# Find the least $n$ such that the expression is divisible by $700$.

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

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.

• 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. – 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.

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