29
$\begingroup$

The following conjecture is one I have made today with the aid of computer software.

Conjecture:

Let $s(\cdot)$ denote the sum of the digits of $\cdot$ in base $10$. Then the only integer values $a,b>1$ that satisfy $$s(a^b)=ab$$ are $(2,2),(3,3),(3,6),(3,9)$ and $(3,27)$.

Remarks:

  • The number of digits of an integer $n$ is $1+\lfloor\log_{10}n\rfloor$ where $\lfloor\cdot\rfloor$ denotes the floor function. This means that $s(a^b)<9+9\lfloor b\log_{10} a\rfloor$ as each digit takes a value of at most $9$. From this plot it is evident that the equality will never hold for $a\ge 9$, after checking the first $21$ values of $b$. Therefore it suffices to consider $2\le a\le 8$.

  • I have excluded the case $b=1$ as it is trivial - it forces $a$ to be a single-digit integer.

  • In PARI/GP the code is given by mfun(b)={for(i=2,8,for(j=2,b,if(sumdigits(i^j,10)==i*j,print(i," ",j))));} and tests up to $b\le 10^6$ verify the above conjecture.

Aside:

  • The equation $s(ab)=a+b$ is much simpler to solve. From here, the upper bound $a\le 23$ is apparent and for these values of $a$, we can form the upper bound $b\le 22$ as there will be no solutions when the red line lies above the blue lines. This $23\times22$ grid can be computed through software and it is found that the only solutions are $$(a,b)=(2,2),(3,6),(6,3).$$ The similarity in the first two solutions with the conjectured solutions may only be coincidental.

Advances on this will be appreciated.

$\endgroup$
9
  • $\begingroup$ So, you are only considering base-10? If such a conjecture is true, you would expect it to be true for any base. $\endgroup$
    – alex811
    Feb 24, 2019 at 17:17
  • 1
    $\begingroup$ Though, in light of @alex811's comment, one wonders what the set of pairs would be in other bases. For example, $(a,a)$ will work in any base where $(a,a)$ is represented by a single digit. I also wonder if the fact that $3^2 = 9 = 10-1$ is of any relevance. $\endgroup$
    – Xander Henderson
    Feb 24, 2019 at 22:00
  • 1
    $\begingroup$ If $3\mid a$ then $9\mid s(a^b)$ so you need either $9\mid a$ or $3\mid b$. Don't know if that helps. I prepared a largely useless plot here related to the case $a=2$. I found this somewhat related post the most delightful contribution around this theme. $\endgroup$ Feb 25, 2019 at 10:31
  • 2
    $\begingroup$ Showing that $s_2(2^b) < 2b$ for $b$ large is probably very difficult. I don't know how to show, for instance, that a power of $2$ can't have all 7's,8's, or 9's in its decimal expansion. $\endgroup$ Mar 2, 2019 at 20:29
  • 1
    $\begingroup$ @TheSimpliFire I posted a new question regarding my effort to solve your conjecture, to get advice about some asymptotic relation for s(a^b). Any help would be appreciated. I am not a professional mathematician, but I think it's worth reading! math.stackexchange.com/questions/3370010/… $\endgroup$ Sep 26, 2019 at 15:47

3 Answers 3

13
$\begingroup$

I like to crack numbers but this time I'll pass. It's highly unlikely that there is any other solution. I have created a few plots of function:

$$f(a)=s(a^b)-ab$$

...for $a=2,3,4,5,6,7,8$ and $b\in [1,4000]$. All these plots look the same. The function slides towards negative infinity in a pretty linear fashion, with very little variation from the straight line. If you accept bets, I can bet a house that there are no other solutions except small ones that you already listed.

Mathematica code:

For[a=2,a<=8,a++,DiscretePlot[Total[IntegerDigits[a^b,10]]-a*b,{b,2,4000},Filling->f,Joined->False] // Print]

$a=2$

enter image description here

$a=3$

enter image description here

$a=4$

enter image description here

$a=5$

enter image description here

$a=6$

enter image description here

$a=7$

enter image description here

$a=8$

enter image description here

$\endgroup$
1
  • 6
    $\begingroup$ It looks like there may be hope in providing a linear upper bound for $f$, which would prove the result. A weak one would suffice. $\endgroup$
    – YiFan
    Apr 21, 2019 at 7:54
11
$\begingroup$

Here is a heuristic argument (not a proof sadly, but too long for a comment).

The sequences $2^n,3^n,\dots,9^n$ are known to follow Benford's law, i.e., their first digit has a probability $\log_{10}\left(1+\frac{1}{d}\right)$ of being $d$; their second digit has a probability $\log_{10}\left(1+\frac{1}{10+d}\right)+\log_{10}\left(1+\frac{1}{20+d}\right)+\dots+\log_{10}\left(1+\frac{1}{90+d}\right)$ of being $d$, etc.

Here, "probability" is to be understood as asymptotic density. For instance, for the first digit of the sequence $2^n$, we have: $$\lim_{n\to\infty}\frac{|\{k\in\{1,\dots,n\}:2^k\text{ has $d$ as first digit}\}|}{n} = \log_{10}\left(1+\frac{1}{d}\right)$$ and more generally, for all $a \in \{2,\dots,9\}$: $$\lim_{n\to\infty}\frac{|\{k\in\{1,\dots,n\}:a^k\text{ has $d$ as $j$-th digit}\}|}{n} = \sum_{i=10^{j-2}}^{10^{j-1}-1}\log_{10}\left(1+\frac{1}{10i+d}\right)$$ and as the last quantity approaches $\frac 1 {10}$ as $j \to \infty$, the $j$-th digit of $a^n$ are almost equidistributed.

If we heuristically take these probabilities at face value, the sum of the digits of $a^n$ should match the sum of their expected values, and, apart from the first figures, this expected value is close to $4.5$. Thus, we should expect to have: $$s(a^n) \approx 4.5(1+\lfloor n\log_{10}a\rfloor) \approx 4.5 n\log_{10}a$$ and thus: $$s(a^n) - an \approx (\underbrace{4.5\log_{10} a - a}_{<0 \textrm{ for all } a})n \to_{n \to\infty} -\infty$$ This aligns with the graphs provided by @Oldboy. For instance, for $a=6$, we should expect a slope of $4.5\log_{10}6 - 6 \approx -2.50$, which matches the observed slope of $-10000/4000$.

$\endgroup$
3
$\begingroup$

Also not an answer, but it greatly reduces the set of possible solutions... We know for a fact that the difference between $n$ and $s\left(n\right)$ is always a multiple of $9$: $$ \left(n-s\left(n\right)\right)\mod9=0 $$ So if $s\left(a^{b}\right)=ab$, then: $$\left(a^{b}-ab\right)\mod9=0$$ Therefore, for $2\leq a \leq 8$, $b$ must be of the following forms: $$ b\begin{cases} 3n & \iff a=3~or~6\\ 9n+1 & \iff a=4~or~7\\ 18n+1~or~18n+2 & \iff a=2\\ 18n+1~or~18n+8 & \iff a=8\\ 18n+1~or~18n+14 & \iff a=5\\ \end{cases} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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