Henry Ernest Dudeney/Puzzles and Curious Problems/113 - A Complete Skeleton/Solution

by : $113$

 * A Complete Skeleton

Solution
300324   -- 333)100007892      999        1078         999        -          799          666          1332          1332

10356  --- 29)300324    29     103      87      162      145       174       174       ---

Proof
Let us first consider the first division.

Let $d$ denote the divisor.

Let $q$ denote the quotient.

Let $n$ denote the dividend.

Let $n_1$ to $n_4$ denote the partial dividends which are subject to the $1$st to $4$th division operations respectively.

Let $p_1$ to $p_4$ denote the partial products generated by the $1$st to $4$th division operations respectively.

We see that the difference $n_2 - p_2$ is a $2$-digit number.

Therefore $n_2$ must begin with $1$ and $p_2$ with $9$.

So the difference $n_1 - p_1$ is exactly $1$.

Since $n_1$ has $4$ digits and $p_1$ has $3$, we have:
 * $n_1 = 1000$
 * $p_1 = 999$

$d$ is a $3$-digit divisor of $999$.

Since $p_4$ has $4$ digits, $d$ cannot be $111$.

Since $n_3$ has only $3$ digits and $n_3 - p_3 \ge 10$, $p_3 \ne 999$ and $d \ne 999$.

Hence we get $d = 333$.

Now $p_1 = p_2 = 999$ and thus $q$ must start with $3003$.

Since $p_3$ has $3$ digits but is not $999$, the tens digit of $q$ must be $1$ or $2$.

Therefore $300 \, 310 \le q \le 300 \, 329$.

Now we turn our attention to the second division.

Since $n_2$ has $3$ digits, $p_2$ has only $2$ digits and so does $n_2 - p_2$, $n_2$ must begin with a $1$.

Thus $p_1 = 30 - 1 = 29$.

Since $29$ is prime, the divisor must be $29$.

Using this information, we now know that the quotient $q$ of the first division is divisible by $29$.

Among the numbers $300 \, 310 \le q \le 300 \, 329$, the only multiple of $29$ is:
 * $q = 300 \, 324$

As we already found the divisors of both divisions, the rest of the numbers can be filled in accordingly.