What is the sum of the digits of the smallest positive integer $n^4 + 6n^3 + 11n + 6$ is divisible by $700$.

Hints please.

I got that $P(n) = n(n+1)(n+2)(n+3) \equiv 0 \pmod{700}$

I cannot seem to do anything else, what now?

Hints only.

  • $\begingroup$ I think $n = 25$ is the smallest such $n$. Since you asked for only hints, I'm giving you this as a hint and not submitting my solution as an answer. $\endgroup$ – user98186 Nov 30 '15 at 17:26

You have the product of four numbers in a row, which must be a multiple of $2^2 5^2 7$. At most one of $n$, $n+1$, $n+2$, or $n+3$ can be a multiple of 5...


Try solving the equation modulo $25$, $4$ and $7$, and then use the Chinese remainder theorem.

  • $\begingroup$ (+1) Even modulo $7$, how can I 'solve' this equation? $\endgroup$ – Amad27 Nov 11 '15 at 15:13
  • $\begingroup$ @Amad27: working $\bmod 7$, there are only $7$ choices, so you can just try them all. $\endgroup$ – Ross Millikan Nov 11 '15 at 15:20
  • $\begingroup$ @RossMillikan, which 7 choices? $\endgroup$ – Amad27 Nov 11 '15 at 15:25
  • $\begingroup$ @Amad27: $0$ through $6$ $\endgroup$ – Ross Millikan Nov 11 '15 at 16:33
  • $\begingroup$ Ah - I see, for this, $n=4$ does work as the least. Modulo $4$, $n=1$ works. Modulo $25$ seems tricky. $\endgroup$ – Amad27 Nov 11 '15 at 16:38

As observed earlier $700=2^25^27$ and you want four consecutive numbers that that give you those prime factors, so what about$700=25\cdot 28$?

Another observation: the problem is asking for the sum of the digits of the smallest $n$ that gives us divisibility by $700$ rather than $n$ itself. I wonder if there is something like an approach to "casting of nines" that could be used here?


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.