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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]

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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]

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

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]

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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]

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30 Jan 2014, 02:26
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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, 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]

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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]

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

More about deriving general formula for such problems at: manhattan-remainder-problem-93752.html#p721341
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Kudos [?]: 29 [1] , given: 36

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

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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

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

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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]

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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]

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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

GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

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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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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Kudos [?]: 1935 [1] , given: 193

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

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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

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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, 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.

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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, 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]

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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]

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

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26 Aug 2015, 13:41
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, 13:44
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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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Re: When positive integer n is divided by 5, the remainder is 1. [#permalink]

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10 May 2016, 08:05
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, 08:45
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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.

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When positive integer n is divided by 5, the remainder is 1.   [#permalink] 03 Aug 2016, 08:45

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