Consider the collection of all 5-digit numbers whose sum of digits is 43. One of these numbers is chosen at random. What is the probability that it is a multiple of 11?

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Solution

The maximum sum of the digits of a five-digit number is 45, which corresponds to the sum of the digits of 99999. For a number to have a digit sum of 43, there can be two cases: either it will have three digits equal to 9 and two equal to 8, or it will have four digits equal to 9 and one equal to 7. In the first case, we can choose the position of the first digit 8 in 5 ways and the position of the second in 4 ways. However, since the two digits are indistinguishable, each configuration has been counted twice. The digits 9 must be placed in the remaining positions, which gives us $\frac{5 \cdot 4}{2}=10$ numbers. In the second case, we can choose the position of the digit 7 in 5 ways. Consequently, there are $10+5=15$ five-digit numbers such that the sum of their digits is 43. We now need to recall the divisibility rule for 11:

A number of $n$ digits $\overline{a_{n} a_{n-1} a_{n-2} \ldots a_{1}}$ is divisible by 11 if and only if the alternating sum $a_{n}-a_{n-1}+a_{n-2}-a_{n-3}+\ldots+(-1)^{n} a_{1}$ is divisible by 11.

By the previous criterion, the five-digit number $\overline{a_{5} a_{4} a_{3} a_{2} a_{1}}$ is divisible by 11 if 11 divides $a_{5}-a_{4}-a_{3}+a_{2}-a_{1}=\left(a_{5}+a_{4}+a_{3}+a_{2}+a_{1}\right)-2\left(a_{3}+a_{1}\right)$. Assuming that the sum of its digits is 43 and knowing that $43=11 \cdot 3+10$, it follows that 11 must divide $11 \cdot 3+2\left(5-a_{3}-a_{1}\right)$. Therefore, $a_{3}+a_{1}$ leaves a remainder of 5 when divided by 11, and the only possibilities are $\left(a_{3}, a_{1}\right)=$ $(7,9),(9,7)$ or $(8,8)$. Thus, of the 15 numbers in the collection, only 3 of them are multiples of 11, and consequently, the desired probability is $\frac{3}{15}=\frac{1}{5}$.