Henry Ernest Dudeney/Modern Puzzles/71 - A Complete Skeleton/Solution
Modern Puzzles by Henry Ernest Dudeney: $71$
- A Complete Skeleton
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Solution
1011.1008 --------- 625)631938 625 --- 693 625 --- 688 625 --- 63 0 62 5 ---- 5000 5000 ----
Proof
Let $d$ denote the divisor.
Let $q$ denote the quotient.
Let $n$ denote the dividend.
Let $n_1$ to $n_5$ denote the partial dividends which are subect to the $1$st to $5$th division operations respectively.
Let $p_1$ to $p_5$ denote the partial products generated by the $1$st to $5$th division operations respectively.
As Dudeney suggested, the $3$ zeroes are dropped down into $n_5$ and $p_5$.
Another zero is dropped into $n_4$.
This also gives us the last two but one digits of $q$.
Thus $d$ is a divisor of a multiple of $1000$.
The divisors of $d$ can only be $5$ or $2$ up to a multiplicity of $3$ each, or $x$ where $x < 10$.
Because $d$ has $3$ digits, one divisor must be $5$, and the last digit must therefore be $5$ or $0$.
The subtraction from the final $0$ in $n_4$ giving a digit in the least significant digit of the difference means it must be $5$.
$d$ cannot have $2$ as a divisor or $d$ will end in $0$.
It follows immediately that $d = 5^4 = 625$, and that all of $p_1$ to $p_4$ are also $625$.
The remaining digits of $q$ are now obviously either $1$ or $0$:
1011.1008 --------- 625)****** 625 --- *** 625 --- *** 625 --- ** 0 62 5 ---- 5000 5000 ----
and $n$ follows by multiplying $d$ by $q$.
$\blacksquare$
Sources
- 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $71$. -- A Complete Skeleton
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $145$. A Complete Skeleton