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

[BTGModeratorVI]

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

Answer: B
Source: Official guide

[Brent@GMATPrepNow]

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

Answer: B
Source: Official guide

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

[Scott@TargetTestPrep]

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

Answer: B
Source: Official guide

Solution:

Since the remainder is 1 when n is divided by 5, n will be a [(multiple of 5) + 1], and thus, n 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, has a remainder of 3.

1/7 = 0 remainder 1

6/7 = 0 remainder 6

11/7 = 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.

Alternate Solution:

Since n produces a remainder of 1 when divided by 5, we can express n = 5t + 1 where t is some positive integer. Similarly, since n produces a remainder of 3 when divided by 7, we can express n = 7s + 3 where x is some positive integer. Notice that n + 4 = 5t + 5 = 7s + 7 is a multiple of both 5 and 7; therefore n + 4 is a multiple of LCM(5, 7) = 35 as well. Thus, the smallest positive integer value of k is 4.

Answer: B