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Positive integer n leaves a remainder of 4 after division by

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

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New post 13 Apr 2014, 21:37
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. :-D

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

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New post 13 Apr 2014, 23:08
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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. :-D


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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Re: Manhattan Remainder Problem [#permalink]

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New post 17 Apr 2014, 11:12
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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.



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.
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Re: Manhattan Remainder Problem [#permalink]

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New post 18 May 2014, 11:55
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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Re: Manhattan Remainder Problem [#permalink]

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New post 19 May 2014, 07:00
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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New post 31 May 2014, 13:12
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

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

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

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New post 15 Aug 2014, 21:17
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? :o

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

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New post 18 Aug 2014, 01:33
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? :o


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

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New post 31 Aug 2014, 23:28
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.

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

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New post 31 Aug 2014, 23:41
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.
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]

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New post 05 Feb 2015, 07:02
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. :-D


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

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

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New post 06 Feb 2015, 05:37
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 :-D 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
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]

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New post 11 Feb 2015, 08:02
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 .
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]

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New post 09 Mar 2015, 09:24
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!

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

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New post 09 Mar 2015, 19:13
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khl52 wrote:
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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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]

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New post 29 Mar 2015, 07:38
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.

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

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New post 30 Mar 2015, 00:20
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.
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Re: Positive integer n leaves a remainder of 4 after division by [#permalink]

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New post 02 May 2015, 00:03
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

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Re: Positive integer n leaves a remainder of 4 after division by   [#permalink] 02 May 2015, 00:03

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