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While driving to work I thought about a rule, problem is I don't know how to prove it.

The rule is:

For any base $B$ there will always be two palindrome numbers $A$ and $C$, whose values are: $A=(B+1), C=(B-1)$

$A$ is always $11$ - a palindrome, and $C$ will always have one digit - a palindrome.


Examples:

Base Ten: (B=10, A=B+1=11, C=B-1=9); they both are palindrome.

Binary: (B=2,A=11,C=1)


My Question: How can I formally prove that? am I wrong?

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  • $\begingroup$ There is nothing to show there. The number $B+1$ is always $11_B$ as you noted, a palindrome in base $B$. Note that this works much easier: Take any palindrome but read it as a polynomial, i.e. $101 \hat= x^2 + 1 =: p(x)$. Then the number $p(B)$ will always be a palindrome in base $B$ as long as $p$s highest coefficient is at most $B-1$. $\endgroup$ – AlexR Mar 10 '15 at 20:50
  • $\begingroup$ Examples: $(b+1)^2 = b^2 + 2b + 1$ will be a base-$b$-palindrome for $b \ge 3$, because $(b+1)^2 = 121_b$ as long as $b \ge 3$. $b^n+1 = 1\!\underbrace{0\ldots0}_{n-1\text{ zeros}}\!1_b$ will be a palindrome in any base. $\endgroup$ – AlexR Mar 10 '15 at 20:52
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The most probably cause of your confusion is the simplicity of the proof. Indeed, there is not much more to show than:

In base $B$, the number $B-1$ is written as a single digit: $(B-1)_B$. The number $B+1$ is written as $11_B$. Therefore, $B-1$ and $B+1$ are palindromes in base $B$ for all choices of $B$.

And indeed, that is just a more dressed up way of what you wrote in your question :).

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  • $\begingroup$ By looking at your reputation I guess you right :) $\endgroup$ – yossico Mar 12 '15 at 15:54

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