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# What is the smallest natural number, n, having exactly 100 factors?

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- stephen mLv 41 decade agoFavorite Answer
If you write n in its prime factorisation:

2^(a1) * 3^(a2) * 5^(a3) * ...

then the factors are formed by choosing a power of 2 between 0 and a1 (which is a1+1 choices), a power of 3 between 0 and a2 (which is a2+1 choices).. etc.

So the total number of factors is (a1+1)(a2+1)(...

We need that to be 100. So lets look at how we can factorise 100.

Leaving it as 100 gives a1 = 99, or 2^99, which is big.

50*2 means a1 = 49, a2 = 1, which gives 2^49 * 3.

You'll see we really want to split 100 into quite a few factors.

100 is 5*5*2*2. That means a1 = 4, a2 = 4, a3 = 1, a4 = 1, which gives 2^4 * 3^4 * 5^1 * 7^1 = 45360.

If you try other combinations, you'll see you can't beat that.

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