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Old 04-28-2007   #2
Temperal
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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...