17 Consecutive Integers each with Common Factor with Product of other 16

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Theorem

The $17$ consecutive integers from $2184$ to $2200$ have the property that each one is not coprime with the product of the other $16$.


Proof

We obtain the prime decomposition of all $17$ of these integers:

\(\displaystyle 2184\) \(=\) \(\displaystyle 2^3 \times 3 \times 7 \times 13\)
\(\displaystyle 2185\) \(=\) \(\displaystyle 5 \times 19 \times 23\)
\(\displaystyle 2186\) \(=\) \(\displaystyle 2 \times 1093\)
\(\displaystyle 2187\) \(=\) \(\displaystyle 3^7\)
\(\displaystyle 2188\) \(=\) \(\displaystyle 2^2 \times 547\)
\(\displaystyle 2189\) \(=\) \(\displaystyle 11 \times 199\)
\(\displaystyle 2190\) \(=\) \(\displaystyle 2 \times 3 \times 5 \times 73\)
\(\displaystyle 2191\) \(=\) \(\displaystyle 7 \times 313\)
\(\displaystyle 2192\) \(=\) \(\displaystyle 2^4 \times 137\)
\(\displaystyle 2193\) \(=\) \(\displaystyle 3 \times 17 \times 43\)
\(\displaystyle 2194\) \(=\) \(\displaystyle 2 \times 1097\)
\(\displaystyle 2195\) \(=\) \(\displaystyle 5 \times 439\)
\(\displaystyle 2196\) \(=\) \(\displaystyle 2^2 \times 3^2 \times 61\)
\(\displaystyle 2197\) \(=\) \(\displaystyle 13^3\)
\(\displaystyle 2198\) \(=\) \(\displaystyle 2 \times 7 \times 157\)
\(\displaystyle 2199\) \(=\) \(\displaystyle 3 \times 733\)
\(\displaystyle 2200\) \(=\) \(\displaystyle 2^3 \times 5^2 \times 11\)

It can be seen by inspection that each of the integers in this sequence shares at least one prime factor with at least one other.


It is then worth noting that:

\(\displaystyle 2183\) \(=\) \(\displaystyle 37 \times 59\)
\(\displaystyle 2201\) \(=\) \(\displaystyle 31 \times 71\)

and it can be seen that the sequence can be extended neither upwards nor downwards.

$\blacksquare$


Sources