What you did is correct. The reason you did it will tell you whether you got lucky or used good reasoning.

The question tells you that "n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5"

The options are not the values of n; they are the values of remainder that is leftover after you divide n by 30. You divided the options by 6 and 5 and got the remainder as 4 and 3 respectively. Was that a mistake you made? If yes, then you got lucky.

Though, if you had used logic and said, "Ok, so when n is divided by 30, groups of 30 are made. What is leftover is given in the options. 30 is completely divisible by 6 and by 5 hence the groups of 30 can be evenly divided into groups of 6 as well as groups of 5. So whatever is leftover after division by 30, we need to split that into further groups of 6 and 5. When we split it into groups of 6, we must have remainder as 4 since n leaves remainder 4. When we split it into groups of 5, we must have remainder as 3 since n leaves remainder 3. And, if that is the reason you divided the options by 6 and 5, checked their remainders and got your answer, then well done! You got the logic!"
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Bunuel – very good explanation.

First case - i). given statement in a question is - Remainder is 7 when positive integer n is divided by 18. And ii). if we are asked to find out the remainder when n is divided by 6,

Then since 18 is completely divisible by 6 or 6 is a factor of 18, we can find out the solution easily by above given statement. As, n = 18 q + 7; 18 is completely divisible by 6, thus no remainder exists when 18q/6 and when 7 is divided by 6, it would yield 1 as the remainder.

Second case i). given statement in a question is - Remainder is 7 when positive integer n is divided by 18. And ii). if we are asked to find out the remainder when n is divided by 8,

Then since 18 is not divisible by 8 or 8 is not a factor of 18, we cannot find out the solution by above given statement.

Third case i). given statement in a question is - Remainder is 7 when positive integer n is divided by 6. And ii). if we are asked to find out the remainder when n is divided by 12,

Then since 6 is not divisible by 12, however it true the other way round. We cannot find out the solution by above given statement.

Fourth case, which you have explained – “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?”

We have two given statements – since one statement will not yield us the answer as explained in my earlier three cases. But, since we have two given statements which derive – n = 6Q1 + 4 & n = 8Q2 + 2. Q1, Q2 are quotients respectively.
And if we look at 6Q1 and 8 Q2, the LCM yields 24. Since we are asked the remainder when division is done by 12 and since 12 is completely divisible by 24. Thus, we can come up with a solution. Otherwise we cannot.

I hope I was able to explain what I wanted to and the above conclusion described in the 4 cases is correct.

Hi Bunuel,

All of this makes sense but I would like to challenge the last statement highlighted above.

In this case, we can obviously write n=12(2k)+10 as you highlighted in the later posts. If for some reason, let's say it asked for a number that wasn't divisible by 24, wouldn't that make the equation n=24k+10 invalid? Meaning, it asked what is the remainder that n leaves after division by 11?

Additionally, what if it asked "what is the remainder that n leaves after division by 48. Could we still apply the same logic and say 10?

Thanks!

If the question were what is the remainder when n is divided by 11, then the answer would be "cannot be determined". The same if the question asked about the remainder when n is divided by 48. See, according to this general formula valid values of \(n\) are: 10, 34, 58, ... These values give different remainders upon division by 11, or 48.
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Hi,
Probably a silly question....
But can some one explain why Remainder r would be the first common integer in the two patterns
n=6p+4
n=5q+3

Many Thanks

N is of the form 6p+4 which means it is one of 4, 10, 16, 22, 28, ...
N is also of the form 5q+3 which means it must also be one of 3, 8, 13, 18, 23, 28, ...

Since N must be a value in both the lists, N can take the values common to both. The first such value is 28.
So N can be 28.

Now what other values can N take? 28 is a number that will leave a remainder of 4 when divided by 6 and a remainder of 3 when divided by 5. The next such number will be (LCM of 6 and 5) + 28. Why? because whatever you add to 28, that should be divisible by 6 as well as 5. Then whatever you add will have no relevance to the remainder and the remainders will remain the same.
That is why some other values of N will be 30+28, 60+28 ...etc.

After you divide any of these numbers by 30, the remainder will be 28.

