(n\) can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer \(q\).

So we should derive general formula (based on both statements) that will give us only valid values of \(n\).

How can these two statement be expressed in one formula of a type \(n=kx+r\)? Where \(x\) is divisor and \(r\) is a remainder.

Divisor \(x\) would be the least common multiple of above two divisors 6 and 8, hence \(x=24\).

Remainder \(r\) would be the first common integer in above two patterns, hence \(r=10\).

Therefore general formula based on both statements is \(n=24k+10\). Thus according to this general formula valid values of \(n\) are: 10, 34, 58, ...

Now, \(n\) divided by 12 will give us the reminder of 10 (as 24k is divisible by 12).
Hope it helps.


Hi Bunuel,

All of this makes sense but I would like to challenge the last statement highlighted above.

In this case, we can obviously write n=12(2k)+10 as you highlighted in the later posts. If for some reason, let's say it asked for a number that wasn't divisible by 24, wouldn't that make the equation n=24k+10 invalid? Meaning, it asked what is the remainder that n leaves after division by 11?

Additionally, what if it asked "what is the remainder that n leaves after division by 48. Could we still apply the same logic and say 10?

Thanks!
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Re: Manhattan Remainder Problem [#permalink] New post   19 May 2014, 08:00
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russ9 wrote:
Bunuel wrote:
To elaborate more.

Suppose we are told that:
Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 2 after division by 8. What is the remainder that n leaves after division by 12?

The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: \(n=6p+4\). Thus according to this particular statement \(n\) could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ...

The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: \(n=8q+2\). Thus according to this particular statement \(n\) could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ...

The above two statements are both true, which means that the only valid values of \(n\) are the values which are common in both patterns. For example \(n\) can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer \(q\).

So we should derive general formula (based on both statements) that will give us only valid values of \(n\).

How can these two statement be expressed in one formula of a type \(n=kx+r\)? Where \(x\) is divisor and \(r\) is a remainder.

Divisor \(x\) would be the least common multiple of above two divisors 6 and 8, hence \(x=24\).

Remainder \(r\) would be the first common integer in above two patterns, hence \(r=10\).

Therefore general formula based on both statements is \(n=24k+10\). Thus according to this general formula valid values of \(n\) are: 10, 34, 58, ...

Now, \(n\) divided by 12 will give us the reminder of 10 (as 24k is divisible by 12).
Hope it helps.


Hi Bunuel,

All of this makes sense but I would like to challenge the last statement highlighted above.

In this case, we can obviously write n=12(2k)+10 as you highlighted in the later posts. If for some reason, let's say it asked for a number that wasn't divisible by 24, wouldn't that make the equation n=24k+10 invalid? Meaning, it asked what is the remainder that n leaves after division by 11?

Additionally, what if it asked "what is the remainder that n leaves after division by 48. Could we still apply the same logic and say 10?

Thanks!


If the question were what is the remainder when n is divided by 11, then the answer would be "cannot be determined". The same if the question asked about the remainder when n is divided by 48. See, according to this general formula valid values of \(n\) are: 10, 34, 58, ... These values give different remainders upon division by 11, or 48.
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink] New post   31 May 2014, 14:12
Hi,
Probably a silly question....
But can some one explain why Remainder r would be the first common integer in the two patterns
n=6p+4
n=5q+3

Many Thanks
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Positive integer n leaves a remainder of 4 after division by [#permalink] New post   Updated on: 17 Oct 2022, 00:46
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bhatiavai wrote:
Hi,
Probably a silly question....
But can some one explain why Remainder r would be the first common integer in the two patterns
n=6p+4
n=5q+3

Many Thanks


N is of the form 6p+4 which means it is one of 4, 10, 16, 22, 28, ...
N is also of the form 5q+3 which means it must also be one of 3, 8, 13, 18, 23, 28, ...

Since N must be a value in both the lists, N can take the values common to both. The first such value is 28.
So N can be 28.

Now what other values can N take? 28 is a number that will leave a remainder of 4 when divided by 6 and a remainder of 3 when divided by 5. The next such number will be (LCM of 6 and 5) + 28. Why? because whatever you add to 28, that should be divisible by 6 as well as 5. Then whatever you add will have no relevance to the remainder and the remainders will remain the same.
That is why some other values of N will be 30+28, 60+28 ...etc.

After you divide any of these numbers by 30, the remainder will be 28.

Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5. If n is greater than 30, what is the remainder that n leaves after division by 30?

