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Question: What are the smallest and largest multiples of 11 that are made up of the nine digits 1,2,3,4,5,6,7,8,9?

123456789

123456798

123456879

123456897

123457689

123457698

123457869

123457896

123457968

123457986

..................

..................

..................

987654312

987654321

Answer:

If $A=a_1a_2a_3a_4a_5a_6a_7a_8a_9$ is a multiple of 11, then we know the difference $(a_1+a_3+a_5+a_7+a_9)-(a_2+a_4+a_6+a_8)$ is also a multiple of 11, say $11x$.

Note that $a_1+a_2+\cdots+a_9=1+2+\cdots+9=45$ and hence $a_1+a_3+a_5+a_7+a_9=(45+11x)/2$ and $a_2+a_4+a_6+a_8=(45-11x)/2$.

Note that if $x=3$, then $a_2+a_4+a_6+a_8=6$ which is impossible. Likewise $x\ne -3$. Therefore $x$ can only be $1$ or $-1$.

Therefore either $a_1+a_3+a_5+a_7+a_9=28, a_2+a_4+a_8+a_9=17$ or $a_1+a_3+a_5+a_7+a_9=17, a_2+a_4+a_8+a_9=28$.

To get the smallest multiple of 11, we would like to see $A=12345\ldots$. Both $1+3+5+a_7+a_9=28$ and $1+3+5+a_7+a_9=17$ are impossible (the latter one is possible only if $\{a_7, a_9\}=\{2,6\}$ but we already have $a_2=6$.).

Try $A=1234\ldots$. Now $1+3+a_5+a_7+a_9=28$ is possible if $\{a_5,a_7,a_9\}=\{7,8,9\}$ (and so $\{a_6,a_8\}=\{5,6\}$) but $1+3+a_5+a_7+a_9=17$ is impossible. Therefore the smallest multiple of 11 is $A=123475869$.

To get the largest multiple of 11, we would like to see $A=98765\ldots$. Now $9+7+5+a_7+a_9=28$ is possible if $\{a_7,a_9\}=\{3,4\}$ (and so $\{a_6,a_8\}=\{1,2\}$). Thus the largest multiple of 11 is $A=987652413$.