For more on divisibility and remainders, check:
http://www.veritasprep.com/blog/2011/04 ... unraveled/
http://www.veritasprep.com/blog/2011/04 ... y-applied/
http://www.veritasprep.com/blog/2011/05 ... emainders/
http://www.veritasprep.com/blog/2011/05 ... s-part-ii/

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?
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Is this a sub-600 level question?

No. It is 650-700 level question. It seems quite do-able only because we have done the concept of remainders in great detail. For a newbie, this problem can be quite challenging.
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What will be the answer of they ask us what the remainder is if n is divided by something other than 30, lets say 11 or 12. any random number which is not a factor of 30.

In this case we wouldn't be able to determine the remainder. For example, if we were asked to find the remainder of n divided by 11, then we would get different answers: if n = 28, then remainder would be 6 but if n = 58, then the remainder would be 3.
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Hi Karishma, could you kindly explain the bold part? I have trouble understanding why it was the case..

You might want to check out this post first on divisibility and remainders:
http://www.veritasprep.com/blog/2011/04 ... unraveled/

This tells you why when you divide a number by 30 and get, say 4 as remainder, you will get 4 as remainder when you divide the same number by 6 or 5 or 10 or 15 (factors of 30 greater than 4).

After that, you should check out this post which discusses this particular question in detail:
http://www.veritasprep.com/blog/2014/05 ... -the-gmat/

I think these two together will help you get a very clear picture.
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Awesome I have to admit I have a little bit of troubles imagining the picture for this problem in the beginning, but I just figured it out now with a bit more logics I were also able to prove the theory with algebra as follow: :D

We have:
x = q. y + r (y is divisor)

y = a.b (two factors of y)

We need to prove:

Modulo[ (r/a) ]= Modulo[ (x/a) ]
<=> Modulo[ (x - q.y)/a] = Modulo[ (x/a)]
<=> Modulo[ x/a - q. a. b/a] = Modulo[ (x/a)]
<=> Modulo [x/a] - Modulo [q.a.b/a] = Modulo[ (x/a)]
<=> Modulo [x/a] - 0 = Modulo[ (x/a)] (as q.a.b is divisible by a)
<=> Modulo [x/a] = Modulo[ (x/a)]
<=> True

Interesting theory, indeed :D

N= 6x + 4 --1
N= 5y + 3 --2
from 1 and 2
6x+1 = 5y
this seems simple now , a multiple of 6 to which when 1 is added it turns into a multiple of 5 ; 24+1=25

x=4, N=28 ; 28%30 = 28 .

Hey this is a great explanation, thank you for the help. My question was WHY would it be that "divisor would be the LCM of two divisors 6 and 8" and WHY would it be that the remainder would be first of the common integers? I am having some trouble making the jump in that logic and would appreciate your help. Thanks again!

To understand the 'why' of divisibility and remainders, check out these posts in this order:

http://www.veritasprep.com/blog/2011/04 ... unraveled/
http://www.veritasprep.com/blog/2011/04 ... y-applied/
http://www.veritasprep.com/blog/2011/05 ... emainders/
http://www.veritasprep.com/blog/2011/05 ... s-part-ii/
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Hi Karishma,

Thanks for the post. I just want to ask, whether we can form a general formula of a number which satisfies 2 equations: divided by "a", remainder b (x=a.n + b) and divided by "c", remainder d (x=c.m + d). Then decide the remainder of x when divided by a*b.

For example, without plugging in number, how to solve the problem: find remainder when x divided by 32 if x = 8n + 3 = 6m + 4?

I find 1 solution but it can only apply for a-b=1 (for eg: x=a.n +b = (a+1)m +d)
Then we have:
(a+1)x = a(a+1)n +(a+1)b
ax = a(a+1)m + ad

Minus 2 equation => x = a(a+1) (n-m) + (a+1)b - ad => remainder when x divided by a(a+1) is (a+1)b-ad.

Did you check out the last link provided above? Here it is again: http://www.veritasprep.com/blog/2011/05 ... s-part-ii/

With a case such as this: x = 8n + 3 = 6m + 4, for the first number of this form, you need to use hit and trial.
Note that this example is incorrect since 8n+3 is an odd number and 6m+4 is an even number. But check the post to see how to apply the method to correct questions.
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Dear All,


Its difficult that on what basis we took "First common integer in two patterns" as a remainder for the formula?


Can someone please explain,

Thanks