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Let $m,n, N$ be $3$ natural numbers. I want to know how many ordered pairs $(m,n)$ are possible such that $2^m\cdot 3^n< 10^N$.
My attempt: Taking $\log$ base $10$ on both sides, we have $m \log2 + n\log 3 < N$ Which is kind of a Diophantine inequality. How do I proceed now?
Also, can this approach be generalised for more than 2 primes?
Assuming your natural numbers start at $0$, an exact expression would be $$\sum_{m=0}^{\left\lfloor\dfrac N{\log 2}\right\rfloor}\left\lfloor1+\dfrac {N-m\log 2}{\log 3}\right\rfloor.$$
We're basically letting $m$ assume all possible values, then for each $m$ calculating the range of possible $n$ values. This approach can be extended to multiple primes, e.g. $$\sum_{x_2=0}^{\left\lfloor\dfrac N{\log 2}\right\rfloor}\sum_{x_3=0}^{\left\lfloor\dfrac {N-x_2\log2}{\log 3}\right\rfloor}\sum_{x_5=0}^{\left\lfloor\dfrac {N-x_2\log2-x_3\log3}{\log 5}\right\rfloor}\left\lfloor1+\dfrac {N-x_2\log 2-x_3\log3-x_5\log5}{\log 7}\right\rfloor.$$
This is all a bit messy. If only $\log 2$ and $\log 3$ were integers, then we could use Pick's Theorem!
A fairly accurate approximation is $\dfrac12 \left(\left\lfloor\dfrac N{\log 2}\right\rfloor+1.5\right)\left(\left\lfloor\dfrac N {\log 3}\right\rfloor+1.5\right)$. This approximates the number of lattice points (points with integer coordinates) under the line $m \log 2 + n \log 3 = N$ by halving an (almost) average of what we'd get if we rounded down the intercepts vs. rounding up.
