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

Source Work

1997: David Wells: Curious and Interesting Numbers (2nd ed.):

The Dictionary

$2185$

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:

\(\ds 2185\)

\(=\)

\(\ds 5 \times 19 \times 23\)

\(\ds 2186\)

\(=\)

\(\ds 2 \times 1093\)

\(\ds 2187\)

\(=\)

\(\ds 3^7\)

\(\ds 2188\)

\(=\)

\(\ds 2^2 \times 547\)

\(\ds 2189\)

\(=\)

\(\ds 11 \times 199\)

\(\ds 2190\)

\(=\)

\(\ds 2 \times 3 \times 5 \times 73\)

\(\ds 2191\)

\(=\)

\(\ds 7 \times 313\)

\(\ds 2192\)

\(=\)

\(\ds 2^4 \times 137\)

\(\ds 2193\)

\(=\)

\(\ds 3 \times 17 \times 43\)

\(\ds 2194\)

\(=\)

\(\ds 2 \times 1097\)

\(\ds 2195\)

\(=\)

\(\ds 5 \times 439\)

\(\ds 2196\)

\(=\)

\(\ds 2^2 \times 3^2 \times 61\)

\(\ds 2197\)

\(=\)

\(\ds 13^3\)

\(\ds 2198\)

\(=\)

\(\ds 2 \times 7 \times 157\)

\(\ds 2199\)

\(=\)

\(\ds 3 \times 733\)

\(\ds 2200\)

\(=\)

\(\ds 2^3 \times 5^2 \times 11\)

\(\ds 2201\)

\(=\)

\(\ds 31 \times 71\)

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.

Sources

• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $2185$