Why are multiples of 11 palindromes?

Question

One thing I noticed is that for any integer $-10<a<10$, $11a$ is always a palindrome. I'm assuming this is because 11 is the first row of Pascal's triangle. For that same reason, for any nonnegative integer $n<5$, $(11a)^n$ is a palindrome.

Is there a proof for this? Is there a proper formula that proves that multiples of 11 are palindromes?

But that's not true: put $a=15$, e.g. Then $11\cdot 15 = 165$ — amWhy, May 02 '17 at 18:41
$;11\cdot132=1452;$ ...doesn't look too palindromic to me... — DonAntonio, May 02 '17 at 18:41
It is true, however, that all palindromes with an even number of digits are divisible by $11$. — Marcus Andrews, May 02 '17 at 18:41
For which multiples? Aside from $1$ to $9$, for which integer $n$ is $11n$ a palindrome? — Mark Viola, May 02 '17 at 18:44
@Dr.MV $11\cdot 11$ is a palindrome. As is $11^3$ and $11^4$. — D Wiggles, May 02 '17 at 18:48
@Dr.MV Like I said in the question - (11a)^n for -10<a<10 and -1<n<5. — DonielF, May 02 '17 at 18:59
@MarcusStuhr 121 doesn't contain an even number of digits. Neither does 14641. — DonielF, May 02 '17 at 19:00
@DonielF Sometimes a palindrome with an odd number of digits can be divisible by $11$, but not all of them. My claim was that all palindromes with an even length are divisible by $11$. — Marcus Andrews, May 02 '17 at 19:06

score 1 · Answer 1 · answered May 02 '17 at 19:15

Since $11 = 10 + 1$, the binomial theorem gives $$ 11^{n} = (10 + 1)^{n} = \sum_{k=0}^{n} \binom{n}{k} 10^{n - k}. $$ If $0 \leq n < 5$, then $\binom{n}{k} < 10$, i.e., is represented base $10$ by a single digit. Concretely, \begin{align*} 11^{2} &= 121, \\ 11^{3} &= 1331, \\ 11^{4} &= 14641, \\ 11^{5} &= 161051. \end{align*} (At the fifth power, there's carrying from $\binom{5}{2} = \binom{5}{3} = 10$.)

The question about $11a$ is similar but trivial: If $0 < a < 10$ is a single digit base $10$, then $11a = 10a + a$ has decimal representation $aa$.

It's not true that $(11a)^{n}$ is a palindrome, e.g., $22^{5} = 5153632$.

Why are multiples of 11 palindromes?

1 Answers1