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.

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.

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.

Kudos?

How do we solve a question which has different remainder ? because in this case remainder is -2 , ie common to both the cases? What happens to the case when both the case have different remainder ?

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

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.

Thanks for the amazing concept. I just have one more query:

How will we apply this approach for questions such as:

What is the remainder when the positive integer n is divided by 12?

1. When n is divided by 6, the remainder is 1.

2. When n is divided by 12, the remainder is greater than 5.

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.

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

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.

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

Is there a quick way to come with this value without listing the nos out. sometime the common nos in both the series would come after 6 or 7 term hence sometimes consumes time. Is there a quicker way?