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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

Problem Solving Question: 68 Category:Arithmetic Properties of numbers Page: 70 Difficulty: 650

Each 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.

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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, 01:51

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SOLUTION

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

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 --> \(n+4=35k+31+4=35(k+1)\).

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) 35

Sol: Given n=5a+1 where a is any non-negative integer and also n=7b+3 where b is any non-negative 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 _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

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.

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

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SOLUTION

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

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 --> \(n+4=35k+31+4=35(k+1)\).

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]

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

Problem Solving Question: 68 Category:Arithmetic Properties of numbers Page: 70 Difficulty: 650

Each 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

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).

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

Re: When positive integer n is divided by 5, the remainder is 1. [#permalink]

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26 Aug 2015, 14:41

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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01 Dec 2015, 14:44

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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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