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I want to know whether octal palindromes with even number digits (11 or 1221, but not 121) are all composite numbers, and a general proof if so or a counterexample if not.

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closed as off-topic by Scientifica, Gibbs, Xander Henderson, Parcly Taxel, ArsenBerk Nov 4 '18 at 8:24

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  • $\begingroup$ I m not sure what do you mean by octal palindromes, btw eany palindrome with even difit is multiple of 11, you can see it be the criteria of divisibility by 11, i.e. the sum of digit in even position must be equal to the sum of digit of odd position $\endgroup$ – ALG Nov 3 '18 at 19:20
  • $\begingroup$ Does "octal" here mean that you are writing in base $8$? $\endgroup$ – lulu Nov 3 '18 at 19:20
  • $\begingroup$ Yes, it's a number in base 8 which happens to be a palindrome. $\endgroup$ – bbayu Nov 3 '18 at 19:42
  • $\begingroup$ Let me clarify something, I want to know why octal palindromes, for example 1221, when converted into decimal, so 657, will result in it being a composite number. $\endgroup$ – bbayu Nov 3 '18 at 19:49
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let us consider any number in base 8, $a_n 8^n +,\dots, + a_0$ observe that if $n$ is even then $ a^n \equiv 1 \;\text{mod} 9$ and if $n$ is odd then $ a^n \equiv -1\; \text{mod} 9$ then write the number mod 9, it became $-a_n + a_{n-1}+ \dots + a_0$ if $n$ is even ( or with different sign for $n$ odd). In any case we get argue as in the case of base 10 to show that any octal palindrome number with enev number of digit is multiple of $9$. In general if you consider a number in base $b$ palindrome and with a even number of digits it will be a multiple of $b+1$

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  • $\begingroup$ I have difficulty understanding it coherently. Do you mind explaining in more details? $\endgroup$ – bbayu Nov 3 '18 at 19:53
  • $\begingroup$ Let us consider first the case in base 10, so let's pick a number let's say 1245, now this number can be written as $1\times 10^3+ 2\times 10^2 + 4\times 10 +5 $, now we are going to divide the number by 11 and study what it the reminder, observe that if we divide $10^{2n}$ by $11$ the rest is 1 and if we divide $10^{2n+1}$ by 11 the remaninder is 10 or since it is more useful -1, in other words you can write $10^{2n+1}=11\times k +1$ and $10^{2n+1} = 11\times h -1$. At the end when we divide our number 1245 by 11 we obtain 1*(-1)+2*(1) + 4*(-1) + 5, and we want this number to be 0. $\endgroup$ – ALG Nov 4 '18 at 0:20

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