Products of 2-Digit Pairs which Reversed reveal Same Product

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Theorem

The following positive integers can be expressed as the product of $2$ two-digit numbers in $2$ ways such that the factors in one of those pairs is the reversal of each of the factors in the other:

$504, 756, 806, 1008, 1148, 1209, 1472, 1512, 2016, 2208, 2418, 2924, 3024, 4416$


Proof

Let $n \in \Z_{>0}$ such that:

$n = \sqbrk {a b} \times \sqbrk {c d} = \sqbrk {b a} \times \sqbrk {d c}$

where $\sqbrk {a b}$ denotes the two-digit positive integer:

$10 a + b$ for $0 \le a, b \le 9$

from the Basis Representation Theorem.


We have:

\(\ds \paren {10 a + b} \paren {10 c + d}\) \(=\) \(\ds \paren {10 b + a} \paren {10 d + c}\)
\(\ds \leadsto \ \ \) \(\ds 100 a c + 10 \paren {a d + b c} + b d\) \(=\) \(\ds 100 b d + 10 \paren {b c + a d} + a c\)
\(\ds \leadsto \ \ \) \(\ds 99 a c\) \(=\) \(\ds 99 b d\)
\(\ds \leadsto \ \ \) \(\ds a c\) \(=\) \(\ds b d\)


Thus the problem boils down to finding all the sets of one-digit integers $\set {a, b, c, d}$ such that $a c = b d$, and so that:

$n = \sqbrk {a b} \times \sqbrk {c d} = \sqbrk {b a} \times \sqbrk {d c}$

and also:

$n = \sqbrk {a d} \times \sqbrk {b c} = \sqbrk {d a} \times \sqbrk {c b}$


Thus we investigate all integers whose divisor count is $3$ or more, and find all those which have the product of single-digit integers in $2$ ways, as follows:

\(\ds \map {\sigma_0} 4\) \(=\) \(\ds 3\) $\sigma_0$ of $4$
\(\ds \leadsto \ \ \) \(\ds 4\) \(=\) \(\ds 1 \times 4\)
\(\ds \) \(=\) \(\ds 2 \times 2\)
\(\ds \leadsto \ \ \) \(\ds 12 \times 42\) \(=\) \(\ds 21 \times 24\)
\(\ds \) \(=\) \(\ds 504\)


\(\ds \map {\sigma_0} 6\) \(=\) \(\ds 4\) $\sigma_0$ of $6$
\(\ds \leadsto \ \ \) \(\ds 6\) \(=\) \(\ds 1 \times 6\)
\(\ds \) \(=\) \(\ds 2 \times 3\)
\(\ds \leadsto \ \ \) \(\ds 12 \times 63\) \(=\) \(\ds 21 \times 36\)
\(\ds \) \(=\) \(\ds 756\)
\(\ds 13 \times 62\) \(=\) \(\ds 31 \times 26\)
\(\ds \) \(=\) \(\ds 806\)


\(\ds \map {\sigma_0} 8\) \(=\) \(\ds 4\) $\sigma_0$ of $8$
\(\ds \leadsto \ \ \) \(\ds 8\) \(=\) \(\ds 1 \times 8\)
\(\ds \) \(=\) \(\ds 2 \times 4\)
\(\ds \leadsto \ \ \) \(\ds 12 \times 84\) \(=\) \(\ds 21 \times 48\)
\(\ds \) \(=\) \(\ds 1008\)
\(\ds 14 \times 82\) \(=\) \(\ds 41 \times 28\)
\(\ds \) \(=\) \(\ds 1148\)


\(\ds \map {\sigma_0} 9\) \(=\) \(\ds 3\) $\sigma_0$ of $9$
\(\ds \leadsto \ \ \) \(\ds 9\) \(=\) \(\ds 1 \times 9\)
\(\ds \) \(=\) \(\ds 3 \times 3\)
\(\ds \leadsto \ \ \) \(\ds 13 \times 93\) \(=\) \(\ds 31 \times 39\)
\(\ds \) \(=\) \(\ds 1209\)


\(\ds \map {\sigma_0} {10}\) \(=\) \(\ds 4\) $\sigma_0$ of $10$
\(\ds \leadsto \ \ \) \(\ds 10\) \(=\) \(\ds 1 \times 10\) and so does not lead to a solution
\(\ds \) \(=\) \(\ds 2 \times 5\)


Further integers $n$ such that $\map {\sigma_0} n \le 4$ need not be investigated, as one of the pairs of factors will be greater than $9$.


