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Re: Positive integer n leaves a remainder of 4 after division by
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 03 May 2015, 21:27
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. Also, I think you did not check out the 4 links I gave here: positive-integer-n-leaves-a-remainder-of-4-after-division-by-93752-40.html#p1496547They are very useful in understanding divisibility fundamentals.
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Positive integer n leaves a remainder of 4 after division by
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27 Aug 2015, 01:43
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. n=6p+4, n=5q+3 --> 6p+4=5q+3 -> 6p+1=5q -> 5p +(p+1)=5q, so p+1 must be a multiple of 5, means P=4,9,14 etc... if p=4: 6p+4=28 -> q=5:5*5+3=28 => 28 is the first common number 28/30=0+28(R) Answer (E) Question 2n=6p+4, n=8q+2 /divide by 2 both expressions => 3q+(q-1)=3p q-1 must be a multiple of 3: 4,10,16 etc... let's pick 4 for q 8*4+2=34 --> 34/12=2*12+10(R)
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Positive integer n leaves a remainder of 4 after division by
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14 Dec 2015, 05:31
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 for the great explanation Bunuel. I have a doubt though. This seems to be valid when the final divisor mentioned in the question is a factor of LCM of the previous divisors. Are there any cases, where this would not be true? As in what if the question asked for divisibility by 14? i.e remainder of 24k+10 when divided by 14?
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Re: Positive integer n leaves a remainder of 4 after division by
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09 Apr 2016, 07:26
How do we solve a question which has different remainder ? because in this case remainder is -2 , ie common to both the cases? What happens to the case when both the case have different remainder ? 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.
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Positive integer n leaves a remainder of 4 after division by
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Updated on: 09 Sep 2018, 12:09
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
n/6=q+4
n/5=p+3
6q+4=5p+3➡
5p-6q=1
least value of p that will make 6q an integer=5
least value of n=28
28+6*5=58=next least value of n
58/30 leaves a remainder of 28
E
Originally posted by gracie on 31 Aug 2016, 20:11.
Last edited by gracie on 09 Sep 2018, 12:09, edited 1 time in total.
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Re: Positive integer n leaves a remainder of 4 after division by
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31 Aug 2016, 21:07
happy19 wrote: How do we solve a question which has different remainder ? because in this case remainder is -2 , ie common to both the cases? What happens to the case when both the case have different remainder ?
Check out this post. It discusses how to handle questions in which the remainders are different. 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
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05 Oct 2016, 12:23
n = 6d + 4
n = 5d + 3
if we multiply the first equation by 5 and the second by 6 we get
5n = 30d + 20 (1)
6n = 30d + 18 (2)
if we make (1) - (2)
-n = 2; n = -2
so -2/30 has a remainder of 28
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Positive integer n leaves a remainder of 4 after division by
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03 Dec 2016, 07:59
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 \(N=4\) (mod 6)\(N= 3\) (mod 5)The difference between divisor and remainder is the same for both linear congruences \(6 -4 = 5 - 3 = 2 = D\) Hence our N can b represented as: N = LCM (6, 5)*X - D, where X is >=0 and D is our common differnce. \(N = 30X - 2\) X=0 ---> N=28 and for N>30 \(\frac{30*X - 2}{30} = \frac{0 - 2}{30}\) = \(-2\) (mod 30) = \(28\) (mod 30)Answer E.
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Positive integer n leaves a remainder of 4 after division by
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16 Jan 2017, 13:08
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 for the amazing concept. I just have one more query: How will we apply this approach for questions such as: What is the remainder when the positive integer n is divided by 12? 1. When n is divided by 6, the remainder is 1. 2. When n is divided by 12, the remainder is greater than 5.
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Re: Positive integer n leaves a remainder of 4 after division by
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17 Jan 2017, 00:54
ashikaverma13 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. Thanks for the amazing concept. I just have one more query: How will we apply this approach for questions such as: What is the remainder when the positive integer n is divided by 12? 1. When n is divided by 6, the remainder is 1. 2. When n is divided by 12, the remainder is greater than 5. This question is discussed here: what-is-the-remainder-when-the-positive-integer-n-is-divided-161170.html
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Re: Positive integer n leaves a remainder of 4 after division by
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27 Feb 2017, 06:45
6k+4 => 6k-2
5q+3 => 5q-2
Hence, 30m-2 => -2+ 30 = 28
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Re: Positive integer n leaves a remainder of 4 after division by
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03 Jun 2017, 16:51
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. Hi Bunuel, I understood everything up to this point "Hence remainder when positive integer n is divided by 30 is 28" I was wondering could you please explain how you got 28 from this statement? Would greatly appreciate it!! Thank You!
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Re: Positive integer n leaves a remainder of 4 after division by
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03 Jun 2017, 20:58
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 assume difference of 1 between n/6 and n/5 quotients (n-3)/5-(n-4)/6=1 n=28 28+5*6=58 58/30 gives a remainder of 28 E
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Re: Positive integer n leaves a remainder of 4 after division by
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11 Sep 2017, 01:38
I think one easy way to solve this is as follows:
Given: n=6q+4 and n=5p+3
multiply first eq by 5 and 2nd by 6(in order to get 30 as factor in quotient)
therefore we have:
5n=30q+20 (a)
6n=30p+18 (b)
b-a: n=30(p-q)-2
since remainder is -2, positive remainder =30-2=28 (option E)
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Positive integer n leaves a remainder of 4 after division by
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19 May 2018, 03:00
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 ..Thanks for the explanation. Can you plz help with the logic of this formulae. I am not able to understand the logic behind the formulae. Thanks in advance..!!!
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Positive integer n leaves a remainder of 4 after division by
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19 May 2018, 11:33
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 let q and p be quotients 6q+4=5p+3➡ 5p-6q=1 because 6q can't have digits unit of 9, 5p must have units digit of 5, and therefore 6q must have digits unit of 4 thus, 5p=25 and 6q=24, least values of p and q are 5 and 4 respectively, and least value of n=28 28/30 leaves a remainder of 28 E
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Re: Positive integer n leaves a remainder of 4 after division by
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09 Sep 2018, 06:54
n can be 4, 10,16,...
or 4 + (n1- 1)*6
n can be 3, 8 , 13,...
or 3 + (n2-1)*5
solving n2/n1= 6/5
substituting n2 or n1 , n= 28.
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Positive integer n leaves a remainder of 4 after division by
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11 Oct 2018, 07:43
#MY10SECAPPROACH : Just see the options , the answer options are remainders given , just divide them with 5 & 6 to see which one satisfies the condition of leaving remainder 3 and 4 respectively. Only 1 , answer option E. Time taken : 10 Seconds. Kudos ?
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Positive integer n leaves a remainder of 4 after division by
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11 Oct 2018, 19:42
Jp27 wrote:
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?
hi Jp27, sometimes you can can save time this way first, do the two equations just as Bunuel suggests: n=6p+4 n=8q+2 but then put them in this form: 8q-2=6p (it works faster for me with the greatest coefficient on the left) then plug in successive values of q until you find the least value that will make p an integer: (you can do this very quickly in your head) 0 won't work, but 1 will 8*1-2=6*1 plugging values into original equations, least value of n=10 hope this helps, gracie
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