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I've been doing a bit of study in number theory and modulo operations to prove certain things. This one in particular is a question I came across recently:

$$\text{Find the largest and smallest positive values of n for which} (n^{300}+100)\text{is divisible by} (n+10)$$

This is what I was able to come up with:

Take $d=n+10 \rightarrow n=d-10$

$\begin{align} n^{300}+100 &= (d-10)^{300} \\ &\equiv 10^{300}+100 \quad \bmod d \end{align}$

$\begin{align} 10^{300}+100 &= 10^2(10^{298}+1) \\ &= 100(100^{149}+1) \\ &= 100((101-1)^{149}+1) \end{align}$

So from 100 the smallest factor one can produce from d is 1, but that would lead to a negative value for n($-9$) which should probably also work but still.

The next one thus is 10 which gives $n=0$, something that could've been guessed from the beginning when you think about it. But I'm not entirely sure if 0 would be counted as a positive integer so I'd go for $20 \implies n=10$ and from the expression in brackets the $\bmod 101$ from both are -1 and 1 respectively so they cancel out. Thus the expression is divisible by 101.

The greatest factor you can produce then should be the greatest from $100$ and $100^{149}+1$ which is $101 \cdot 100 = 10100$

$n=10100-10=10090$

Am I right or can you still get a bigger factor? Or is my logic flawed at any step?

Arjun
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Nεo Pλατo
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1 Answers1

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We have $$n^{300}+100\equiv (-10)^{300}+100=10^{300}+100\mod (n+10)$$

as you found out. Hence $$n+10\mid n^{300}+100$$ if and only if $$n+10\mid 10^{300}+100$$ Hence the possible solutions are the divisors of $\ 10^{300}+100\ $ minus $\ 10\ $. Since $\ n\ $ must be positive, we need the smallest divisor of $\ 10^{300}+10\ $ that exceeds $\ 10\ $. Partial factorization reveals $$10^{300}+100=2^2\cdot 5^2\cdot 101\cdot N$$ where $\ N\ $ has no prime factor upto $\ 101\ $. Hence the smallest divisor exceeding $\ 10\ $ is $\ 20\ $, which gives $\ n=10\ $ as the minimum value. The maximum value for $n$ is $$10^{300}+100-10=10^{300}+90$$

Peter
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