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

Source Work

 * The Dictionary
 * $2185$
 * $2185$

Mistake

 * The start of a sequence of $17$ consecutive integers, each of which has a common factor, greater than $1$, with the product of the remaining $16$.

It is easiest to see this is wrong if one is first to obtain the prime decomposition of all $17$ of these integers:

We have that:
 * $2197 = 13^3$, and that none of the other $16$ have $13$ as a divisor.


 * $2201 = 31 \times 71$: none of the other $16$ has either $31$ or $71$ as a divisor.

Thus neither $2197$ nor $2201$ have a common factor, greater than $1$, with the product of the remaining $16$.

The key number here is that the sequence of $17$ starts at $2184$ and not $2185$:
 * $2184 = 2^3 \times 3 \times 7 \times 13$

which then provides that necessary common factor with $2197$, and allows $2201$ to be removed from the other end of the sequence.