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Positive integer n leaves a remainder of 4 after division by
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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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Originally posted by bchekuri on 05 May 2010, 23:49.
Last edited by Bunuel on 20 Apr 2012, 03:05, edited 1 time in total.
Edited the question and added the OA




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Re: Manhattan Remainder Problem
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06 May 2010, 04:18
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.
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Re: Manhattan Remainder Problem
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29 Nov 2010, 04:40
Friends,
IT IS VERY COMMON IN GMAT to solve this kind of qtns using "NEGATIVE REMAINDER" theory.
The theory says:
if a # x is devided by y and leave the positive # r as the remainder then it can also leave negative # (ry) as the remainder.
e.g:
9 when devided by 5 leves the remainder 4 : 9=5*1+4 it can also leave the remainder 45 = 1 : 9=5*2 1
back to the original qtn: n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5 ==> n leaves a remainder of 2 (i.e. 46) after division by 6 and a remainder of 2 (i.e. 35) after division by 5 ==> n when devided by 5 and 6 leaves the same remainder 2. what is n? LCM (5,6)2 = 302 = 28 CHECK: 28 when devided by 6 leaves the remainder 4 and when devided by 5 leaves the remainder 3
However, the qtn says n > 30
so what is the nex #, > 28, that can give the said remainders when devided by 6 and 5 nothing but 28 + (some multiple of 6 and 5) as this "some multiple of 6 and 5" will not give any remainder when devided by 5 or 6 but 28 will give the required remainders.
hence n could be anything that is in the form 28 + (some multiple of 6 and 5) observe that "some multiple of 6 and 5" is always a multiple of 30 as LCM (5,6) = 30.
hence when n (i.e. 28 + some multiple of 6 and 5) is devided by 30 gives the remainder 28.
ANSWER "E"
Regards, Murali.
Kudos?




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Re: Manhattan Remainder Problem
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06 May 2010, 00:22
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.
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Re: Manhattan Remainder Problem
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06 May 2010, 12:09
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? Once we get the concept, We can also use the options to answer faster. Since the number leaves the remainder 4 after division by 6 we know it is an even number. Only even number that leaves a remainder of 3 after being divisible by 5 has to end with 8. So narrows the options to 18 and 28. since 30+18 = 48 is divisible by 6 the answer is 30+28 = 54
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Re: Manhattan Remainder Problem
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22 Jun 2010, 16:19
i would have thought the answer is c: 78. why am i wrong? i had 28 at first, but doesn't n have to be greater than 30?
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Re: Manhattan Remainder Problem
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28 Nov 2010, 21:14
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. Thanks Bunuel  you are a gem.
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Re: Manhattan Remainder Problem
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28 Nov 2010, 23:29
Possible values of n are:
n = 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, ...
If R = 3 after division of 5, then possible values of n are
n = 3, 8, 13, 18, 23, 28, 33, 38, 43, 48, 53, 58, ...
Therefore according to give conditions, some possible values of n are
n = 28, 58, ...
n > 30
Let's say n = 58
58/30 gives us a remainder of : 28.



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Re: Manhattan Remainder Problem
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10 Jun 2012, 12:21
muralimba wrote: Friends,
IT IS VERY COMMON IN GMAT to solve this kind of qtns using "NEGATIVE REMAINDER" theory.
The theory says:
if a # x is devided by y and leave the positive # r as the remainder then it can also leave negative # (ry) as the remainder.
e.g:
9 when devided by 5 leves the remainder 4 : 9=5*1+4 it can also leave the remainder 45 = 1 : 9=5*2 1
back to the original qtn: n leaves a remainder of 4 after division by 6 and a remainder of 3 after division by 5 ==> n leaves a remainder of 2 (i.e. 46) after division by 6 and a remainder of 2 (i.e. 35) after division by 5 ==> n when devided by 5 and 6 leaves the same remainder 2. what is n? LCM (5,6)2 = 302 = 28 CHECK: 28 when devided by 6 leaves the remainder 4 and when devided by 5 leaves the remainder 3
However, the qtn says n > 30
so what is the nex #, > 28, that can give the said remainders when devided by 6 and 5 nothing but 28 + (some multiple of 6 and 5) as this "some multiple of 6 and 5" will not give any remainder when devided by 5 or 6 but 28 will give the required remainders.
hence n could be anything that is in the form 28 + (some multiple of 6 and 5) observe that "some multiple of 6 and 5" is always a multiple of 30 as LCM (5,6) = 30.
hence when n (i.e. 28 + some multiple of 6 and 5) is devided by 30 gives the remainder 28.
ANSWER "E"
Regards, Murali.
Kudos? Hello Murali, Can you please explain how will you solve following example using "NEGATIVE REMAINDER" theory discussed above? "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?"



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Re: Manhattan Remainder Problem
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11 Jun 2012, 21:15
kuttingchai wrote: Hello Murali,
Can you please explain how will you solve following example using "NEGATIVE REMAINDER" theory discussed above? "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?"
Question: 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? Solution: n = 6a + 4 (or remainder = 2) n = 8b + 2 (or remainder = 6) The negative remainder is not the same in the two cases. So don't use negative remainders here. Here are some links that discuss divisibility and remainders. The third link discusses negative remainders but you will need to go through the first two posts to understand it properly. Also, the links discuss how to solve such questions. Go through them and get back if there is a doubt. http://www.veritasprep.com/blog/2011/04 ... unraveled/http://www.veritasprep.com/blog/2011/04 ... yapplied/http://www.veritasprep.com/blog/2011/05 ... emainders/
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Re: Manhattan Remainder Problem
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12 Sep 2012, 22:24
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. This is awesome I was searching for this logic for a long time. Just one question  you said, i quote, Remainder \(r\) would be the first common integer in above two patterns, hence \(r=10\). Is there a quick way to come with this value without listing the nos out. sometime the common nos in both the series would come after 6 or 7 term hence sometimes consumes time. Is there a quicker way? My sincere thanks. cheers



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Re: Manhattan Remainder Problem
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13 Sep 2012, 04:59
Jp27 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. This is awesome I was searching for this logic for a long time. Just one question  you said, i quote, Remainder \(r\) would be the first common integer in above two patterns, hence \(r=10\). Is there a quick way to come with this value without listing the nos out. sometime the common nos in both the series would come after 6 or 7 term hence sometimes consumes time. Is there a quicker way? My sincere thanks. cheers In some cases we can use some other approaches, though I think simple listing is the easiest way.
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Re: Manhattan Remainder Problem
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15 Sep 2012, 21:32
Bunuel wrote: In some cases we can use some other approaches, though I think simple listing is the easiest way. Thanks for your response. Yes that easy. Sorry this might be a silly questions.... I have a basic doubt here Lets assume p = 9 n = 11 and what is the reminder when pn / 15? so 99 / 15 give me a reminder of 9 but if i split the denominator for ex 9 * 11 / 3 * 5 = 9/3 * 11 / 5 => I can always mutiply reminders as long as i can correct the excess so here 9/3 reminder 0 11/5 reminder 1 = 0 * 1 = 0 = reminder. So i should never split the denominator or the numertator when calculating for root or can I? something is wrong here! pls help.



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Re: Manhattan Remainder Problem
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03 Nov 2012, 01:46
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



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Re: Manhattan Remainder Problem
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03 Nov 2012, 01:50
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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Re: Manhattan Remainder Problem
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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
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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
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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
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18 Dec 2012, 01:40
Ans: 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
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08 Jul 2013, 01:10
Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE
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