Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
$9657 = f(7569) = f(87 \times 87) \ne f(87) \times f(87) = 78 \times 78 = 6084$. –  Aeryk May 6 '13 at 19:18
    
$f(87 \times 87) \neq f(87) \times f(87)$, so we have "$f(87 \times 87) = f(87) \times f(87) \implies 8,7 \in \{0,1,2,3\}$". –  user12806 May 6 '13 at 19:24
    
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? –  ARupinski May 6 '13 at 23:28
    
@ARupinski, OP alleges A implies B, you present a counterexample to B implies A. –  Gerry Myerson May 7 '13 at 6:17
    
@Gerry Myerson: thanks for clearing that up. Obviously I read the formulation through too quickly without thinking about what it was asserting. –  ARupinski May 8 '13 at 0:04

2 Answers 2

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.

share|improve this answer
    
Thanks for the answer. But why are the digits lower or equal than $3$ ? –  user12806 May 8 '13 at 19:18

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.

share|improve this answer
1  
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 –  Gerhard Paseman May 7 '13 at 6:54
    
Thanks for the answer. –  user12806 May 8 '13 at 19:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.