Enigma 1742 - Chip-chop

[Chappy]

Enigma 1742 - Chip-chop
New Scientist magazine, 30 March 2013.
By Adrian Somerfield.

I have a seven-figure number that uses seven
consecutive digits in some order. Starting at
the left and deleting a digit leaves a six-
digit number, then deleting the right-end digit
leaves a five-digit number, then deleting the
left-hand end digit leaves a four-digit number,
and so on, alternating sides until a single
digit is left. Looking at the list of seven
numbers obtained in this way I see that they
are all odd, and that only the six-digit number
is divisible by 3 (but not 9). Surprisingly,
if I had carried out the process starting by
deleting the right-hand end digit and then the
left, and so on down to a single digit, all the
above facts would still be true.

What number did I start with?

Ciao,
Chappy.

[Ted Schuerzinger]

The seven-digit number has to end with four odd digits, meaning either
1234567 is used, or 3456789. Except that 3456789 sums to 42, which is
divisible by 3, and so 1234567 must be the digits used.

Only subtracting 4 from the left yields a sum divisible by 3, and
subtracting 7 from the right leaves a 6-digit number divisible by 3 but
not 9. So far we've got:

4XXXXX7

To make numbers not divisible by 3, we with a bit of trial and error get
4625317.

Praise be to modular arithmetic!

--
Ted S.
fedya at hughes dot net
Now blogging at http://justacineast.blogspot.com