Help with PS#68 from Quant Review 2nd edition.

jeremy8 wrote:"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+1 is a multiple of 35?"

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

Solution B

The explanation in the guide is not very clear to me and seems to not be the optimal way to get the answer.

What I did instinctively under time pressure was: n=5k+1 and n=7k+3.
5k+1=7k+3 and k=-1
Therefore, substituting, n=-4. Since 0 is a multiple of every number; for k=4, n+k=0, which is also the right answer.

I'm really curious. Is what I did a correct other way to get the rigth answer or is it completely wrong and it just happens to lead to the right answer?

jeremy8 wrote:"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+1 is a multiple of 35?"

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

Solution B

The explanation in the guide is not very clear to me and seems to not be the optimal way to get the answer.

What I did instinctively under time pressure was: n=5k+1 and n=7k+3.
5k+1=7k+3 and k=-1
Therefore, substituting, n=-4. Since 0 is a multiple of every number; for k=4, n+k=0, which is also the right answer.

I'm really curious. Is what I did a correct other way to get the rigth answer or is it completely wrong and it just happens to lead to the right answer?

outreach wrote:is the Q properly worded
there are two unrelated variable k and n

selango wrote:I think the question must be n+k is a multiple of 35.

n is different when divided by 5 and 7.So they need to be expressed by 2 diff variables.

n=5a+1

n=6,11,16,21,26,31......

n=7b+3

n=10,17,24,31,38..............

n+k is a multiple of 35.So n+k=35q

Try substituting value of k from options.Let start with the least one.

k=3,n=35q-3

n=32,67,102

But n does not satisfy both the equations.

k=4,n=35q-4

n=31,66....31,66 is available in both the equations and satisfy.If u check,all values satisfy both the equations.

Hence k=4

Pick B