Henry Ernest Dudeney/Modern Puzzles/59 - The Two Digits/Solution

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Modern Puzzles by Henry Ernest Dudeney: $59$

The Two Digits
Write down any $2$-figure number (different figures and no $0$)
and then express that number by writing the same figures in reverse order,
with or without arithmetical signs.


Solution

Examples given by Dudeney:

\(\ds 15\) \(=\) \(\ds 5 \div \sqrt {\cdotp \dot 1}\)
\(\ds 4^2\) \(=\) \(\ds 2^4\)
\(\ds 24\) \(=\) \(\ds \paren {\sqrt 4 + 2}!\)
\(\ds 25\) \(=\) \(\ds 5^2\)
\(\ds \) \(=\) \(\ds 5 \div {\cdotp 2}\)
\(\ds 28\) \(=\) \(\ds \dbinom 8 2\) Definition of Binomial Coefficient
\(\ds 36\) \(=\) \(\ds 6 \times 3!\)
\(\ds 64\) \(=\) \(\ds \paren {\sqrt 4}^6\) \(\ds \text {or } \sqrt {4^6}\)
\(\ds 71\) \(=\) \(\ds \sqrt {1 + 7!}\) see Brocard's Problem


It is worth pointing out that the second inadequate example quoted by Dudeney himself:

$81 = \paren {1 + 8}^2$

which is ineligible because of the $2$, leads us on neatly into:

$\sqrt {81} = 1 + 8$

which Dudeney unaccountably missed.


There may be more.


Matt Westwood offers up:

$94 \equiv 4 \pmod 9$

which may or may not be eligible.

And indeed:

\(\ds 90\) \(\equiv\) \(\ds 0\) \(\ds \pmod 9\)
\(\ds 91\) \(\equiv\) \(\ds 1\) \(\ds \pmod 9\)
\(\ds 92\) \(\equiv\) \(\ds 2\) \(\ds \pmod 9\)

and so on.


Sources

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