Quote:
Originally Posted by Archive
a) To solve this, we need to find all numbers x where 2a=x, 3b=x and 5c=x.
These are all primes, so we could try multiplying.
2*3*5=30
4*3*5=60
6*3*5=90
8*3*5=120
We see a pattern here. Since it must end in 0 or 5, it must be even, and divisible by 3, this is easy. The intersection is I=(30,60,90,120,150...) all the way up to 1,000.
Is it right, somedude? I'll do the rest later...

You just divide 1000 by 30 and round down.