Henry Ernest Dudeney/Puzzles and Curious Problems/110 - An Absolute Skeleton/Solution/Initial Deductions

by : $110$

 * An Absolute Skeleton

Declarations
It is apparent from the structure of the long division sum presented that no digit of $Q$ may be zero.

We also have, from the additional constraint on $Q$, that:

This also gives us that:

The rule about repeated digits offers similar constraints on $n_2$ to $n_8$ and $p_1$ to $p_8$, and will be used without further comment in the following.

However, note that this does not apply to $n_1$, which in this context is the first $4$ digits of $N$, which has no constraint on the uniqueness of its digit.

We have:

This gives us a firm upper bound on $N$, and we can immediately state:


 * $(1): \quad 10 \, 000 \, 000 \, 000 \le N \le 10 \, 859 \, 999 \, 999$

Hence also:

Thus we have:
 * $902 \le p_1 \le 987$

and thus we have the first digit of $p_1$.

Hence we have:

******** ***)10*********     9**      ---       ***

We have that $p_1$, $p_2$ and $p_4$ each have $3$ digits.

But each of $q_1$, $q_2$ and $q_4$ are distinct.

Hence:

We have that $p_3$, $p_5$, $p_6$, $p_7$ and $p_8$ each have $4$ digits.

But each of $q_3$, $q_5$, $q_6$, $q_7$ and $q_8$ are distinct

Hence:

That is:
 * $247 \le D \le 329$

Thus we have established bounds on $D$ and $N$.

Hence we can now establish bounds on $Q$:

So we immediately see that $q_1 = 3$ or $q_1 = 4$.

Suppose $q_1 = 4$.

Then:

But we already have that $p_1 \le 987$.

So $q_1 \ne 4$.

This directly gives us that $q_1 = 3$.

We also have that:

Hence by the same reasoning:
 * $q_2 < 4$ and $q_4 < 4$

and so either:
 * $q_2 = 1$ and $q_4 = 2$

or:
 * $q_2 = 2$ and $q_4 = 1$

Now let us consider the lower bound and upper bound for $Q$.

Recall:

and that $q_1$ to $q_8$ are unique.

We also have that:

This gives us the limits on $Q$:


 * $31 \, 427 \, 586 \le Q \le 32 \, 918 \, 675$

Now:

Similarly:

Thus we have revised our bounds on $D$:
 * $304 \le D \le 329$

This in turns gives:

and we have so far:

3******* 3**)10*********     9**      ---       ***

Suppose $q_2 = 2$.

Then by the mechanics of a long division:

and using the same mechanics:

Otherwise, that is if $q_2 = 1$ and $q_4 = 2$: