Henry Ernest Dudeney/Modern Puzzles/52 - The Five Cards/Solution
Modern Puzzles by Henry Ernest Dudeney: $52$
- The Five Cards
- I have $5$ cards bearing the figures $1$, $3$, $5$, $7$ and $9$.
- How can I arrange them in a row so that the number formed by the $1$st pair multipied by the number formed with the last pair,
- with the central number subtracted,
- will produce a number composed of repetitions of one figure?
Solution
- $\boxed 3 \boxed 9 \ \boxed 1 \ \boxed 5 \boxed 7$
or:
- $\boxed 5 \boxed 7 \ \boxed 1 \ \boxed 3 \boxed 9$
Proof
Let $d_1$ and $d_2$ be the two $2$-f numbers at either end.
Let $s$ be the single-digit subtrahend.
Let $n$ be the repdigit that results from $d_1 \times d_2 - s$.
Suppose $n$ is odd.
Then $n + s = d_1 \times d_2$ is even.
But because $d_1$ and $d_2$ are both odd, this is impossible.
So $n$ is even.
Notice that:
- $15 \times 37 \le d_1 d_2 \le 93 \times 75$
that is:
- $555 \le d_1 d_2 \le 6975$
Hence we must have that $n$ is one of:
- $666, 888, 2222, 4444, 6666$
This leaves a small enough domain to perform an exhaustive search.
We factorize all $30$ possibilities of $n + s$, and filter the results using these criteria:
- $(1)$ Those with a prime factor greater than $100$ cannot be expressed as $d_1 \times d_2$.
- $(3)$ Numbers divisible by $89$ should also be removed as no multiple of $89$ is expressible as $d_1$ or $d_2$.
We have:
| \(\ds 667\) | \(=\) | \(\ds 23 \times 29\) | fails $(2)$ | |||||||||||
| \(\ds 669\) | \(=\) | \(\ds 3 \times 223\) | fails $(1)$ | |||||||||||
| \(\ds 671\) | \(=\) | \(\ds 11 \times 61\) | fails $(2)$ | |||||||||||
| \(\ds 669\) | \(\text {is}\) | \(\ds \text {prime}\) | fails $(1)$ | |||||||||||
| \(\ds 675\) | \(=\) | \(\ds 3^3 \times 5^2\) | ||||||||||||
| \(\ds 889\) | \(=\) | \(\ds 7 \times 127\) | fails $(1)$ | |||||||||||
| \(\ds 891\) | \(=\) | \(\ds 3^4 \times 11\) | ||||||||||||
| \(\ds 893\) | \(=\) | \(\ds 19 \times 47\) | fails $(2)$ | |||||||||||
| \(\ds 895\) | \(=\) | \(\ds 5 \times 179\) | fails $(1)$ | |||||||||||
| \(\ds 897\) | \(=\) | \(\ds 3 \times 13 \times 23\) | ||||||||||||
| \(\ds 2223\) | \(=\) | \(\ds 3^2 \times 13 \times 19\) | ||||||||||||
| \(\ds 2225\) | \(=\) | \(\ds 5^2 \times 89\) | fails $(3)$ | |||||||||||
| \(\ds 2227\) | \(=\) | \(\ds 17 \times 131\) | fails $(1)$ | |||||||||||
| \(\ds 2229\) | \(=\) | \(\ds 3 \times 743\) | fails $(1)$ | |||||||||||
| \(\ds 2231\) | \(=\) | \(\ds 23 \times 97\) | fails $(2)$ | |||||||||||
| \(\ds 4445\) | \(=\) | \(\ds 5 \times 7 \times 127\) | fails $(1)$ | |||||||||||
| \(\ds 4447\) | \(\text {is}\) | \(\ds \text {prime}\) | fails $(1)$ | |||||||||||
| \(\ds 4449\) | \(=\) | \(\ds 3 \times 1483\) | fails $(1)$ | |||||||||||
| \(\ds 4451\) | \(\text {is}\) | \(\ds \text {prime}\) | fails $(1)$ | |||||||||||
| \(\ds 4453\) | \(=\) | \(\ds 61 \times 73\) | fails $(2)$ | |||||||||||
| \(\ds 6667\) | \(=\) | \(\ds 59 \times 113\) | fails $(1)$ | |||||||||||
| \(\ds 6669\) | \(=\) | \(\ds 3^3 \times 13 \times 19\) | ||||||||||||
| \(\ds 6671\) | \(=\) | \(\ds 7 \times 953\) | fails $(1)$ | |||||||||||
| \(\ds 6673\) | \(\text {is}\) | \(\ds \text {prime}\) | fails $(1)$ | |||||||||||
| \(\ds 6675\) | \(=\) | \(\ds 3 \times 5^2 \times 89\) | fails $(3)$ |
This leaves:
- $675, 891, 897, 2223, 6669$
For $675$, it is divisible by $25$.
Since there is only one $5$ available, only one of $d_1, d_2$ can be divisible by $5$, and thus $25$.
The only multiple of $25$ not involving even digits, $75$, would result in:
- $675 = 75 \times 9$
so $675$ has no such expression.
For $891$, it is divisible by $11$.
All $2$-digit multiples of $11$ have repeated digits.
Therefore $891$ has no such expression.
For $897$, both $23$ and $23 \times 3 = 69$ have even digits.
Therefore $897$ has no such expression.
For $2223$, since:
- $13 \times 19 > 3^2 \times 19 > 3^2 \times 13 > 100$
we are forced to use:
- $\set {d_1, d_2} = \set {3 \times 13, 3 \times 19} = \set {39, 57}$
from which we derive our solutions:
- $\boxed 3 \boxed 9 \ \boxed 1 \ \boxed 5 \boxed 7$
or:
- $\boxed 5 \boxed 7 \ \boxed 1 \ \boxed 3 \boxed 9$
For $6669$, $13$ and $19$ cannot be multiplied together as their product is larger than $100$.
By Pigeonhole Principle, one of $13$ or $19$ must receive at least $3^2$ during multiplication.
But as shown before:
- $3^2 \times 19 > 3^2 \times 13 > 100$
so this is not possible.
Therefore $6667$ has no such expression.
We have exhausted all possibilities, so there are no more solutions.
$\blacksquare$
Sources
- 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $52$. -- The Five Cards
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $103$. The Five Cards