Henry Ernest Dudeney/Modern Puzzles/66 - Another 37 Division/Solution

by : $66$

 * Another $37$ Division

Solution
The chances are $1$ in $40$ that the number will be divisible by $37$.

Proof
Suppose $n = \sqbrk {abcdefghi}$ is a haphazard arrangement of the $9$ digits divisible by $37$.

Since the sum of digits of $n$ is $45$, which is divisible by $9$:
 * $9 \divides n$ by Divisibility by $9$.

By Divisibility by $37$:
 * $37 \divides n \iff 37 \divides \paren {\sqbrk {abc} + \sqbrk {def} + \sqbrk {ghi} }$

Thus $\sqbrk {abc} + \sqbrk {def} + \sqbrk {ghi}$ would be divisible by $9$ and $37$, and hence divisible by their product, $333$.

Note that:

Thus $\sqbrk {abc} + \sqbrk {def} + \sqbrk {ghi}$ is one of:
 * $999, 1332, 1665, 1998, 2331$

In any case of $\sqbrk {abc} + \sqbrk {def} + \sqbrk {ghi} = \sqbrk {wxyz}$, we have:

For each of the cases above, we determine the possible values of $c_1, c_2$ and find the sets of digits $\set {a, d, g}, \set {b, e, h}, \set {c, f, i}$ that satisfies the relations above.

Also the sum of any three distinct digits cannot exceed $7 + 8 + 9 = 24$ or be less than $1 + 2 + 3 = 6$.

Case $1$: $999$
In this case, we have:
 * $45 = 9 \paren {c_1 + c_2} + 0 + 9 + 9 + 9$

which leads to:
 * $c_1 + c_2 = 2$.

Neither $c_1$ nor $c_2$ could be equal to $2$ since the sum of three distinct digits cannot reach $27$.

Hence:
 * $c_1 = c_2 = 1$
 * $a + d + g = 8$
 * $b + e + h = 18$
 * $c + f + i = 19$

For $\set {a, d, g} = \set {1, 2, 5}$, the possibilities for $\set {b, e, h}$ are:
 * $\set {3, 6, 9}$
 * $\set {3, 7, 8}$
 * $\set {4, 6, 8}$

For $\set {a, d, g} = \set {1, 3, 4}$, the possibilities for $\set {b, e, h}$ are:
 * $\set {2, 7, 9}$
 * $\set {5, 6, 7}$

so there are $5$ total possibilities.

Case $2$: $1332$
In this case, we have:
 * $45 = 9 \paren {c_1 + c_2} + 10 + 3 + 3 + 2$

which leads to:
 * $c_1 + c_2 = 3$.

Neither $c_1$ nor $c_2$ could be equal to $3$ since the sum of three distinct digits cannot exceed $30$.

Hence:
 * $\set {c_1, c_2} = \set {1, 2}$

so either:
 * $a + d + g = 12$
 * $b + e + h = 22$
 * $c + f + i = 11$

or:
 * $a + d + g = 22$
 * $b + e + h = 11$
 * $c + f + i = 12$

the difference between these two cases is simply an exchange of place values, so we only need to consider the second case.

For $\set {a, d, g} = \set {5, 8, 9}$, the possibilities for $\set {b, e, h}$ are:
 * $\set {1, 3, 7}$
 * $\set {1, 4, 6}$
 * $\set {2, 3, 6}$

For $\set {a, d, g} = \set {6, 7, 9}$, the possibilities for $\set {b, e, h}$ are:
 * $\set {1, 2, 8}$
 * $\set {2, 4, 5}$

so there are $5 \times 2 = 10$ total possibilities.

Case $3$: $1665$
In this case, we have:
 * $45 = 9 \paren {c_1 + c_2} + 10 + 6 + 6 + 5$

which leads to:
 * $c_1 + c_2 = 2$.

Since $5$ and $25$ cannot be a sum of distinct digits, we must have $c_1 = 1$.

Thus $c_2 = 1$ as well.

Hence:
 * $a + d + g = b + e + h = c + f + i = 15$

The possible sets of three distinct digits that sums to $15$ are:
 * $\set {1, 5, 9}$
 * $\set {1, 6, 8}$
 * $\set {2, 4, 9}$
 * $\set {2, 5, 8}$
 * $\set {2, 6, 7}$
 * $\set {3, 4, 8}$
 * $\set {3, 5, 7}$
 * $\set {4, 5, 6}$

There is only two groups of digits that satisfies $a + d + g = b + e + h = c + f + i = 15$:
 * $\set {\set {a, d, g}, \set {b, e, h}, \set {c, f, i} } = \set {\set {1, 5, 9}, \set {2, 6, 7}, \set {3, 4, 8} }$
 * $\set {\set {a, d, g}, \set {b, e, h}, \set {c, f, i} } = \set {\set {1, 6, 8}, \set {2, 4, 9}, \set {3, 5, 7} }$

The three sets of digits can freely swap between themselves.

So there are $2 \times 3! = 12$ total possibilities.

Case $4$: $1998$
In this case, we have:
 * $45 = 9 \paren {c_1 + c_2} + 10 + 9 + 9 + 8$

which leads to:
 * $c_1 + c_2 = 1$.

Hence:
 * $\set {c_1, c_2} = \set {0, 1}$

so either:
 * $a + d + g = 18$
 * $b + e + h = 8$
 * $c + f + i = 19$

or:
 * $a + d + g = 8$
 * $b + e + h = 19$
 * $c + f + i = 18$

By our analysis in Case $1$, there are $5 \times 2 = 10$ total possibilities.

Case $5$: $2331$
In this case, we have:
 * $45 = 9 \paren {c_1 + c_2} + 20 + 3 + 3 + 1$

which leads to:
 * $c_1 + c_2 = 2$.

Neither $c_1$ nor $c_2$ could be equal to $0$ since the sum of three distinct digits cannot be less than $4$.

Hence:
 * $c_1 = c_2 = 1$
 * $a + d + g = 11$
 * $b + e + h = 12$
 * $c + f + i = 22$

By our analysis in Case $1$, there are $5$ total possibilities.

We see that in all the above cases, there are $42$ total possibilities.

All the digits within those sets can be freely exchanged.

Therefore the number of possibities of $n$ is:
 * $42 \times 3! \times 3! \times 3! = 9072$

And thus the probability that a haphazard arrangement of the $9$ digits would be divisible by $37$ is:
 * $\dfrac {9072} {9!} = \dfrac 1 {40}$