S96-03

Question

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

Bunuel · Answer

Official Solution:

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

The simplest way to approach this problem is to work backwards from the answer choices. Let's construct a possible value of  n  for each choice, and then test those values against the given constraints.
 
Since we are asked for the remainder after division by 30, the easiest possible value of  n  for each choice is 30 more than the choice.
 
(A)  3 + 30  gives us  n = 33 
 
(B)  12 + 30  gives us  n = 42 
 
(C)  18 + 30  gives us  n = 48 
 
(D)  22 + 30  gives us  n = 52 
 
(E)  28 + 30  gives us  n = 58 
 
Now test those values of  n  against the constraints.
 
(A)  n = 33  divided by 6 gives remainder 3 - FAIL
 
(B)  n = 42  divided by 6 gives remainder 0 - FAIL
 
(C)  n = 48  divided by 6 gives remainder 0 - FAIL
 
(D)  n = 52  divided by 6 gives remainder 4 - PASS
 
(E)  n = 58  divided by 6 gives remainder 4 - PASS
 
We can now just test the surviving choices for how they behave upon division by 5. To leave remainder 3 after division by 5, a number must end in either 3 or 5:
 
(D)  n = 52  divided by 5 gives remainder 2 - FAIL
 
(E)  n = 58  divided by 5 gives remainder 3 - PASS
 
Another way to approach this problem is to translate the given language of remainders into the language of multiples. If  n  leaves a remainder of 4 after division by 6, then  n  is 4 more than a multiple of 6. Leaving aside the size requirement for a moment, we can see that  n  could be 4, 10, 16, 22, 28, 34, etc.
 
Likewise, if  n  leaves a remainder of 3 after division by 5, then  n  is 3 more than a multiple of 5. Again leaving aside the size requirement, we can see that  n  could be 3, 8, 13, 18, 23, 28, 33, etc. As we noted earlier,  n  must end in 3 or 8.
 
We might now spot 28 on both lists. Although  n  is not actually allowed to be 28 (because  n  must be larger than 30), we might try adding 30 to it to get 58. Since 30 is a multiple of 6, adding 30 to 28 won't change the fact that after division by 6, we'll get 4 as the remainder. The same idea holds true for 5: since 30 is a multiple of 5, adding 30 to 28 won't change the fact that after division by 5, we'll get 3 as the remainder. This way, we have constructed a possible  n  without using the answer choices.
 
Finally, the remainder after dividing 58 by 30 is 28.

Answer: E

lucky star · Answer

why is the remainder 28? 58/30 is not 28. I don't really get it.

Bunuel · Answer

The remainder upon division 58 by 30 IS 28: 58 = 1*30 + 28.

Check the links on remainders below:

Divisibility and Remainders on the GMAT

Theory on remainders problems

Tips on remainders

Cyclicity on the GMAT

Units digits, exponents, remainders problems

Hope it helps.

moulayabdesslam · Answer

Hi,

I had a very similar question in the real GMAT few weeks ago and was able to solve the question. In fact, in this type of problems, we can deduct the same number - here 2 - from the multiples given in the question. Here we can deduct 2 from the multiples of 6 and from the multiples of 5. Hence I deduced that the answer was be divisible by a multiple of 6*5 minus 2.
However, is there any algebraic way for solving this problem in case we can't deduct the same number from each?

--
Need Kudos

RohitLohani001 · Answer

Lucky Star you need to study a lot.

rsd1991 · Answer

we can use a formula to solve this...

if we check the question, the difference of the remainder and the divisor is constant in both cases, that is 2.

so n=k*lcm(divisor1, divisor2)- common difference

in this case n=k*lcm(6,5)-2
n=30k-2

now the condition on n is that it should be greater than 30, so put k=2 . this gives n=58...
we can then find the remainder to be 28...

derhamabbelah · Answer

Positive integer n leaves a remainder of 4 after division by 6, therefore n=6p+4 where p is a positive integer
Positive integer n leaves a remainder of 3 after division by 5, hence n=5k+3 where k is a positive integer
Then n=6p+4=5k+3, or 6p+1=5k. When 6p=54, we have 5k=55. Therefore, n=58 and the remainder is 28
 Answer: E

akadiyan · Answer

Remainder of 4 when divided by 6 = 34,40,46,52,58,64,70
Remainder of 3 when divided by 5 = we know that the number has to be either ends in 3 or 8 , so we have 33,38,43,48,53,58,63,68,73,78,83,88

We need to use the answer choices to come to conclusion quickly

A. 3 --> 30+3 = 33 - Not in the list
B. 12 --> 30+12=42 - Not in the list
C. 18 --> 30+18=48 - Not in the first , but in second list -- Not correct
D. 22 --> 30+22=52 - In the first list but not in second -- Not correct
E. 28 --> 30+28=58 -- 58 is both the list - Correct

Ans:  E

Censach · Answer

Remainder = 5 q + 3 = 6 k + 4
5 q + 3 last digit can be either 8 or 5 -> 18 or 28
Only 28 respects those criteria

sividela · Answer

Hi, another aproach.

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

1) n = 6*k + 4
2) n = 5*p + 3

k, p postives intergers.

1) /*5 --> 5n = 30*k + 20
2) /*6 --> 6n = 30*p + 18

2)-1) --> n = 30* (p - k) - 2

Then you add and substract 30 in the ecuation

n = 30* (p-k-1) -2 +30
n = 30* (p-k-1) +28

hence, p, k, 1 are integers, when n is divided by 20 leaves a reminder of 28.

Hope that helps.

bumpbot · Answer

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S96-03

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