Puzzles

15, 18 and 45 are elastic.

15's factors are 5 and 3. 105, 1005, 10005, etc will all be multiples of 5 (because they end in 5) and multiples of 3 (as their digits add to 6). Hence they are all multiples of 15.

Similarly, 108, 1008, 10008, etc are all multiples of 9 (adding digits) and 2 (they are even), so they are multiples of 18; and 405, 4005, 40005, etc are all multiples of 9 (adding digits) and 5 (last digits are 5), so they are multiples of 45.

How many elastic numbers are there in other bases?

Yes: 8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9, 16.

It is clearly possible for the numbers 1 to 15 (remove the 16 from the end of the sequence above) and 1 to 17 (add 17 to the start of the sequence above).

The OEIS sequence A090461 gives other numbers for which this is possible. It starts 15, 16, 17, 23, then includes every number from 25 onwards. It is conjectured, but not proven, that it is possible for every number above 25.

Yes. It can be shown by induction:

First it's easy to check that \(1\times1!=2!-1\). Next assume that \(1\times1!+2\times2!+...+k\times k!=(k+1)!-1\). Now we try to show the pattern holds for \(k+1\).

Hence, by induction, the pattern holds for all \(k\geq1\).

The first player can always win by first placing a plate on the exact centre of the table. Then the first player can copy what the second player does, but rotated 180°, and hence can always place a plate if the second player could.

What if the two players play on a regular hexagonal table? Or a regular octagonal table? Or a regular pentagonal table? Or a regular \(n\)-gonal table?

The murderer was Ms. Trois (3). The motive was To worry others (2). The location was Network room (2). The weapon was Antique candlestick (7).

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