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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 $a,b,c,d,e,f \ge 0$?

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  • $\begingroup$ Given that each exponent is allowed to be $0$, it is true that every sufficiently small integer ($\ge 2$) can be so represented. $\endgroup$ – Keith Backman Apr 2 at 18:19
  • $\begingroup$ @KeithBackman Are you trolling? $\endgroup$ – Paolo Leonetti Apr 2 at 18:21
  • $\begingroup$ 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. $\endgroup$ – Keith Backman Apr 2 at 18:25
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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 $o(N)$.

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No numbers congruent to $71$ or $119$ mod $120$ can be represented in this way.

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