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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.