Here's a fun problem to chew on.
I assert that all 4digit palindromic numbers are multiples of 11. Can you prove or disprove this assertion?
5 Replies  7219 Views  Last Post: 21 April 2013  12:53 AM
#1
Number Theory Challenge: 4digit palindromic numbers
Posted 18 April 2013  08:37 AM
Replies To: Number Theory Challenge: 4digit palindromic numbers
#2
Re: Number Theory Challenge: 4digit palindromic numbers
Posted 18 April 2013  10:39 AM
Since the set of 4digit palindromic numbers is finite (and not even very large), this could easily proven or disproven by exhaustive search, but here's a different approach:
Spoiler
#3
Re: Number Theory Challenge: 4digit palindromic numbers
Posted 18 April 2013  10:58 AM
Nice one, that's exactly the logic I used. But then I started working at it, and I believe it's further provable that this is true for any base b. That is, xFAAF/x11 = xebf exactly.
Oddly enough, this seems to be related to the fact that the digital sum of a multiple of (b1) in base b always sums to b1.
Oddly enough, this seems to be related to the fact that the digital sum of a multiple of (b1) in base b always sums to b1.
#4
Re: Number Theory Challenge: 4digit palindromic numbers
Posted 18 April 2013  12:43 PM
POPULAR
x*10^3 + y*10^2 + y*10 + x = 11 * (x*91 + y*10)
This post has been edited by Nikitin: 18 April 2013  12:45 PM
#5
Re: Number Theory Challenge: 4digit palindromic numbers
Posted 19 April 2013  09:15 PM
jon.kiparsky, on 18 April 2013  08:37 AM, said:
Here's a fun problem to chew on.
I assert that all 4digit palindromic numbers are multiples of 11. Can you prove or disprove this assertion?
I assert that all 4digit palindromic numbers are multiples of 11. Can you prove or disprove this assertion?
Let x = a*10^3 + b*10^2 + b*10 + a, where 0 < a,b < 10, a and b are integers.
x = a*1001 + b*110
Since 111001 and 11110, 11(a*1001 + b*110) for any integer a,b (and specifically for 0<a,b<10)
(This is from rule 7. http://primes.utm.ed...ge/divides.html )
Thus 11x, which means x is a multiple of 11.
By construction of x, x is any 4digit palindromic number, which means any 4digit palindromic number is a multiple of 11.
#6
Re: Number Theory Challenge: 4digit palindromic numbers
Posted 21 April 2013  12:53 AM
While your proof is good, your range is off. It should be 0 < a < 10 and 0 <= b < 10.
Page 1 of 1
