Henry Ernest Dudeney/Modern Puzzles/64 - Dividing by Eleven/Solution

by : $64$

 * Dividing by Eleven

Solution

 * $115$ to $11$ against a haphazard arrangement being divisible by $11$.

Proof
Let $\sqbrk {abcdefghi}$ be a haphazard arrangement of the $9$ digits.

We set:
 * $m = a + c + e + g + i$
 * $n = b + d + f + h$

From Divisibility by 11, $\sqbrk {abcdefghi}$ is divisible by $11$ :
 * $m - n = 11k$

for some integer $k$.

We have:

Moreover, since $m + n = 45$ is odd, so is $m - n = \paren {m + n} - 2n$.

Hence we can only have $m - n = \pm 11$, which give the following cases when solved:
 * Case $1$: $m = 28, n = 17$
 * Case $2$: $m = 17, n = 28$

For Case $1$, the following $9$ quadruples of distinct digits give $n = 17$:
 * $1, 2, 5, 9$
 * $1, 2, 6, 8$
 * $1, 3, 4, 9$
 * $1, 3, 5, 8$
 * $1, 3, 6, 7$
 * $1, 4, 5, 7$
 * $2, 3, 4, 8$
 * $2, 3, 5, 7$
 * $2, 4, 5, 6$

For Case $2$, the following $2$ quintuples of distinct digits give $m = 17$:
 * $1, 2, 3, 4, 7$
 * $1, 2, 3, 5, 6$

Hence there are exactly:
 * $\paren {9 + 2} \times 4! \times 5! = 31680$

numbers, out of $9! = 362880$ possibilities that our number is divisible by $11$.

This gives a probability of:
 * $\dfrac {31680} {362880} = \dfrac {11} {126}$

which is exactly $115$ to $11$.