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# S96-03

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Math Expert
Joined: 02 Sep 2009
Posts: 51301

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16 Sep 2014, 00:50
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Difficulty:

35% (medium)

Question Stats:

70% (01:43) correct 30% (01:32) wrong based on 126 sessions

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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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Joined: 02 Sep 2009
Posts: 51301

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16 Sep 2014, 00:50
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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.

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Joined: 20 Sep 2016
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21 Sep 2016, 02:36
why is the remainder 28? 58/30 is not 28. I don't really get it.
Math Expert
Joined: 02 Sep 2009
Posts: 51301

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21 Sep 2016, 02:44
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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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Status: Pursuit of Happiness
Joined: 10 Sep 2016
Posts: 30
Location: United States (IL)
Concentration: Finance, Economics
Schools: HBS '19, CBS '19
GMAT 1: 590 Q44 V27
GMAT 2: 690 Q50 V34
GPA: 3.94

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17 Oct 2016, 17:36
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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Joined: 01 Jan 2017
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20 Mar 2017, 00:48
Mispo
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Joined: 11 Jun 2017
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16 Aug 2017, 11:47
Lucky Star you need to study a lot.
Intern
Joined: 15 Apr 2018
Posts: 3

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18 Jul 2018, 09:34
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...
Re: S96-03 &nbs [#permalink] 18 Jul 2018, 09:34
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# S96-03

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