by RonPurewal Sun Sep 30, 2007 3:57 am
Remember the prime box concept: if the number you want is divisible by X, then all the prime factors of X have to go into the prime box. If it's also divisible by Y, and some of the factors of Y are also factors of X, then remember that those factors DON'T have to appear in the prime box twice.
The first seven positive integer multiples of 5, when factored into primes, are
5
2x5
3x5
2x2x5
5x5
2x3x5
7x5
Using the principle outlined above, the prime box for the desired number must contain:
two 2's (because of the 2x2x5)
one 3
two 5's (because of 5x5)
one 7
so the number is 2x2x3x5x5x7 = 10x10x3x7 = 2100.
Answer: D
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Note that if you're really fast at long division, you can just try dividing by all 7 numbers. I've seen students perform long division many orders of magnitude faster than I can, so it's always a possibility...