# Why are multiples of 11 palindromes?

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?

• $11a$ is not always a palindrome. e.g. $11 \cdot 10 = 110$ – Marcus Andrews May 2 '17 at 18:40
• But that's not true: put $a=15$, e.g. Then $11\cdot 15 = 165$ – Namaste May 2 '17 at 18:41
• $\;11\cdot132=1452\;$ ...doesn't look too palindromic to me... – DonAntonio May 2 '17 at 18:41
• You're correct - let me update this. – DonielF May 2 '17 at 18:41
• It is true, however, that all palindromes with an even number of digits are divisible by $11$. – Marcus Andrews May 2 '17 at 18:41

## 1 Answer

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., $22^{5} = 5153632$.