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Find 100 Consecutive Composite Numbers

Outwardly, this may seem a bit difficult to accomplish, but think about the sequence

    \[101! + 2, 101!+3, \stackrel{100}{\ldots}, 101!+100, 101!+101\]

Here, we can easily show that k|(101!+k)\ \forall\ 2\le k \le 101.

    \begin{align*} 101! &= 2 \times 3 \times \ldots \times k \times \ldots \times 100 \times 101 \\ &= \left(2 \times 3 \times \ldots \times (k-1) \times (k+1) \times \ldots \times 100 \times 101\right) k \\ &\iff k|101! \end{align*}

Since k|101! and k|k trivially, then it must be the case that k|(101!+k) and all one hundred terms of the sequence \{101! + n\}_{n=2}^{101} are composite. \blacksquare

In fact, this can be generalized. To find k consecutive composite numbers, the same line of reasoning can be used to show that the sequence \{(k+1)! + n\}_{n=2}^{k+1} contains all composite numbers. This also naively bounds the rate of growth of the spread of primes to be at least \mathbb{O}(n!) for n the spread of primes..

Posted By Lee

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