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# Positive integer n leaves a remainder of 4 after division by

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03 Nov 2012, 01:50
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breakit wrote:
Bunuel wrote:
To elaborate more.

Suppose we are told that:
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?

The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: $$n=6p+4$$. Thus according to this particular statement $$n$$ could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ...

The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: $$n=8q+2$$. Thus according to this particular statement $$n$$ could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ...

The above two statements are both true, which means that the only valid values of $$n$$ are the values which are common in both patterns. For example $$n$$ can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer $$q$$.

So we should derive general formula (based on both statements) that will give us only valid values of $$n$$.

How can these two statement be expressed in one formula of a type $$n=kx+r$$? Where $$x$$ is divisor and $$r$$ is a remainder.

Divisor $$x$$ would be the least common multiple of above two divisors 6 and 8, hence $$x=24$$.

Remainder $$r$$ would be the first common integer in above two patterns, hence $$r=10$$.

Therefore general formula based on both statements is $$n=24k+10$$. Thus according to this general formula valid values of $$n$$ are: 10, 34, 58, ...

Now, $$n$$ divided by 12 will give us the reminder of 10 (as 24k is divisible by 12).

Hope it helps.

Sorry , i didn't understand the last statement written marked in red... instead of 12 if we have other values like 14 or 20 then how remainder would vary.. just lil but curious to understand the gravity

$$n=24k+10=12(2k)+10$$ --> $$n$$ can be: 10, 34, 58, ... $$n$$ divided by 12 will give us the reminder of 10.

As, you can see, n divided by 14 can give different remainders. If n=10, then n divided by 14 yields the remainder of 10 but if n=34, then n divided by 14 yields the remainder of 6.
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03 Nov 2012, 14:14
Bunuel wrote:
To elaborate more.

Suppose we are told that:
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?

The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: $$n=6p+4$$. Thus according to this particular statement $$n$$ could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ...

The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: $$n=8q+2$$. Thus according to this particular statement $$n$$ could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ...

The above two statements are both true, which means that the only valid values of $$n$$ are the values which are common in both patterns. For example $$n$$ can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer $$q$$.

So we should derive general formula (based on both statements) that will give us only valid values of $$n$$.

How can these two statement be expressed in one formula of a type $$n=kx+r$$? Where $$x$$ is divisor and $$r$$ is a remainder.

Divisor $$x$$ would be the least common multiple of above two divisors 6 and 8, hence $$x=24$$.

Remainder $$r$$ would be the first common integer in above two patterns, hence $$r=10$$.

Therefore general formula based on both statements is $$n=24k+10$$. Thus according to this general formula valid values of $$n$$ are: 10, 34, 58, ...

Now, $$n$$ divided by 12 will give us the reminder of 10 (as 24k is divisible by 12).

Hope it helps.

Hi Bunuel, I had a quick question with this explanation:

Do we have to find the LCM? I just multiplied 6 x 8 and got 48 => n = 48k + 10 which also leads to a remainder of 10.
My question is is finding the LCM necessary?

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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]

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16 Dec 2012, 14:22
n=24k+10=12(12k)+10 --> n can be: 10, 34, 58, ... n divided by 12 will give us the reminder of 10.

As, you can see, n divided by 14 can give different remainders. If n=10, then n divided by 14 yields the remainder of 10 but if n=34, then n divided by 14 yields the remainder of 6.

Bunuel - Can you please explain this? how does 24k+10 = 12(12k)+10

and can you help me to visualize how you would divide 24k+10 by 12? thanks!

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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]

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16 Dec 2012, 23:29
jmuduke08 wrote:
n=24k+10=12(12k)+10 --> n can be: 10, 34, 58, ... n divided by 12 will give us the reminder of 10.

As, you can see, n divided by 14 can give different remainders. If n=10, then n divided by 14 yields the remainder of 10 but if n=34, then n divided by 14 yields the remainder of 6.

Bunuel - Can you please explain this? how does 24k+10 = 12(12k)+10

and can you help me to visualize how you would divide 24k+10 by 12? thanks!

It's n=24k+10=12*2k+10, not n=24k+10=12*12k+10.
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]

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18 Dec 2012, 01:40
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since n is greater than 30 we check for the number which gives a remainder of 4 after dividing by 6 and 3 after dividing by 5 , the number comes out to 58. So it will give a remainder of 28 after dividing by 30. Answer (E).
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]

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08 Jul 2013, 01:10
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]

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13 Apr 2014, 22:37
Maybe I did something wrong, or I just got lucky, so please inform me. I also had 28 as the solution, by just plugging in the values.

a. 3 / 6 = 0 r 6 INCORRECT
b. 12 / 6 = 2 r 0 INCORRECT
c. 18 / 6 = 3 r 0 INCORRECT
d. 22 / 6 = 3 r 4 & 22 / 5 = 4 r 2 INCORRECT
e. 28 / 6 = 4 r 4 & 28 / 5 = 4 r 3 CORRECT

Super-simple math, so there must be something wrong here, haha.

