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When positive integer n is divided by 5, the remainder is 1. [#permalink]
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The Official Guide For GMAT® Quantitative Review, 2ND EditionWhen positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35 ? (A) 3 (B) 4 (C) 12 (D) 32 (E) 35 Problem Solving Question: 68 Category: Arithmetic Properties of numbers Page: 70 Difficulty: 650 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you!
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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink]
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SOLUTIONWhen positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35 ?(A) 3 (B) 4 (C) 12 (D) 32 (E) 35 Positive integer n is divided by 5, the remainder is 1 > \(n=5q+1\), where \(q\) is the quotient > 1, 6, 11, 16, 21, 26, 31, ... Positive integer n is divided by 7, the remainder is 3 > \(n=7p+3\), where \(p\) is the quotient > 3, 10, 17, 24, 31, .... There is a way to derive general formula for \(n\) (of a type \(n=mx+r\), where \(x\) is divisor and \(r\) is a remainder) based on above two statements:Divisor \(x\) would be the least common multiple of above two divisors 5 and 7, hence \(x=35\). Remainder \(r\) would be the first common integer in above two patterns, hence \(r=31\). Therefore general formula based on both statements is \(n=35m+31\). Thus the smallest positive integer k such that k+n is a multiple of 35 is 4 > \(n+4=35k+31+4=35(k+1)\). Answer: B. More about deriving general formula for such problems at: manhattanremainderproblem93752.html#p721341
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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink]
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30 Jan 2014, 03:07
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When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35 ?
(A) 3 (B) 4 (C) 12 (D) 32 (E) 35Sol: Given n=5a+1 where a is any nonnegative integer and also n=7b+3 where b is any nonnegative integer.....so n is of the form Possible values of n in case 1 : 1,6,11,16,21,26, 31.... Possible value of n in case 2 : 3,10,17, 24, 31... So, n=35C+ 31....Now for K+ n to be multiple of 35 K needs to be 4 so that k+n = 35C+31+4 or 35(c+1) Ans B. 650 level is okay
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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink]
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30 Jan 2014, 03:26
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Ans.B n=5p+1=>n=6,11,16,21,26,31,36... n=7q+3=>n=10,17,24,31,... First no. common to both the series=31.We have to add 4. I know there is a better way of doing this kind of question.



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When positive integer n is divided by 5, the remainder is 1. [#permalink]
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31 Jan 2014, 10:06
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When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35 ?
(A) 3 (B) 4 (C) 12 (D) 32 (E) 35
Method 1
n is divided by 5, the remainder is 1 > \(n= 5x + 1\) or, n + k = 5x + (1 + k) So, n + k is divisible by 5, when (1+ k) is a multiple of 5. Or, Possible values of k are 4, 9, 14,19, 24, 29, 33,.....
n is divided by 7, the remainder is 3 > \(n=7y + 3\) Or, n + k = 7y +(3 + k) So, n + k is divisible by 7, when (3+ k) is a multiple of 7. Or, Possible values of k are 4, 11, 18, 25, 32, 39,.....
As the lowest common value is 4, the answer is (B).
Last edited by arunspanda on 29 Jan 2015, 22:26, edited 1 time in total.



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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink]
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31 Jan 2014, 20:34
is there any other way, than number plugging? What if instead of 5 and 7, the question is changed to some scary numbers like 263 and 911?
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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink]
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01 Feb 2014, 08:49
SOLUTIONWhen positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35 ?(A) 3 (B) 4 (C) 12 (D) 32 (E) 35 Positive integer n is divided by 5, the remainder is 1 > \(n=5q+1\), where \(q\) is the quotient > 1, 6, 11, 16, 21, 26, 31, ... Positive integer n is divided by 7, the remainder is 3 > \(n=7p+3\), where \(p\) is the quotient > 3, 10, 17, 24, 31, .... There is a way to derive general formula for \(n\) (of a type \(n=mx+r\), where \(x\) is divisor and \(r\) is a remainder) based on above two statements:Divisor \(x\) would be the least common multiple of above two divisors 5 and 7, hence \(x=35\). Remainder \(r\) would be the first common integer in above two patterns, hence \(r=31\). Therefore general formula based on both statements is \(n=35m+31\). Thus the smallest positive integer k such that k+n is a multiple of 35 is 4 > \(n+4=35k+31+4=35(k+1)\). Answer: B. More about deriving general formula for such problems at: manhattanremainderproblem93752.html#p721341
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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink]
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14 Feb 2014, 09:59
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According to me simplest way to solve this problem is given below:
Let N=5P+1 or N=7Q+3
So 5P+1=7Q+3 => P=Q+((2Q+2)/5) .. Check for value of 2Q+2 divisible by 5 i.e. 4 and hence P=6 So smallest value of N=31 for P=6 and Q=4.
So, N is 31 and smallest number K , which add to N to make it multiple of 35 is 4. so K=4
Hence answer is B



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When positive integer n is divided by 5, the remainder is 1. [#permalink]
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24 Jun 2014, 02:27
n+k is a multiple of 35 then n+k = 35x => n = 35x  k. When positive integer n is divided by 5, the remainder is 1 => when integer k is divided by 5, the remainder is 4 => B
[or When n is divided by 7, the remainder is 3 => when integer is divided by 7, the remainder is 4 => B or C, but we do not need to check this case]



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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink]
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15 Aug 2014, 03:14
Bunuel or anyone else: Could you post links to practice similar problems? I am having severe problems with this type of question at the moment..



