I want to prove that exist a natural number $n > 1$, which is not a power of $2$ and $$\frac{3^{a_1}n}{2^{b_1}} + \frac{3^{a_2}}{2^{b_2}} + \ldots + \frac{3^{a_k}}{2^{b_k}} \neq 1$$ for all possible non-negative integers $a_1, \ldots, a_k, b_1, \ldots, b_k$. Note that you can also choose the value $k$.

Honestly, I don't know if this result is true or not. It's a conjecture of mine. Any ideas and good insights about this problem are welcome. Thanks.

  • 1
    $\begingroup$ not true , take $a_1=0 , a_2=2,a_3=3$ and $b_1=4,b_2=6,b_3=6$ and $n=7$. $\endgroup$
    – Ahmad
    Sep 25, 2017 at 18:43
  • $\begingroup$ @Ahmad Maybe I didn't make this as clear as I wanted. But I want a $n$ which works for all possible choices of non-negative integers $a_1, \ldots, a_k, b_1, \ldots, b_k$. Not just an instance of them. $\endgroup$
    – Integral
    Sep 25, 2017 at 19:00

1 Answer 1


I already got an answer to this problem. Given any $n \in \mathbb{N}$, there is a number $p \in \mathbb{N}$ such that $2^p > n$. Now consider the expression for $2^p-n$ in base 3, i.e., there are numbers $c_0,c_1, \ldots, c_k \in \{0,1,2\}$ such that

$$2^p - n = \sum_{i=0}^k 3^i c_i.$$

This is equivalent to

$$\frac{n}{2^p} + \sum_{i=0}^k \frac{3^i c_i}{2^p} = 1.$$

Note that each number $\frac{3^i c_i}{2^p}$ is of the form $\frac{3^a}{2^b}$, since each $c_i$ is 0, 1 or 2 (in the case $c_i = 0$ there is no factor to consider, which is also ok). Therefore the result I was trying to prove is false.

  • $\begingroup$ Ah yes? So your "conjecture" was known to be false before you posted your question? I'm waiting for you to delete your downvote before I delete it. Regards. $\endgroup$
    – Piquito
    Sep 26, 2017 at 14:41
  • $\begingroup$ @Piquito My conjecture wasn't known to be false. I got this proof after. In fact, 20 hours after. About your "answer", you surely should erase it. Just because it is not an answer but a suggestion to consider another problem. You should leave it as a comment and nothing more. $\endgroup$
    – Integral
    Sep 26, 2017 at 15:32
  • $\begingroup$ So when I put my answer in which it is implicitly say your conjecture was banal. Best regards anyway. $\endgroup$
    – Piquito
    Sep 26, 2017 at 15:35

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