## Puzzle of the week [26]

* Le Monde* weekend puzzle for the past week (I have not peeked at the solution yet!) was quite straightforward: find those

*n*‘s such that

*2*divides

^{n}*n!*and those

*n*‘s for which

*2*divides

^{n-1}*n*. Looking at the problem in the plane to Montpellier, I think that the solution is that no positive

*n*exists such that

*2*divides n! and powers of 2 are those numbers for which

^{n}*2*divides

^{n-1}*n*.

**M**y reasoning is

- that the numbers with the highest potential is a power of 2,
- that
*2*divides n! when^{n-1}*n*is of the form*2*, and^{m} - therefore that no integer n can satisfy the harder constraint.

Proving that *2 ^{n-1}* divides

*n!*when

*n*is of the form

*n=2*can be done by induction: it works for

^{m}*n=2*and if it works for

*2*, then it works for

^{m}*n=2*by considering a separation of

^{m+1}*n!*into

and by using the induction assumption that *(2 ^{m})!* can be divided by

Recycling the dividers for the second part leads to its being divisible by

because the very last term in the factorial is by *2 ^{m+1}*, which can be divided by

*2*… Proving that integers other than the

^{m+1}*2*‘s cannot be divided by

^{m}*2*again works by an induction proof on

^{n-1 }*m*.

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