Hi KSHITIJ205,

You could approach it in two ways:

Method 1:
List numbers that leave a remainder of 4 after division by 6: 4, 10, 16, 22, 28, 34, ....
List numbers that leave a remainder of 3 after division by 5: 3, 8, 13, 18, 23, 28, 33, ....

Notice 28 is the first number that satisfies both of the conditions, and 28 divided by 30 leaves a remainder of 28. I know the problem says n has to be greater than 30, but it is not necessary. You can go on and you will find the next common number is 58.

Method 2: Use the remainder algorithm.
n = 6q + 4
n = 5p + 3

we are looking for integer values of p and q(quotients) that satisfy the two conditions. Equate them and we obtain 6q+4 = 5p +3
Simplify to 5p = 6q + 1

Rewrite it as:
p = (6q+1)/5 = (5q+q+1)/5 = q + (q+1)/5

This is a standard technique to solve linear equations where variables are constrained to be integers.
The least value of q that makes p an integer is q=4(Why?) and the next one is q=9.

If q = 9, then n = 6(9)+4=58 and 58 divided by 30 leaves a remainder of 28.

Cheers,
Dabral

KSHITIJ205 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

We can PLUG IN THE ANSWERS, which represent the REMAINDER when n is divided by 30.
Implication:
n = (multiple of 30) + (correct remainder).

Thus, if we add 30 to the answer choices, the results will be possible values for n:
3+30 = 33.
12+30 = 42.
18+30 = 48.
22+30 = 52.
28+30 = 58.

Of the resulting options for n, only 52 and 58 yield a remainder of 4 when divided by 6.
Eliminate A, B and C.
Between 52 and 58, only 58 yields a remainder of 3 when divided by 5.
Eliminate D.

The correct answer is E.

search for numbers which leaves remainder 4 when n is divided by 6
search for numbers which leaves remainder 3 when n is divided by 5 - these numbers could end with 3 or 8.

28 , 58 fits it criteria.

since n is > 30 so 58 is answer.

With remainder questions, first try direct substitution, find the first overlapping number, and then add the least common multiple (LCM) of the dividing numbers.

In other words,

From n=6p+4=4,10,16,22,28... and n=5p+3=3,8,13,18,23,28...
the first overlapping number is 28, and the least common multiple of the two dividing numbers 6 and 5 is 30, so just add this to make a sequence like below:

n=28, 58, 88, 118,...

Dividing any of the number above by 30 will give 28=30*0+28, 58=30*1+28, 88=30*2+28.., meaning they will all have the remainder of 28, so the answer is E. 28

As you can see, remainder questions are solved by 1) direct substitution, 2) finding the first overlapping number and adding the LCM of the dividing numbers.

GMATGuruNY wrote:

KSHITIJ205 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

We can PLUG IN THE ANSWERS, which represent the REMAINDER when n is divided by 30.
Implication:
n = (multiple of 30) + (correct remainder).

Thus, if we add 30 to the answer choices, the results will be possible values for n:
3+30 = 33.
12+30 = 42.
18+30 = 48.
22+30 = 52.
28+30 = 58.

Of the resulting options for n, only 52 and 58 yield a remainder of 4 when divided by 6.
Eliminate A, B and C.
Between 52 and 58, only 58 yields a remainder of 3 when divided by 5.
Eliminate D.

The correct answer is E.

Great way of Elimination.

We know that to have a remainder of 3 after being divided by 5, the number must end in a 3 or 8.... As we get an even number when dividing by 6, we know that the number must end in an 8. Let's look at a few numbers that end in 8 and if they are 4 above a multiple of 6:


8 -- no
18 -- no
28 -- yes


When 28 is divided by 30, the remainder is 28...

E