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

Question

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

Bunuel · Accepted Answer

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: https://gmatclub.com/forum/manhattan-rem ... ml#p721341

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Bunuel · Answer

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ashueureka · Answer

The rule is good to solve such problems but sometime we may just solve GMAT problems simply by observation. In this case we see that number n leaves a remainder 1 from 5 so if we add 4 to n, then the number will be divisible by 5.
Similarly, the number n leaves remainder of 3 from 7, again adding 4 to n makes it divisible by 7. So in both the cases adding 4 makes the number n divisible by both 5 and 7 and hence it should also be divisible by LCM of 5,7 i.e.35. So 4 is the answer.
It is better to remember the rule but just in case you don't then simply observe.
Thanks!

KarishmaB · Answer

First, let us say I have a number n which is divisible by 5 and by 7. We all agree that it will be divisible by 35, the LCM of 5 and 7.

Now, if I have a number n which when divided by 5 gives a remainder 1 and when divided by 7 gives a remainder 1, we can say the number is of the form
n = 5a + 1 e.g. 5 + 1, 10 + 1, 15 + 1, 20 + 1, 25 + 1, 30 + 1, 35 + 1 etc
and 
n = 7b + 1 e.g. 7 + 1, 14 + 1, 21 + 1, 28 + 1, 35 + 1 etc
So when it is divided by the LCM, 35, it will give 1 as remainder (as is apparent above)

Next, if I have a number n which when divided by 5 gives a remainder 1 and when divided by 7 gives a remainder 3, we can say the number is of the form
n = 5a + 1
and
n = 7b + 3
Now, the only thing you should try to understand here is that when n is divided by 5 and if I say the remainder is 1, it is the same as saying the remainder is -4. e.g. When 6 is divided by 5, remainder is 1 because it is 1 more than a multiple of 5. I can also say it is 4 less than the next multiple of 5, can't I? 6 is one more than 5, but 4 less than 10. 
Therefore, we can say n = 5x - 4 and n = 7y - 4 (A remainder of 3 when divided by 7 is the same as getting a remainder of -4)
Now this question is exactly like the question above. So when you divide n by 35, remainder will be -4 i.e. n will be 4 less than a multiple of 35. So you must add 4 to n to make it a multiple of 35

A trickier version is: If I have a number n which when divided by 5 gives a remainder 1 and when divided by 7 gives a remainder 5, what is the remainder when n is divided by 35?
n = 5a + 1 = 5x - 4
n = 7b + 5 = 7y -2
Nothing common! Now, I will need to check for the smallest such number.
I put b = 1. n = 12. Is it of the form 5a + 1? No.
Put b = 2. n = 19. Is it of the form 5a + 1? No.
Put b = 3. n = 26. Is it of the form 5a + 1? Yes.
When 26 is divided by 5, it gives a remainder of 1. When it is divided by 7, it gives a remainder if 5.
Next such number will be 35 + 26. 
Next will be 35*2 + 26
and so on...
The remainder when n is divided by 35 will be 26 (or we can say it will be -9). If we want to find the number that must be added to n to make it divisible by 35, that number will be 9.

kp1811 · Answer

here n is divided by 5 and 7 and remainders are 1 and 3. There is a rule wherein if the difference b/w and remainder is same then the number of obtained from LCM of 2 (here 2) numbers and the constant difference.

Here constant difference is 5-1 = 4 and 7-3 = 4 
so the required number if of the form A(LCM of 5 and 7) - constant difference = 35A - 4

So to obtain a multiple of 35 we would need to add 4 to 35A - 4.
Hence B - 4

BrentGMATPrepNow · Answer

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.

Answer = B

RELATED VIDEO 
 https://www.youtube.com/watch?v=XCxgIR042tE

WoundedTiger · Answer

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

arunspanda · Answer

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) .

PareshGmat · Answer

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

BrainLab · Answer

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

JeffTargetTestPrep · Answer

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   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
 
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

avigutman · Answer

Video solution from Quant Reasoning starts at 26:43: Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=_uwR9AZas9M

CrackverbalGMAT · Answer

When a number N is divided by X , the remainder is a and when N is divided by Y , the remainder is b.
if the difference between the divisor and the remainder is constant throughout, then  N = LCM ( X,Y ) - difference

Lets try this logic here,
When positive integer n is divided by 5, the remainder is 1. 
The difference between divisor and remainder = D = 5-1 = 4

When n is divided by 7, the remainder is 3.
The difference between divisor and remainder = D = 7-3= 4

Since Difference D is constant through out , Least possible value of N = LCM( 5 , 7) - D = LCM( 5 , 7) - 4
 = 35 -4 = 31.

When you divide 31 by 5, the remainder is 1 and when you divide by 7, the remainder is 3.

What is the smallest positive integer k such that k + N is a multiple of 35 ?
Lowest possible value of N = 31 and its given that ( 31 + K ) is a multiple of 35. So smallest integer K = 4.
Such that 35 will be multiple of 35.

Option B is the correct answer.

Thanks,
Clifin J Francis,
GMAT SME

BrushMyQuant · Answer

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

Theory: Dividend = Divisor*Quotient + Remainder

n -> Dividend
5 -> Divisor
a -> Quotient (Assume)
1 -> Remainders
=> n = 5*a + 1 = 5a + 1

When n is divided by 7, the remainder is 3

Let quotient be b
=> n = 7b + 3

n = 7b + 3 = 5a + 1
=> 5a = 7b + 2
=> a =  \frac{7b + 2}{5}

Now, only those values of "b" which will give "a" also as an integer will give us the common values of n which satisfy both n = 7b + 3 = 5a + 1 the conditions

b=4 => a =  \frac{7*4 + 2}{5}  = 6 => Integer
=> n = 7*4 + 3 = 28 + 3 = 31

What is the smallest positive integer k such that k + n is a multiple of 35

31 + 4 = 35 => k = 4

So,  Answer will be B 
Hope it helps!

Playlist on Solved Problems on Remainders  here

Watch the following video to MASTER Remainders

https://www.youtube.com/watch?v=P0S6j2uTaj4

Schachfreizeit · Answer

I understand everything until the last step, where you add 4 on both sides of the equation. Can you elaborate this?

Bunuel · Answer

The question asks: what is the smallest  positive  integer k such that k + n is a multiple of 35 ?

We got that  n=35m+31 . 35m IS a multiple of 35 so we should add something to 31, so that 31 + k also becomes a multiple of 35. The smallest positive integer we can add to 31 such that the sum is a multiple of 35 is 4.

onlymalapink · Answer

hown is this.questiin 555 am i missing something

Bunuel · Answer

The difficulty level of a question on the site, after sufficient attempts, is determined automatically based on various parameters collected from users' attempts via timer, such as the percentage of correct answers and the time taken to answer the question. https://gmatclub.com/forum/download/file.php?id=84707 GMAT-Club-Forum-317wsvuq.png

minionbaba · Answer

If n divided by 5 leaves remainder 1, then you can add 4 to get the next number that will be divisible by 5

if n divided by 7 leaves remainder 3, then you can add 4 to get the next number that will be divisible by 7

i.e. n+ 4 will be divisible by both 5 & 7 & the ONLY number divisible by both 5 & 7 (i.e. without leaving a remainder) is 35 (5*7), so n+4 = 35

therefore, k = 4

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

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