a0 + a1 . 10 + a2 . 102 + ... + an-1 . 10n-1 + an . 10n
Then
an + an-1 . 10 + an-2 . 102 + ... + a1 . 10n-1 + a0 . 10n
is "the same number written backwards." The difference is—
(an - a0)(10n - 1) + (an-1 - a1)(10n-2 - 1) . 10 + ...
+ (an/2+1 - an/2-1)(102-1) . 10n/2-1 if n be even, but
+ (a(n+1)/2 - a(n-1)/2)(10-1) . 10(n-1)/2 if n be odd.
And every term of this difference, as involving a factor of the form (1 - 10n), is divisible by 9; and therefore the difference is divisible by 9.
C. Mansfield Ingleby.
Birmingham.