# Find the lowest number that's $\geq N$ and that multiplying it with a set of numbers results in natural numbers

Given a set of numbers, I need to find the lowest number that multiplying it with each of the numbers in the set results in a natural number, while being bigger or equal to $N$.

For example, for the numbers $2.2$, $3$ and $4.2$, the result could be $5$, $10$, $15$, etc as they can multiply $2.2$ and $4.2$ and result in natural numbers. Then, I need the lowest one that's higher than a given number.

I'm trying to implement it as a software algorithm in an efficient way. How would I go about doing that?

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First find the least common multiple of the denominators. You have to multiply by that to clear the fractions, so do so. Then if $a$ is the smallest number after multiplying by the LCM, divide $N$ by $a$ and round up to the next whole number. Call that $b$. The multiplier for the original set is $b*LCM$
@shesek: It would be clearer to make it so. I was thinking that $a$ was the original smallest number, which is why I said to divide by $a * lcm$. Now it will just be $a$. I will update – Ross Millikan Oct 21 '12 at 22:28
@shesek: the LCM is taken over denominators. So if you have 3 and 7 and $N=50$, the LCM of the denominators is 1, then $b=17$ – Ross Millikan Oct 21 '12 at 23:02