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18 May 2014, 12:55
Bunuel wrote:
To elaborate more.

Suppose we are told that:
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?

The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: $$n=6p+4$$. Thus according to this particular statement $$n$$ could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ...

The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: $$n=8q+2$$. Thus according to this particular statement $$n$$ could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ...

The above two statements are both true, which means that the only valid values of $$n$$ are the values which are common in both patterns. For example $$n$$ can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer $$q$$.

So we should derive general formula (based on both statements) that will give us only valid values of $$n$$.

How can these two statement be expressed in one formula of a type $$n=kx+r$$? Where $$x$$ is divisor and $$r$$ is a remainder.

Divisor $$x$$ would be the least common multiple of above two divisors 6 and 8, hence $$x=24$$.

Remainder $$r$$ would be the first common integer in above two patterns, hence $$r=10$$.

Therefore general formula based on both statements is $$n=24k+10$$. Thus according to this general formula valid values of $$n$$ are: 10, 34, 58, ...

Now, $$n$$ divided by 12 will give us the reminder of 10 (as 24k is divisible by 12).
Hope it helps.

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!

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19 May 2014, 08:00
russ9 wrote:
Bunuel wrote:
To elaborate more.

Suppose we are told that:
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?

The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: $$n=6p+4$$. Thus according to this particular statement $$n$$ could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ...

The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: $$n=8q+2$$. Thus according to this particular statement $$n$$ could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ...

The above two statements are both true, which means that the only valid values of $$n$$ are the values which are common in both patterns. For example $$n$$ can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer $$q$$.

So we should derive general formula (based on both statements) that will give us only valid values of $$n$$.

How can these two statement be expressed in one formula of a type $$n=kx+r$$? Where $$x$$ is divisor and $$r$$ is a remainder.

Divisor $$x$$ would be the least common multiple of above two divisors 6 and 8, hence $$x=24$$.

Remainder $$r$$ would be the first common integer in above two patterns, hence $$r=10$$.

Therefore general formula based on both statements is $$n=24k+10$$. Thus according to this general formula valid values of $$n$$ are: 10, 34, 58, ...

Now, $$n$$ divided by 12 will give us the reminder of 10 (as 24k is divisible by 12).
Hope it helps.

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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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]

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31 May 2014, 14:12
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

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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]

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15 Aug 2014, 22:17
jmuduke08 wrote:
n=24k+10=12(12k)+10 --> n can be: 10, 34, 58, ... n divided by 12 will give us the reminder of 10.

As, you can see, n divided by 14 can give different remainders. If n=10, then n divided by 14 yields the remainder of 10 but if n=34, then n divided by 14 yields the remainder of 6.

Bunuel - Can you please explain this? how does 24k+10 = 12(12k)+10

and can you help me to visualize how you would divide 24k+10 by 12? thanks!

Is this a sub-600 level question?

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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]

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18 Aug 2014, 02:33
alphonsa wrote:
jmuduke08 wrote:
n=24k+10=12(12k)+10 --> n can be: 10, 34, 58, ... n divided by 12 will give us the reminder of 10.

As, you can see, n divided by 14 can give different remainders. If n=10, then n divided by 14 yields the remainder of 10 but if n=34, then n divided by 14 yields the remainder of 6.

Bunuel - Can you please explain this? how does 24k+10 = 12(12k)+10

and can you help me to visualize how you would divide 24k+10 by 12? thanks!

