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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 $m \leq 10^8$.

Thanks in advance.

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  • $\begingroup$ $9657 = f(7569) = f(87 \times 87) \ne f(87) \times f(87) = 78 \times 78 = 6084$. $\endgroup$
    – Aeryk
    Commented May 6, 2013 at 19:18
  • $\begingroup$ $f(87 \times 87) \neq f(87) \times f(87)$, so we have "$f(87 \times 87) = f(87) \times f(87) \implies 8,7 \in \{0,1,2,3\}$". $\endgroup$
    – user12806
    Commented May 6, 2013 at 19:24
  • $\begingroup$ You claim that this is true for $m < 10^8$, but why does for example $m = 32$ work? I have that $f(32\times 32) = 4201 \neq f(32)\times f(32) = 529$ even though for 32 one has $a_0,a_1\in\{0,1,2,3\}$. Is there an extra assumption that is missing here? $\endgroup$
    – ARupinski
    Commented May 6, 2013 at 23:28
  • $\begingroup$ @ARupinski, OP alleges A implies B, you present a counterexample to B implies A. $\endgroup$
    – Gerry Myerson
    Commented May 7, 2013 at 6:17
  • $\begingroup$ @Gerry Myerson: thanks for clearing that up. Obviously I read the formulation through too quickly without thinking about what it was asserting. $\endgroup$
    – ARupinski
    Commented May 8, 2013 at 0:04

2 Answers 2

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At OEIS we have "Skinny numbers: numbers $n$ such that there are no carries when $n$ is squared by long multiplication." It gives as an equivalent formulation, $R(n^2) = R(n)^2$, where $R(n)$ means the digit reversal of $n$, and it says, "The decimal expansion of a skinny number $n$ may contain only 0's, 1's, 2's and 3's." However, it niether gives nor cites a proof.

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    $\begingroup$ That's because no carries is a stronger condition which gives an almost immediate characterzation of the digits: any digit bigger than 3 involves a carry in the decimal system when squared. Gerhard "Ask Me About System Design" Paseman, 2013.05.06 $\endgroup$
    – Gerhard Paseman
    Commented May 7, 2013 at 6:54
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Way back when I was in school, I had investigated number with these properties. These numbers are a special case of a more general property of numbers satisfying $R(mn) =R(m)R(n)$ where $R(n)$ means the digits reversal of $n$.

Examples

12*13 = 156, 21*31 = 651

101*102 = 10302, 101*103 = 20301

12012*11212 = 134678544, 21021*21211 = 445876431

11013*10212 = 112464756, 31011*21201 = 657464211

Infinitely many such numbers can be constructed but as Gerry Myerson said, the digits will be 0's, 1's, 2's or 3's.

A more challenging problem would be to study numbers with the property that

$R(n_1 n_2 \ldots n_k) =R(n_1)R(n_2)\ldots R(n_k)$.

Some interesting question in this direction:

  1. What is the largest $k$ for which we will find a solution?

  2. What is the largest $k$ for which we will find a solution if all $n_i$'s are equal? i.e. what is the largest $k$ for which $R(n^k) =R(n)^k$ has a solution?

For some reason I stopped at $k=2$. May be someone might want to revisit the problem.

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  • $\begingroup$ Thanks for the answer. But why are the digits lower or equal than $3$ ? $\endgroup$
    – user12806
    Commented May 8, 2013 at 19:18
  • $\begingroup$ In base $b$ , let $m$ be maximum digit of numbers satisfying your $k$ relation. Then $m^k \lt b$. So the general solution is any digit is at most the $k$'th root of $b$. $\endgroup$
    – Jérôme JEAN-CHARLES
    Commented Jun 18 at 7:52

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