Probability of Divisibility

Problem

probability of divisibility

Answer

Answer: $1.$

You may want to consult the solution below.

Solution

The problem is obviously contrived. The last two digits of the longish number are $76,$ implying that the number is divisible by $4$ and $\displaystyle 396/4 = 99 = 9\times 11.$ We need to check that, however the ten digits are distributed over the empty spaces, the resulting number is divisible by $9$ and also by $11.$

A number is divisible by $9$ iff the sum of its digits is divisible by $9.$ The sum of the missing digits is

$0 + 1 + 2 + ... + 9 = 45.$

Which is divisible by $9.$ The sum of the present digits is

$\begin{align}5 + 3 + 8 + 3 + 8 + 2 + 9&+ 3 + 6 + 5\\ &\qquad+ 8 + 2 + 0 + 3 + 9 + 3 + 7 + 6 = 90\end{align}$

Which is also divisible by $9.$ It follows that the resulting number is divisible by $9$ regardless of the placement of the missing digits.

To check whether a number is divisible by $11$ we compute two sums: that of the evenly placed digits and the sum of the oddly placed digits. All the missing digits come together along with $8,$ $3,$ $0,$ and $6,$ which add up to $\displaystyle 45 + 17 = 62.$ The sum of the remaining digits is the $\displaystyle 90 - 17 = 73.$ The difference of the two sums is $\displaystyle 73 - 62 = 11$ which is divisible by $11$, and we are finished.

Reference

  1. C. W. Trigg, Mathematical Quickies, Dover, 1985, #56

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