TITLE: Divisibility question from a very old textbook.
QUESTION [11 upvotes]: I have been looking at a rather old-fashioned book "The Tutorial Arithmetic" (1947), and have been amused by some of the questions in the "Harder Problems" section at the back of the book (some are hilarious by modern standards). Clearly the book was written before any sensible electronic/mechanical aids were available for multi-precision arithmetic, so I was a bit surprised to see the following problem:

"show that the number 9000080000700006000050000400002 is divisible by 41 even though all the digits except the zeros be interchanged". (sic)

I assume he means that any permutation of the non-zero digits are allowed.
I could solve this with the aid of Pari/gp (although it would be tedious), but am at a loss as to any (sensible) approach to this problem that only uses methods available to school pupils back in 1947.
If anyone can provide a neat solution, I would be interested.

REPLY [18 votes]: $$  10^5 \equiv 1 \pmod {41}   $$
$$ 9+8+7+6+5+4+2 = 41 $$