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Edited by Alkis It has been posted twice!
On 12/2/2017 at 8:14 PM, andrei said:
A palindrome is a number which reads the same backward or forward (e.g. 434, 87678, etc.). Could you prove that for any integer n (not divisible by 10) there is a palindrome divisible by n?
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I've checked for all numbers up to 162, it's true:
81* 12345679= 999999999
172839506*162=27999999972
Is there any simple proof for any integer?
Indeed, 81 is the first number that requires a long divisor its palindrome is very long. I didn't have the patience to go up to nine digits, as you did. I stopped when the quest for a palindrome reached 8 digits (It took alreadt 32 secs!). In a PC I checked all numbers 1-1000 and, here are the numbers with >=8-digit result (not sulution!): 81, 162, 243, 324, 405, 486, 567, 648, 655, 729, 891, 972. There are no many and may there is a reason for that. For one thing they are multiples of 81, with 1-11 as multipliers ... So, it happens that 81 is the "mystery" number!
Now, it seems there's indeed a palindrome for every number. But how indeed can it be prooven (if it can)?
On 12/2/2017 at 8:14 PM, andrei said:
(My comment has been posted twice - See previous one
Palindromes by multiplication
in New Logic/Math Puzzles
Posted · Edited by Alkis
It has been posted twice!
Indeed, 81 is the first number that requires a long divisor its palindrome is very long. I didn't have the patience to go up to nine digits, as you did. I stopped when the quest for a palindrome reached 8 digits (It took alreadt 32 secs!). In a PC I checked all numbers 1-1000 and, here are the numbers with >=8-digit result (not sulution!): 81, 162, 243, 324, 405, 486, 567, 648, 655, 729, 891, 972. There are no many and may there is a reason for that. For one thing they are multiples of 81, with 1-11 as multipliers ... So, it happens that 81 is the "mystery" number!
Now, it seems there's indeed a palindrome for every number. But how indeed can it be prooven (if it can)?
(My comment has been posted twice - See previous one