For example, for the numbers 1 to 10, one can just find the necessary factors and multiply them: $5 \times 7 \times 8 \times 9 = 2520$, and all the other numbers in that range follow. But with larger ranges, larger numbers logically result. Is there a simpler way to find this smallest number other than straight multiplication filtering?
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1Sorry; that didn't show up in the related questions as I was asking this. – Lee Sleek May 06 '13 at 14:42
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1It isn't easy to search for unless you know what to look for. I knew of the OEIS page and searched for that. That version has some good answers. – Ross Millikan May 06 '13 at 14:44
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1@Lee No apology is necessary. – MJD May 06 '13 at 14:48
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Notice that for the numbers $1$ through $10$, inclusive, we have that $5\times 7\times 8 \times 9 = 2^3 \times 3^2 \times 5\times 7\times = 2520$ and every number $2$ to $10$ can be written as the product of some combination of the given factors. This is due to the fact that:
What you can say is that the least integer divisible by every element in $\{1, 2, \cdots, n\}$ is equal to the least common multiple $\operatorname{lcm}(1, 2, \cdots, n)$.
amWhy
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I was looking for more of a formula, so this is the best answer. In other words, multiply all the prime numbers less than n, raised to the highest power less than n. – Lee Sleek May 07 '13 at 03:37
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Yes, indeed, you're correct: in the example above, $2^3$ covers $2, 4, 8$; $3^2$ covers $3, 9,$ $5\times 2$ covers 10, and 7 covers 7 – amWhy May 07 '13 at 03:47