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For all $n=\overline{a_k a_{k-1}\ldots a_1 a_0} := \sum_{i=0}^k a_i 10^i\in \mathbb{N}$, where $a_i \in \{0,...,9\}$ and $a_k \neq 0$,

we define $f(n)=\overline{a_0 a_1 \ldots a_{k-1} a_k}= \sum_{i=0}^k a_{k-i}10^i$.

Is it true that, for all $m=\overline{a_k a_{k-1}\ldots a_1 a_0} \in \mathbb{N}$, we have

$f(m\times m)=f(m)\times f(m) \implies$$\forall i \in \{0, \ldots, k\}, a_i \in \{0,1,2,3\}$ ?

Example: $f(201)\times f(201)=102 \times 102=10404=f(40401)=f(201\times 201)$.

It's true for $m \leq 10^8$.

francis-jamet
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1 Answers1

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If $m=...4$, then $m^2=...6$, but $f(m)=4...$ and $f(m)^2=1...$ or $2...$ (because $4^2=16$ and $5^2=25$).

The same can be calculated explicitly for $m$ ending in $5, \ldots, 8$, and only a little bit different for $9$.

If $m=...9$, then $m^2=...1$, but $f(m)=9...$ and $f(m)^2=8...$ or $9...$ not $1...$ (as $9^2=81$, $10^2=100$ and the inequality is strict).

evgeny
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