Question 770320
N-35 to be perfectly divisible by all those numbers:


12: 2*2*3
18: 2*3*3
20: 2*2*5
21: 3*7
28: 2*2*7
30: 2*3*5


This is like looking for a lowest common denominator.  You need 2*2*3*3*5*7.
That product is 1260.

You want {{{(N-35)/P=M}}} where M is a WHOLE number.  The largest number which needs to divide {{{N-35}}} is listed as 30, so,
{{{(N-35)/30=1260/30}}},... intuition partly at work here, so difficult to logically explain better, but find N.