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When positive integer n is divided by 5, the remainder is 1.

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When positive integer n is divided by 5, the remainder is 1. [#permalink] New post 30 Jan 2014, 00:51
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

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


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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink] New post 30 Jan 2014, 00:51
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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 --> \(n+4=35k+31+4=35(k+1)\).

Answer: B.

More about deriving general formula for such problems at: manhattan-remainder-problem-93752.html#p721341
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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink] New post 30 Jan 2014, 02:07
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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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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink] New post 30 Jan 2014, 02:26
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] New post 31 Jan 2014, 09:06
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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, 21:26, edited 1 time in total.
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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink] New post 31 Jan 2014, 19:34
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] New post 01 Feb 2014, 07:49
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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 --> \(n+4=35k+31+4=35(k+1)\).

Answer: B.

More about deriving general formula for such problems at: manhattan-remainder-problem-93752.html#p721341
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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink] New post 14 Feb 2014, 08:59
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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] New post 24 Jun 2014, 01:27
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] New post 15 Aug 2014, 02:14
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] New post 17 Sep 2014, 09:24
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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


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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] New post 02 Oct 2014, 02:10
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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] New post 26 Nov 2014, 11:23
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

30-28 = 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)

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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink] New post 12 Jan 2015, 11:11
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> 35-31
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] New post 12 Jan 2015, 11:15
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] New post 24 Mar 2015, 12:24
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] 24 Mar 2015, 12:24
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