Originally posted by KarishmaB on 01 Jun 2014, 20:56.
Last edited by KarishmaB on 17 Oct 2022, 00:46, edited 1 time in total.
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink] New post   01 Sep 2014, 00:28
Bunuel wrote:
bchekuri wrote:
Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5. If n is greater than 30, what is the remainder that n leaves after division by 30?
(A) 3
(B) 12
(C) 18
(D) 22
(E) 28

How to approach this Problem?


Positive integer n leaves a remainder of 4 after division by 6 --> \(n=6p+4\) --> 4, 10, 16, 22, 28, ...
Positive integer n leaves a remainder of 3 after division by 5 --> \(n=5q+3\) --> 3, 8, 13, 18, 23, 28, ...

\(n=30k+28\) - we have 30 as lcm of 5 and 6 is 30 and we have 28 as the first common integer in the above patterns is 28.

Hence remainder when positive integer n is divided by 30 is 28.

Answer: E.

P.S. n>30 is a redundant information.



What will be the answer of they ask us what the remainder is if n is divided by something other than 30, lets say 11 or 12. any random number which is not a factor of 30.
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink] New post   01 Sep 2014, 00:41
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rohansangari wrote:
Bunuel wrote:
bchekuri wrote:
Positive integer n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5. If n is greater than 30, what is the remainder that n leaves after division by 30?
(A) 3
(B) 12
(C) 18
(D) 22
(E) 28

How to approach this Problem?


Positive integer n leaves a remainder of 4 after division by 6 --> \(n=6p+4\) --> 4, 10, 16, 22, 28, ...
Positive integer n leaves a remainder of 3 after division by 5 --> \(n=5q+3\) --> 3, 8, 13, 18, 23, 28, ...

\(n=30k+28\) - we have 30 as lcm of 5 and 6 is 30 and we have 28 as the first common integer in the above patterns is 28.

Hence remainder when positive integer n is divided by 30 is 28.

Answer: E.

P.S. n>30 is a redundant information.



What will be the answer of they ask us what the remainder is if n is divided by something other than 30, lets say 11 or 12. any random number which is not a factor of 30.


In this case we wouldn't be able to determine the remainder. For example, if we were asked to find the remainder of n divided by 11, then we would get different answers: if n = 28, then remainder would be 6 but if n = 58, then the remainder would be 3.
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink] New post   05 Feb 2015, 08:02
VeritasPrepKarishma wrote:
Saabs wrote:
Maybe I did something wrong, or I just got lucky, so please inform me. I also had 28 as the solution, by just plugging in the values.

a. 3 / 6 = 0 r 6 INCORRECT
b. 12 / 6 = 2 r 0 INCORRECT
c. 18 / 6 = 3 r 0 INCORRECT
d. 22 / 6 = 3 r 4 & 22 / 5 = 4 r 2 INCORRECT
e. 28 / 6 = 4 r 4 & 28 / 5 = 4 r 3 CORRECT

Super-simple math, so there must be something wrong here, haha. :-D


What you did is correct. The reason you did it will tell you whether you got lucky or used good reasoning.

The question tells you that "n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5"

The options are not the values of n; they are the values of remainder that is leftover after you divide n by 30. You divided the options by 6 and 5 and got the remainder as 4 and 3 respectively. Was that a mistake you made? If yes, then you got lucky.

Though, if you had used logic and said, "Ok, so when n is divided by 30, groups of 30 are made. What is leftover is given in the options. 30 is completely divisible by 6 and by 5 hence the groups of 30 can be evenly divided into groups of 6 as well as groups of 5. So whatever is leftover after division by 30, we need to split that into further groups of 6 and 5. When we split it into groups of 6, we must have remainder as 4 since n leaves remainder 4. When we split it into groups of 5, we must have remainder as 3 since n leaves remainder 3. And, if that is the reason you divided the options by 6 and 5, checked their remainders and got your answer, then well done! You got the logic!"


Hi Karishma, could you kindly explain the bold part? I have trouble understanding why it was the case..
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink] New post   05 Feb 2015, 20:56
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tieurongthieng wrote:
Hi Karishma, could you kindly explain the bold part? I have trouble understanding why it was the case..


You might want to check out this post first on divisibility and remainders:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/04 ... unraveled/

This tells you why when you divide a number by 30 and get, say 4 as remainder, you will get 4 as remainder when you divide the same number by 6 or 5 or 10 or 15 (factors of 30 greater than 4).

After that, you should check out this post which discusses this particular question in detail:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/05 ... -the-gmat/

I think these two together will help you get a very clear picture.
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
05 Feb 2015, 20:56
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