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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

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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

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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Updated on: 29 Jan 2015, 22:26

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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).

Originally posted by arunspanda on 31 Jan 2014, 10:06.
Last edited by arunspanda on 29 Jan 2015, 22:26, edited 1 time in total.

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

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24 Jun 2014, 02:27

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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]

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]

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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

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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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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. 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. [#permalink]

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Updated on: 19 Dec 2016, 15:15

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

#1: n=r1+d1(d2-1) n=1+5(7-1)=31 35-31=4 B

#2: let x=difference between quotients (n-1)/5-(n-3)/7=x➡ n+4=35x/2 because x must be even, start with x=2 n+4=35*2/2=35 n=31 k=4 B

#3: n=5q+1 n=7p+3 5q-2=7p the least value of q that will make 5q-2 a multiple of 7=6 n=5*6+1=31 35-31=4 B

Originally posted by gracie on 31 Aug 2016, 16:42.
Last edited by gracie on 19 Dec 2016, 15:15, edited 2 times in total.

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

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19 Dec 2016, 03:01

A number which leaves a remainder x-a when divided by x and leaves a remainder y-a when divided by y can be written as : LCM(x,y)-a Here the number leaves a remainder 1 when divided by 5 , thus x=5 x-a=1 , therfore a=4 the same number leaves a remainder 3 when divided by 7 , thus y-7 and x-a=3 thus a=4

So the number is LCM(5,7)- 4 which is 35-4 =31 , the smallest number required to get a multiple of 35 is 31+4 =35

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

There's a nice rule that says, If, when N is divided by D, the remainder is R, then the possible values of N include: R, R+D, R+2D, R+3D,. . . For example, if k divided by 6 leaves a remainder of 2, then the possible values of k are: 2, 2+6, 2+(2)(6), 2+(3)(6), 2+(4)(6), . . . etc.

When n is divided by 5, the remainder is 1. So, possible values of n are 1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, 66, 71, 76, etc.

When n is divided by 7, the remainder is 3. So, possible values of n are 3, 10, 17, 24, 31, 38, 45, 52, 59, 66, 73, etc.

So, we can see that n could equal 31, or 66, or an infinite number of other values.

Important: Since the Least Common Multiple of 7 and 5 is 35, we can conclude that if we list the possible values of n, each value will be 35 greater than the last value. So, n could equal 31, 66, 101, 136, and so on.

Check the answer choices....

Answer choice A: If we add 3 to any of these possible n-values, the sum is NOT a multiple of 35. ELIMINATE A

Answer choice B: if we take ANY of these possible n-values, and add 4, the sum will be a multiple of 35.

So, the smallest value of k is 4 such that k+n is a multiple of 35.

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

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13 Jan 2018, 23:57

Bunuel wrote:

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.

Hi Bunuel,

Would you please have a look at my solution? I've got the same answer yet a different path. Can you point it out what's wrong with my thought? Many thanks.

Question: What is the smallest value of K, when N+K is the multiple of 35?

My solution process: step 1. n+k=x(35), I was assuming that if x=1(because the hint is 'the smallest integer value'), then n+k=35, so k=35-n step 2. how to get n? (the same process as you've stated above) n=31 step 3. k=35-n, and n=31, therefore k=4

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.

Hi Bunuel,

Would you please have a look at my solution? I've got the same answer yet a different path. Can you point it out what's wrong with my thought? Many thanks.

Question: What is the smallest value of K, when N+K is the multiple of 35?

My solution process: step 1. n+k=x(35), I was assuming that if x=1(because the hint is 'the smallest integer value'), then n+k=35, so k=35-n step 2. how to get n? (the same process as you've stated above) n=31 step 3. k=35-n, and n=31, therefore k=4

Yes, the smallest value of n is 31, adding 4 gives 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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14 Jan 2018, 02:35

Bunuel wrote:

bingzhang wrote:

Bunuel wrote:

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.

Hi Bunuel,

Would you please have a look at my solution? I've got the same answer yet a different path. Can you point it out what's wrong with my thought? Many thanks.

Question: What is the smallest value of K, when N+K is the multiple of 35?

My solution process: step 1. n+k=x(35), I was assuming that if x=1(because the hint is 'the smallest integer value'), then n+k=35, so k=35-n step 2. how to get n? (the same process as you've stated above) n=31 step 3. k=35-n, and n=31, therefore k=4

Yes, the smallest value of n is 31, adding 4 gives a multiple of 35.

thank you! thank you, Bunuel. It's very nice of you.

gmatclubot

Re: When positive integer n is divided by 5, the remainder is 1.
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14 Jan 2018, 02:35