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Positive integer n leaves a remainder of 4 after division by [#permalink]
05 May 2010, 22:49
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Question Stats:
74% (02:23) correct
26% (01:39) wrong based on 1049 sessions
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?
Re: Manhattan Remainder Problem [#permalink]
06 May 2010, 03:18
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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).
Re: Manhattan Remainder Problem [#permalink]
29 Nov 2010, 03:40
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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 # (r-y) as the remainder.
e.g:
9 when devided by 5 leves the remainder 4 : 9=5*1+4 it can also leave the remainder 4-5 = -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. 4-6) after division by 6 and a remainder of -2 (i.e. 3-5) after division by 5 ==> n when devided by 5 and 6 leaves the same remainder -2. what is n? LCM (5,6)-2 = 30-2 = 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.
Re: Manhattan Remainder Problem [#permalink]
06 May 2010, 11:09
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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 _________________
Re: Manhattan Remainder Problem [#permalink]
05 May 2010, 23:22
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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. _________________
Re: Manhattan Remainder Problem [#permalink]
11 Jun 2012, 20:15
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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.
Re: Manhattan Remainder Problem [#permalink]
06 May 2010, 18:51
1
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Silvers 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?
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
Exactly how I did it - we know the number has to have 8 in the ones digit, so then we can test 30+18 or 30+28
Re: Manhattan Remainder Problem [#permalink]
28 Nov 2010, 20:14
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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.
Thanks Bunuel - you are a gem. _________________
The night is at its darkest just before the dawn... never, ever give up!
Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
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.
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!" _________________
Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
01 Jun 2014, 19:56
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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.
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? _________________
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).
Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
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:
Re: Positive integer n leaves a remainder of 4 after division by [#permalink]
03 May 2015, 20:27
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vikasbansal227 wrote:
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
You want a number, n, that satisfies two conditions: "leaves a remainder of 4 after division by 6" and "a remainder of 2 after division by 8"
So you find the first such number by writing down the numbers. You get that it is 10. 10 satisfies both conditions.
What will be the next number? Any number that is a multiple of 6 more than 10 will continue to satisfy the first condition. e.g. 16, 22, 28, 34 etc Any number that is a multiple of 8 more than 10 will continue to satisfy the second condition. e.g. 18, 26, 34 etc
So any number that is a multiple of both 6 and 8 more than 10 will satisfy both conditions. LCM of 6 and 8 is 24. 24 is divisible by both 6 and 8.
So any number that is a multiple of 24 more than 10 will satisfy both conditions. e.g. 34, 58 .. etc These numbers can be written as 10 + 24a.
Re: Manhattan Remainder Problem [#permalink]
10 Jun 2012, 11:21
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 # (r-y) as the remainder.
e.g:
9 when devided by 5 leves the remainder 4 : 9=5*1+4 it can also leave the remainder 4-5 = -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. 4-6) after division by 6 and a remainder of -2 (i.e. 3-5) after division by 5 ==> n when devided by 5 and 6 leaves the same remainder -2. what is n? LCM (5,6)-2 = 30-2 = 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?"
Re: Manhattan Remainder Problem [#permalink]
12 Sep 2012, 21:24
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?
Re: Manhattan Remainder Problem [#permalink]
13 Sep 2012, 03:59
Expert's post
1
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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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