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:
| \(\ds 2184\) | \(=\) | \(\ds 2^3 \times 3 \times 7 \times 13\) | ||||||||||||
| \(\ds 2185\) | \(=\) | \(\ds 5 \times 19 \times 23\) | ||||||||||||
| \(\ds 2186\) | \(=\) | \(\ds 2 \times 1093\) | ||||||||||||
| \(\ds 2187\) | \(=\) | \(\ds 3^7\) | ||||||||||||
| \(\ds 2188\) | \(=\) | \(\ds 2^2 \times 547\) | ||||||||||||
| \(\ds 2189\) | \(=\) | \(\ds 11 \times 199\) | ||||||||||||
| \(\ds 2190\) | \(=\) | \(\ds 2 \times 3 \times 5 \times 73\) | ||||||||||||
| \(\ds 2191\) | \(=\) | \(\ds 7 \times 313\) | ||||||||||||
| \(\ds 2192\) | \(=\) | \(\ds 2^4 \times 137\) | ||||||||||||
| \(\ds 2193\) | \(=\) | \(\ds 3 \times 17 \times 43\) | ||||||||||||
| \(\ds 2194\) | \(=\) | \(\ds 2 \times 1097\) | ||||||||||||
| \(\ds 2195\) | \(=\) | \(\ds 5 \times 439\) | ||||||||||||
| \(\ds 2196\) | \(=\) | \(\ds 2^2 \times 3^2 \times 61\) | ||||||||||||
| \(\ds 2197\) | \(=\) | \(\ds 13^3\) | ||||||||||||
| \(\ds 2198\) | \(=\) | \(\ds 2 \times 7 \times 157\) | ||||||||||||
| \(\ds 2199\) | \(=\) | \(\ds 3 \times 733\) | ||||||||||||
| \(\ds 2200\) | \(=\) | \(\ds 2^3 \times 5^2 \times 11\) |
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:
| \(\ds 2183\) | \(=\) | \(\ds 37 \times 59\) | ||||||||||||
| \(\ds 2201\) | \(=\) | \(\ds 31 \times 71\) |
and it can be seen that the sequence can be extended neither upwards nor downwards.
$\blacksquare$
Sources
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $2185$