
02.4. Eva, Per and Anna play with their pocket calculators. They choose different integers and check, whether or not they are divisible by 11. They only look at nine-digit numbers consisting of all the digits 1, 2, .., 9. Anna claims that the probability of such a number to be a multiple of 11 is exactly 1/11. Eva has a different opinion: she thinks the probability is less than $1 / 11$. Per thinks the probability is more than $1 / 11$. Who is correct?


Solution. We write the numbers in consideration, $n=a_{0}+10 a_{1}+10^{2} a_{2}+\cdots+10^{8} a_{8}$, in the form

$$
\begin{gathered}
a_{0}+(11-1) a_{1}+(99+1) a_{2}+(1001-1) a_{3} \\
+(9999+1) a_{4}+(100001-1) a_{5}+(999999+1) a_{6} \\
\quad+(10000001-1) a_{7}+(99999999+1) a_{8} \\
=\left(a_{0}-a_{1}+a_{2}-a_{3}+a_{4}-a_{5}+a_{6}-a_{7}+a_{8}\right)+11 k \\
=\left(a_{0}+a_{1}+\cdots+a_{8}\right)-2\left(a_{1}+a_{3}+a_{5}+a_{7}\right)+11 k \\
=44+1+11 k-2\left(a_{1}+a_{3}+a_{5}+a_{7}\right)
\end{gathered}
$$

So $n$ is divisible by 11 if and only if $2\left(a_{1}+a_{3}+a_{5}+a_{7}\right)-1$ is divisible by 11 . Let $s=a_{1}+a_{3}+a_{5}+a_{7}$. Then $1+2+3+4=10 \leq s \leq 6+7+8+9=30$ and $19 \leq 2 s-1 \leq 59$. The only multiples of 11 in the desired interval are 33 and 55 , so $s=17$
or $s=28$. If $s=17$, the smallest number in the set $A=\left\{a_{1}, a_{3}, a_{5}, a_{7}\right\}$ is either 1 or 2 $(3+4+5+6=18)$. Checking the cases, we see that there are 9 possible sets $A:\{2,4,5,6\}$, $\{2,3,5,7\},\{2,3,4,8\},\{1,4,5,7\},\{1,3,6,7\},\{1,3,5,8\},\{1,3,4,9\},\{1,2,6,8\}$, and $\{1,2,5,9\}$. If $s=28$, the largest number in $A$ is $9(5+6+7+8=26)$ and the second largest $8(5+6+7+9=27)$. The only possible $A$ 's are $\{4,7,8,9\}$ and $\{5,6,8,9\}$. The number of different ways to choose the set $A$ is $\binom{9}{4}=\frac{9 \cdot 8 \cdot 7 \cdot 6}{2 \cdot 3 \cdot 4}=126$. Of these, the number of choices leading to a number which is a multiple of 11 is $9+2=11$. This means that the probability of picking a number which is divisible by 11 is $\frac{11}{126}<\frac{11}{121}=\frac{1}{11}$. So Eva's opinion is correct.
