OG 12 quant problem, is there an easier solution?

[c210]

I don't quite understand the solution presented by the OG quant book on problem #68 for PS

the question is

when positive integer N is divided by 5, the reminder is 1. when N is divided by 7, the remainder is 3. What is the smaller positive integer K such that K + N is a multiple of 35?


A)3
B)4
C)12
D)32
E)35

whats the best /quickest way to solve this?

best,
c

[Anurag@Gurome]

I don't quite understand the solution presented by the OG quant book on problem #68 for PS

the question is

when positive integer N is divided by 5, the reminder is 1. when N is divided by 7, the remainder is 3. What is the smaller positive integer K such that K + N is a multiple of 35?


A)3
B)4
C)12
D)32
E)35

whats the best /quickest way to solve this?

best,
c

N = 5*a + 1
N = 7*b + 3
Here, a and b are integers.
Note that the difference between divider and remainder (5 - 1 and 7 - 3) is 4 in both the case.
So add 4 on both sides of each of the 2 equations.
So, we get N + 4 = 5*a + 5 = 5 *(a + 1)
N + 4 = 7*b + 7 = 7 * (b + 1)
This means N + 4 is a multiple of both 5 and 7.
Since 5 and 7 are co-prime, N + 4 has to be a multiple of 5 * 7 = 35 as well.
So the smallest possible value of K is 4.

The correct answer is B.

[aneesh.kg]

Number = Divisor*Quotient + Remainder

N = 5x + 1
If you substitute a few integral values for x, you get N = 1, 6, 11, 16, 21, 26, 31, 36 etc.
N = 7y + 3
If you substitute a few integral values for y, you get N = 3, 10, 17, 24, 31, 38 etc.

So, the smallest value common from both the equations, or the smallest postiive value of N is 31.
This is the only value below 35 and K is positive so, You can add 4 to it to get 35. K = 4.

(B) is the answer.

[killer1387]

I don't quite understand the solution presented by the OG quant book on problem #68 for PS

the question is

when positive integer N is divided by 5, the reminder is 1. when N is divided by 7, the remainder is 3. What is the smaller positive integer K such that K + N is a multiple of 35?


A)3
B)4
C)12
D)32
E)35

whats the best /quickest way to solve this?

best,
c

by 5 R=1,
by 7 R=3,
n=31
k=4

Hence B