For this problem your gonna have to learn to manipulate the form D=qd+R where D=dividend q=quotiend d=divisor and R= remainder
Substituting the info given about n using the above equation we have
n=5a+1 (where a is the quotient)
n=7b+3 (where b is the quotient).......(1)
Equating both sides we get
5a+1=7b+3
7b=5a-2
7b=5c+5-2 (where 5c=5a-5....This another manipulation of the above form that i spoke about)
7b=5c+3
Comparing this with the form D=kd+R we have 7b when divided by 5 gives a quotient of c and remainder of 3. When does a number divided by 5 leave a remainder of 3? As all multiples of 5 end in 5 or 10 the only numbers that will leave a remainder of 3 when divided by 5 are numbers ending in 3 or 8. So 7 b is some multiple of 7 ending in 3 or 8. Lets list them down
7b could be 28,63,98,133,203 etc
aka
7b could be 7(4),7(9),7(19),7(29) etc
So b could be 4,9,19,29 etc
We pick the smallest value from among them and put this in eqn1 we get
n=28+3
n=31
Now some number k needs to be added to n that will make it equal some multiple of 35. The smallest one is 4.
The other way as Mitch has shown. List all the possible numbers which when divided by 5 leave a remainder of 1. eg 1,6,11,16,21,26,31,35 etc
List all the possible numbers which when divided by 7 leave a remainder of 3 3,10,17,24,31,33,40 etc
The smallest number common to both lists is 31 and when 4 is added to it, it becomes a multiple of 35.
