Henry Ernest Dudeney/Modern Puzzles/52 - The Five Cards/Solution

by : $52$

 * The Five Cards

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 $d_1 \times d_2$.


 * $(2)$ Any semiprime must be removed if their prime factorization involves even digits.


 * $(3)$ Numbers divisible by $89$ should also be removed as no multiple of $89$ is expressible as $d_1$ or $d_2$.

We have:

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.