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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17799 [0], given: 235 Intern Joined: 22 Aug 2014 Posts: 49 Kudos [?]: 3 [0], given: 25 GPA: 3.6 Re: Positive integer n leaves a remainder of 4 after division by [#permalink] ### Show Tags 01 Sep 2014, 00:28 Bunuel wrote: bchekuri 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 How to approach this Problem? Positive integer n leaves a remainder of 4 after division by 6 --> $$n=6p+4$$ --> 4, 10, 16, 22, 28, ... Positive integer n leaves a remainder of 3 after division by 5 --> $$n=5q+3$$ --> 3, 8, 13, 18, 23, 28, ... $$n=30k+28$$ - we have 30 as lcm of 5 and 6 is 30 and we have 28 as the first common integer in the above patterns is 28. Hence remainder when positive integer n is divided by 30 is 28. Answer: E. P.S. n>30 is a redundant information. 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. Kudos [?]: 3 [0], given: 25 Math Expert Joined: 02 Sep 2009 Posts: 42259 Kudos [?]: 132707 [0], given: 12335 Re: Positive integer n leaves a remainder of 4 after division by [#permalink] ### Show Tags 01 Sep 2014, 00:41 rohansangari wrote: Bunuel wrote: bchekuri 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 How to approach this Problem? Positive integer n leaves a remainder of 4 after division by 6 --> $$n=6p+4$$ --> 4, 10, 16, 22, 28, ... Positive integer n leaves a remainder of 3 after division by 5 --> $$n=5q+3$$ --> 3, 8, 13, 18, 23, 28, ... $$n=30k+28$$ - we have 30 as lcm of 5 and 6 is 30 and we have 28 as the first common integer in the above patterns is 28. Hence remainder when positive integer n is divided by 30 is 28. Answer: E. P.S. n>30 is a redundant information. 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. _________________ Kudos [?]: 132707 [0], given: 12335 Intern Joined: 13 Nov 2014 Posts: 8 Kudos [?]: 2 [0], given: 133 Re: Positive integer n leaves a remainder of 4 after division by [#permalink] ### Show Tags 05 Feb 2015, 08:02 VeritasPrepKarishma wrote: Saabs wrote: Maybe I did something wrong, or I just got lucky, so please inform me. I also had 28 as the solution, by just plugging in the values. a. 3 / 6 = 0 r 6 INCORRECT b. 12 / 6 = 2 r 0 INCORRECT c. 18 / 6 = 3 r 0 INCORRECT d. 22 / 6 = 3 r 4 & 22 / 5 = 4 r 2 INCORRECT e. 28 / 6 = 4 r 4 & 28 / 5 = 4 r 3 CORRECT Super-simple math, so there must be something wrong here, haha. 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!" Hi Karishma, could you kindly explain the bold part? I have trouble understanding why it was the case.. _________________ Please kudo if you find the post helpful Kudos [?]: 2 [0], given: 133 Intern Joined: 13 Nov 2014 Posts: 8 Kudos [?]: 2 [0], given: 133 Positive integer n leaves a remainder of 4 after division by [#permalink] ### Show Tags 06 Feb 2015, 06:37 VeritasPrepKarishma wrote: tieurongthieng wrote: 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: 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: I think these two together will help you get a very clear picture. 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 _________________ Please kudo if you find the post helpful Kudos [?]: 2 [0], given: 133 Director Joined: 07 Aug 2011 Posts: 579 Kudos [?]: 546 [0], given: 75 Concentration: International Business, Technology GMAT 1: 630 Q49 V27 Re: Positive integer n leaves a remainder of 4 after division by [#permalink] ### Show Tags 11 Feb 2015, 09:02 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 . _________________ Thanks, Lucky _______________________________________________________ Kindly press the to appreciate my post !! Kudos [?]: 546 [0], given: 75 Intern Joined: 04 Mar 2015 Posts: 2 Kudos [?]: [0], given: 0 Re: Positive integer n leaves a remainder of 4 after division by [#permalink] ### Show Tags 09 Mar 2015, 10:24 Bunuel wrote: To elaborate more. Suppose we are told that: 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? The statement "positive integer n leaves a remainder of 4 after division by 6" can be expressed as: $$n=6p+4$$. Thus according to this particular statement $$n$$ could take the following values: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, ... The statement "positive integer n leaves a remainder of 2 after division by 8" can be expressed as: $$n=8q+2$$. Thus according to this particular statement $$n$$ could take the following values: 2, 10, 18, 26, 34, 42, 50, 58, 66, ... The above two statements are both true, which means that the only valid values of $$n$$ are the values which are common in both patterns. For example $$n$$ can not be 16 (from first pattern) as the second formula does not give us 16 for any value of integer $$q$$. So we should derive general formula (based on both statements) that will give us only valid values of $$n$$. How can these two statement be expressed in one formula of a type $$n=kx+r$$? Where $$x$$ is divisor and $$r$$ is a remainder. Divisor $$x$$ would be the least common multiple of above two divisors 6 and 8, hence $$x=24$$. Remainder $$r$$ would be the first common integer in above two patterns, hence $$r=10$$. Therefore general formula based on both statements is $$n=24k+10$$. Thus according to this general formula valid values of $$n$$ are: 10, 34, 58, ... Now, $$n$$ divided by 12 will give us the reminder of 10 (as 24k is divisible by 12). Hope it helps. 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! Kudos [?]: [0], given: 0 Intern Joined: 22 Mar 2015 Posts: 1 Kudos [?]: [0], given: 0 Re: Positive integer n leaves a remainder of 4 after division by [#permalink] ### Show Tags 29 Mar 2015, 08:38 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. Kudos [?]: [0], given: 0 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7736 Kudos [?]: 17799 [0], given: 235 Location: Pune, India Re: Positive integer n leaves a remainder of 4 after division by [#permalink] ### Show Tags 30 Mar 2015, 01:20 minhphamvn wrote: 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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