My birthday is 10th of October or 1010 in MMDD format.

I just realized that 1010 contains two copies of the number 10 and if spelled out in binary,


I was wondering how many other numbers have this property

In general we can use the following formula

$N=\sum_{j=0}^n a_j*2^j$ + $\sum_{j=n+1}^{2n+1}a_{j-n-1}*2^j$

$N=\sum_{k=0}^n a_k*10^k$

(In both equations all the $a_i$ are either $0$ or $1$, n is one less than the number of digits in the repeated digit sequence, so for $1010_2$, $n=1$, $a_0=0$, $a_1=1$)

Subtracting the second from the first and factoring coefficients,

$0=\sum_{d=0}^n a_d*(2^{d}+2^{n+d+1}-10^d)$

$0=\sum_{d=0}^n a_d*2^d*(2^{n+1}+1-5^d)$

I am not sure how to solve from here on out

Also would like to know if there are any other solutions besides 0 and 1 in other bases.

  • $\begingroup$ A small comment on notation: you set this up to equate an $(n+1)$-digit decimal numeral with a $(2n+2)$-digit binary numeral. It might be simpler to compare an $n$-digit number with a $2n$-digit number. $\endgroup$ – David K Mar 23 at 20:09
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    $\begingroup$ Requiring a leading $0$ in binary & base $10$ expansions results in relatively few values to check (& I believe the answer is there are no more, but I haven't confirmed) as shown by the answers below. However, you did not explicitly state this is required. Thus, for example, if you allow the use of leading zeros, a valid value to check on could have a base $2$ expansion of $000101000101_2$. This more general case is, I believe, quite interesting. There'll be few values which fit your criteria, but I suspect there may still be infinitely many. If you're interested in this, please specify it. $\endgroup$ – John Omielan Mar 24 at 1:26
  • $\begingroup$ Yes, that is interesting, though I had not considered it initially $\endgroup$ – aman Mar 24 at 3:00

The number of digits in the decimal expression of a number is $\lfloor \log_{10} N \rfloor + 1$, and the number of bits in its binary expansion is $\lfloor \log_2 N \rfloor + 1$, so any number $N$ with the given reduplication property must satisfy $$\lfloor \log_2 N \rfloor + 1 = 2 (\lfloor \log_{10} N \rfloor + 1) .$$ Using twice that $\log_b N \geq \lfloor \log_b N \rfloor > \log_b N - 1$ gives $$\log_2 N < 2 \log_{10} N + 2 = \frac{2}{\log_2 10} \log_2 N + 2.$$ Rearranging gives $$\log_2 N \leq \frac{2 \log_2 10}{\log_2 10 - 2} < 6,$$ so $N < 2^6 ,$ but this leaves only $0, 1, 10, 11$ to check.

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    $\begingroup$ Very nice. Sometimes it's better to work with easier necessary conditions than the exact statement. $\endgroup$ – amsmath Mar 23 at 20:13
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    $\begingroup$ Another way to look at it: setting $n$ as in the original question, the decimal number is at least $10^n,$ and the binary number is less than $4\cdot4^n,$ so you can have a solution only if $10^n<4\cdot4^n,$ or $2.5^n<4.$ Since $2.5^2 = 6.25,$ we must have $n<2,$ so the decimal number can have only $1+1 =2$ digits at most. Basically the same logic but with exponents instead of logarithms. $\endgroup$ – David K Mar 23 at 20:16
  • $\begingroup$ Nicely put, David. $\endgroup$ – Travis Willse Mar 23 at 20:30
  • $\begingroup$ Thanks for the answer. However, N should be used rather than n, that confused me initially $\endgroup$ – aman Mar 24 at 3:04
  • $\begingroup$ You're welcome. I've changed the notation to match that in the question. $\endgroup$ – Travis Willse Mar 24 at 3:28

This is only a partial answer. For $n>1$ there is always a smallest number $d_0\le n$ such that $2^{n+1}+1-5^{d_0} < 0$. So, your equation becomes $$ \sum_{d=0}^{d_0-1}a_d\cdot 2^d(2^{n+1}+1-5^d) = \sum_{d=d_0}^na_d\cdot 2^d(5^d-(2^{n+1}+1)). $$ Assume that one of $a_{d_0+1},\ldots,a_n$ is non-zero. Then the RHS is at least as large as $2^{d_0+1}(5^{d_0+1}-(2^{n+1}+1))$, which by definition of $d_0$ is larger than $2^{d_0+1}(5(2^{n+1}+1)-(2^{n+1}+1)) = 2^{d_0+3}(2^{n+1}+1)$. The LHS can be estimated by \begin{align*} LHS &\le \sum_{d=0}^{d_0-1}2^d(2^{n+1}+1-5^d) = (2^{n+1}+1)(2^{d_0}-1) - \frac{10^{d_0}-1}{9}\\ &\le (2^{n+1}+1)(2^{d_0}-1) - \frac{(2^{n+1}+1)}{9}2^{d_0} + \frac 19\\ &= (2^{n+1}+1)\left(\frac 89\,2^{d_0}-1\right) + \frac 19 < (2^{n+1}+1)\left(2^{d_0}-1\right) + \frac 19\\ &< (2^{n+1}+1)2^{d_0+3} < RHS. \end{align*} Hence, we conclude that $a_{d_0+1}=\cdots=a_n = 0$. This reduces your equation to $$ \sum_{d=0}^{d_0-1}a_d\cdot 2^d(2^{n+1}+1-5^d) = 2^{d_0}(5^{d_0}-(2^{n+1}+1)). $$


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