# Integers which can be written as sum of powers of $2,3$, and $5$

Is it true that every sufficiently large integer can be written in the form $$2^a3^b5^c+2^d3^e5^f$$ for some integer $$a,b,c,d,e,f \ge 0$$?

• Given that each exponent is allowed to be $0$, it is true that every sufficiently small integer ($\ge 2$) can be so represented. – Keith Backman Apr 2 '19 at 18:19
• @KeithBackman Are you trolling? – Paolo Leonetti Apr 2 '19 at 18:21
• Not at all. I saw Robert Israel's answer, and I tried to find any other small integers that could not be so represented, and found I could construct every integer up to $70$ but not $71$. Just a paper and pencil check on the given answer. – Keith Backman Apr 2 '19 at 18:25

No, this is not the case. The number of such possible sums $$\le N$$ for large $$N$$ is far less than $$N$$.
The number of powers of $$2$$ (or of $$3$$ or $$5$$) up to $$N$$ is $$O(\log N)$$. Hence the number of products $$2^a 3^b 5^c$$ up to $$N$$ is $$O((\log N)^3)$$. The number of sums of pairs of such products is $$O((\log N)^6)$$, which is $$o(N)$$.
No numbers congruent to $$71$$ or $$119$$ mod $$120$$ can be represented in this way.