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Re: S96-03 [#permalink]
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lucky star wrote:
why is the remainder 28? 58/30 is not 28. I don't really get it.


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.
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Re: S96-03 [#permalink]
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?

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Re: S96-03 [#permalink]
Lucky Star you need to study a lot.
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Re: S96-03 [#permalink]
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...
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Re: S96-03 [#permalink]
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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
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Re: S96-03 [#permalink]
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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
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Re: S96-03 [#permalink]
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
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Re: S96-03 [#permalink]
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.
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Re: S96-03 [#permalink]
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