\(\ds \map {\sigma_0} {12}\) \(=\) \(\ds 6\) $\sigma_0$ of $12$
\(\ds \leadsto \ \ \) \(\ds 12\) \(=\) \(\ds 1 \times 12\) which does not lead to a solution
\(\ds \) \(=\) \(\ds 2 \times 6\)
\(\ds \) \(=\) \(\ds 3 \times 4\)
\(\ds \leadsto \ \ \) \(\ds 23 \times 64\) \(=\) \(\ds 32 \times 46\)
\(\ds \) \(=\) \(\ds 1472\)
\(\ds 24 \times 63\) \(=\) \(\ds 42 \times 36\)
\(\ds \) \(=\) \(\ds 1512\)


\(\ds \map {\sigma_0} {16}\) \(=\) \(\ds 5\) $\sigma_0$ of $16$
\(\ds \leadsto \ \ \) \(\ds 16\) \(=\) \(\ds 1 \times 16\) which does not lead to a solution
\(\ds \) \(=\) \(\ds 2 \times 8\)
\(\ds \) \(=\) \(\ds 4 \times 4\)
\(\ds \leadsto \ \ \) \(\ds 24 \times 84\) \(=\) \(\ds 42 \times 48\)
\(\ds \) \(=\) \(\ds 2016\)


\(\ds \map {\sigma_0} {18}\) \(=\) \(\ds 6\) $\sigma_0$ of $18$
\(\ds \leadsto \ \ \) \(\ds 18\) \(=\) \(\ds 1 \times 18\) which does not lead to a solution
\(\ds \) \(=\) \(\ds 2 \times 9\)
\(\ds \) \(=\) \(\ds 3 \times 6\)
\(\ds \leadsto \ \ \) \(\ds 23 \times 96\) \(=\) \(\ds 32 \times 69\)
\(\ds \) \(=\) \(\ds 2208\)
\(\ds 26 \times 93\) \(=\) \(\ds 62 \times 39\)
\(\ds \) \(=\) \(\ds 2418\)


\(\ds \map {\sigma_0} {20}\) \(=\) \(\ds 6\) $\sigma_0$ of $20$
\(\ds \leadsto \ \ \) \(\ds 20\) \(=\) \(\ds 1 \times 20\) which does not lead to a solution
\(\ds \) \(=\) \(\ds 2 \times 10\) which does not lead to a solution
\(\ds \) \(=\) \(\ds 4 \times 5\)


\(\ds \map {\sigma_0} {24}\) \(=\) \(\ds 8\) $\sigma_0$ of $24$
\(\ds \leadsto \ \ \) \(\ds 24\) \(=\) \(\ds 1 \times 24\) which does not lead to a solution
\(\ds \) \(=\) \(\ds 2 \times 12\) which does not lead to a solution
\(\ds \) \(=\) \(\ds 3 \times 8\)
\(\ds \) \(=\) \(\ds 4 \times 6\)
\(\ds \leadsto \ \ \) \(\ds 34 \times 86\) \(=\) \(\ds 43 \times 68\)
\(\ds \) \(=\) \(\ds 2924\)
\(\ds 36 \times 84\) \(=\) \(\ds 63 \times 48\)
\(\ds \) \(=\) \(\ds 3024\)


\(\ds \map {\sigma_0} {36}\) \(=\) \(\ds 9\) $\sigma_0$ of $36$
\(\ds \leadsto \ \ \) \(\ds 36\) \(=\) \(\ds 1 \times 36\) which does not lead to a solution
\(\ds \) \(=\) \(\ds 2 \times 18\) which does not lead to a solution
\(\ds \) \(=\) \(\ds 3 \times 12\) which does not lead to a solution
\(\ds \) \(=\) \(\ds 4 \times 9\)
\(\ds \) \(=\) \(\ds 6 \times 6\)
\(\ds \leadsto \ \ \) \(\ds 46 \times 96\) \(=\) \(\ds 64 \times 69\)
\(\ds \) \(=\) \(\ds 4416\)



This theorem requires a proof.
In particular: The challenge remains to prove, without going through all cases exhaustively, that there are no further pairs.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.
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Although this article appears correct, it's inelegant. There has to be a better way of doing it.
In particular: Why not write out the upper triangular part of the multiplication table for $1 \le n \le 9$, $n \ne 5, 7$, and find the common numbers in the table?
$5, 7$ can be eliminated via a simple argument $p x \ge 10$, and there are only $28$ numbers left to compare
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Improve}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.



Sources

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