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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.

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.

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.

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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.

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? _________________

“Confidence comes not from always being right but from not fearing to be wrong.”

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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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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When positive integer n is divided by 5, the remainder is 1.
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03 Aug 2016, 09:45

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