# Representing the integers with powers of 2 and 3

Suppose that I have a number of the form

$$x = \frac{1}{3^m}(2^{h} - \sum\limits_{k=1}^{m}3^{m-k}2^{v_k} )$$

where m is a positive integer, and

$v_1 = 0$

$v_{k+1} = v_k +1$ for $1 \leq k < m-1$

$v_{m} > v_{m-1}$

For a fixed m, can we find a relationship between h and $v_m$, such that x is an integer ?

For $k\leq m-1$, we have $v_k = k-1$. The expression for $x$ is then reduced to $$x=\frac{1}{3^m}\left( 2^h - \sum_{k=1}^{m-1} 3^{m-k}2^{k-1} - 2^{v_m}\right)=\frac{1}{3^m}\left( 2^h - 3^m + 2^m + 2^{m-1} - 2^{v_m} \right)$$ which is integer as soon as $$2^{v_m} \equiv 2^h + 2^m + 2^{m-1} \pmod{3^m}.$$ Taking discrete logarithm (if it exists) modulo $3^m$ will give a solution.