Palindromic Numbers divisible by 11 (Was: Re: computing 20! all the way)
Palindromic Numbers divisible by 11 (Was: Re: computing 20! all the way)
- Subject: Palindromic Numbers divisible by 11 (Was: Re: computing 20! all the way)
- From: "Jonathan Levi, M.D." <email@hidden>
- Date: Sat, 13 Nov 2004 22:42:05 -0500
(Warning: this letter is OT, as was the question from Chris that
inspired it.)
(From Applescript-users Digest, Vol 1, Issue 196)
On Fri, 12 Nov 2004 20:42:44 -0800, Christopher Nebel
<email@hidden> wrote:
... Curiously, all
palindromic 4-, 6-, and 8-digit numbers are divisible by 11-- can
anyone explain why? Is this true for all even lengths?)
Well, for a 4-digit palindromic number p=abba (a x 10^3 + b x 10^2 + b
x 10 + a)
= 1001a + 110b
= 11 x 91 x a + 11 x 10 x b
=11(91a + 10b)
For a 6-digit palindromic number q = abccba
q = 100001a + 1001 x 10 x b + 11 x 100 x c
Since we know that 1001 is divisible by 11, q is divisible by 11 if
100001 is; but 100001 = 11 x 9091, so q is indeed divisible by 11.
More generally, a palindromic number r with an even number of digits is
a sum
r = A0 + A1 + A2 + ... + An,
where each Ai is a product of an integer from 0 to 9, a power of 10,
and a number of the form 11, 1001, 100001 (more generally, a number of
the form 10^(2i+1) + 1). But the latter number equals 11 x (90 90
90...90 + 1), where 90 is repeated i times. Hence each of the Ai is
divisible by 11, therefore so is r, and the answer to your original
question is yes.
Thanks for the opportunity to flex long-unused mathematical muscles,
Jonathan
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