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Question 1035190: Suppose that p, q and r are distinct prime integers such that the product N = pq × qr × rp has a units digit of zero. What is the greatest number of consecutive zeros that can appear to the right of the last nonzero digit of N?
Answer by Edwin McCravy(20062)   (Show Source):
You can put this solution on YOUR website!
There can only be 2 zeros on the end. Here's why.
The only way to get a zero on the end is to multiply by 10.
10 has prime factors 2 and 5. With N = pq × qr × rp, there
are two factors p, two factors q, and two factors r. So one
pair of like factors must be 2's and another pair 5's.
Regardless of what the third prime factor is, we can have 00
on the end only if we have a factor of 100 = 2×2×5×5. The
smallest number that would have two 0's on the end is
N = pq × qr × rp = (2×3) × (3×5) × (5×2) = 900.
All such integers are of the form (2×3) × (3×r) × (r×2) where
r is any prime larger than 3. No factor of 1000 is possible,
for that would take a product of three 2's and three 5's.
Edwin
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