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When positive integer n is divided by 5, the remainder is 1. [#permalink]
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17 Sep 2014, 10:24
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionWhen positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35 ? (A) 3 (B) 4 (C) 12 (D) 32 (E) 35 Problem Solving Question: 68 Category: Arithmetic Properties of numbers Page: 70 Difficulty: 650 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you! Sol: n=5a+1 => Remainder is 1 or 4 n=7b+3 => Remainder is 3 or 4 i.e. if I add +4 to the number the number will be perfect multiple of 35



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When positive integer n is divided by 5, the remainder is 1. [#permalink]
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Picked up 7 & added 3 = 10 10 gives a remainder 3 when divided by 7 Wrote down similar numbers. We are looking for numbers which would either end by 1 or 6 (so they would provide a remainder 1 when divided by 5) 10 17 24 31 ........ stop 31 is the number when divided by 5, provides remainder 1 (Its already tested that it provides remainder 3 when divided by 7) First available multiple of 35 is 35, which is just 4 away from 31 Answer = 4
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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink]
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26 Nov 2014, 12:23
REMAINDER It is suggested to use the division algorithm. By dividing a positive integer N for 5, residue 1 is obtained. The algorithm is associated 5X + 1 = N, where N is the dividend, the divisor, and X is 1 is the remainder. By dividing the same N for 7, remainder 3 is obtained. The associated algorithm is N = 7Y + 3, where N is the dividend and the divisor and 3 is the remainder. You have: N = 5X + 1 and N = 7Y +3 Since N = N, equaling have: 5X +3 +1 = 7Y Grouping variables, with positive results, it: 5X  7Y = 2 It is suggested to start giving positive integers X, then watch multiple of 7 is closer to 5X, and meets equality: Multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40, 35, 50, ..... Multiples of 7: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, .... The condition is: Multiple of 5  multiple of 7 = 2 3028 = 2 Then X = 6 and Y = 4 Thus we obtain N = 31 Smaller value for K, if (K + N) is a multiple of 35? This value is 4, because 4 + 31 = 35, and 35 is a multiple of 35. Correct Answer b) claudio hurtado GMAT GRE SAT math classes part
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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink]
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12 Jan 2015, 12:11
Can't we just do this:
n=5q+1= 1,6,11,16,21,26,31,36,41,46,... n=7q+3= 3,9,17,24,31...
k+31> 35 k> 3531 k> 4
35 is anyway the first multiple of 35 (35*1). Is this wrong..?



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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink]
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12 Jan 2015, 12:15
Sorry, I forgot to mention the reasoning behind k+31> 35.
I thought that since k + n is a multiple of 35, it cannot be smaller. It should actually have a larger or equal sign there. So, this is why it is k+31> 35.
I always do that to remember while solving the problem that a multiple is larger (in comparison to a factor which is smaller).



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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink]
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24 Mar 2015, 13:24
The fastest way here is to realize that 35 is a factor of both 5 and 7. So you need to add a number to the remainder of both situations so that each number a factor of 5 and 7. By adding 4 to 1 it becomes a factor of 5, and by adding 4 to 3 it becomes a factor of 7. So B must be true



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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink]
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Hi all, actually we don't need pluging all those values for x,y... n=5x+1 and n=7y+3 > n+k=> (5x+1+k)/35 so 1+k must be a multiple of 5 if we want this expression to yield an integer so k=4 Use same logic here (7y+3+k)/35 > 3+k must be a multiple of 7, so k=4
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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink]
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N+4 is multiple of both 5 and 7 so N+4 must be an LCM of 5 and 7 i.e. 35. So the smallest value needed to make it multiple of 35 is 4
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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink]
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10 May 2016, 09:05
i dont think that gmat test us the formular just pick some specific numbers. and find 31
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When positive integer n is divided by 5, the remainder is 1. [#permalink]
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03 Aug 2016, 09:45
Quote: When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35 ?
(A) 3 (B) 4 (C) 12 (D) 32 (E) 35 We can find the value of n first by just strategically find values that when divided by 5 have a reminder of 1. For example, since the remainder is 1 when n is divided by 5, n will be a [(multiple of 5) + 1] and thus must be one of the following numbers: 1, 6, 11, 16, 21, 26, 31, … Now we have to find out which of these numbers when divided by 7, have a remainder of 3. 1/7 = 0 remainder 1 6/7 = 0 remainder 6 11/7 = 0 remainder 6 6/7 = 1 remainder 4 16/7 = 2 remainder 2 21/7 = 3 remainder 0 26/7 = 3 remainder 5 31/7 = 4 remainder 3 We can see that 31 is the smallest value of n that satisfies the requirement. So we must determine the value of k such that k + n is a multiple of 35. Obviously, since 4 + 31 = 35 and 35 is a multiple of 35, then the smallest positive integer value of k is 4. Answer: B
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