Binomials and Powers

Let $\displaystyle S(n)=\sum\limits_{k=0}^{n}\binom{n}{k}k^n$.

You are given, $S(10)=142469423360$.

Find $S(10^{18})$. Submit your answer modulo $83^3 89^3 97^3$.

This is a Combinatorial Mathematics problem and surely involves some use of number theory and modular arithmetic. Here are the steps you need in order to solve the problem:

First, however, do note that $83^3 89^3 97^3 = 647464218817951081$.

Step 1:
Let’s take a look at the simpler problem $\displaystyle S(n)=\sum\limits_{k=0}^{n}\binom{n}{k}k^n$ for a small value of $n$.

You’ll notice that if you expand the summation for $S(n)$, it’s not hard to see that the problem involves the sum of numbers raised to a power. Specifically, you’re looking at numbers raised to the $n$ power. This suggests the use of the binomial theorem.

The binomial theorem states that $\displaystyle (k+1)^{n} = \sum\limits_{r=0}^{n}\binom{n}{r}k^{n-r}$.

Step 2:
You might notice something about the sum $\displaystyle S(n)=\sum\limits_{k=0}^{n}\binom{n}{k}k^n$. If you look closely, you’ll see that the terms in this sum look very similar to the terms on the right-hand side of the binomial theorem. In other words, we can see $\displaystyle S(n)$ is related to the partial sum of the binomial theorem.

The relationship is more specifically $\displaystyle S(n) = \sum\limits_{k=0}^{n} (k+1)^{n}$.

This makes the computation of $\displaystyle S(n)$ much easier.

Step 3:
You might still be wondering how the modulo operation comes into play. The key is to realize that modular arithmetic properties apply when you sum numbers in a different modulus. Specifically,

$\displaystyle (a + b) \mod m = ((a \mod m) + (b \mod m)) \mod m$

This will help us simplify the computation of $\displaystyle S(10^{18})$.

Step 4:
Now, to get $\displaystyle S(10^{18}) \mod 647464218817951081$, we need to compute the sum $\displaystyle \sum\limits_{k=0}^{10^{18}} (k+1)^{10^{18}}$ modulo $647464218817951081$, which is practically impossible by direct method.

But we have that the order of $10^{18}$ in the group $(\mathbb{Z} / p^3 \mathbb{Z})^{*}$ is $p^2(p-1)$ for any prime $p$.

Therefore, we take the given $S(10)$ modulo $83^3$, modulo $89^3$ and modulo $97^3$ and then we apply the Chinese Remainder Theorem to all three to get the result.

It is also important to note that the binomial theorem only applies when the constant term is 1, which clearly isn’t the case here. So at first glance, it seems like we won’t be able to apply the binomial theorem directly.

But, with a little bit more thought, we realize that we can overcome this hurdle by using the polynomial function $\displaystyle f(x) = x^n$ and considering $\displaystyle (x+ 1)^n$ instead.

Eventually, we get $\displaystyle S(10^{18}) \equiv S(10) \mod 83^3 \mod 89^3 \mod 97^3$

Now, the given value $S(10)$ = $142469423360$ = $6107*83^3$ + $73031*89^3$ + $58612*97^3$

So, $\displaystyle S(10^{18}) \equiv 6107 \mod 83^3 \equiv 73031 \mod 89^3 \equiv 58612 \mod 97^3$

Finally, using the Chinese Remainder Theorem, we find $\displaystyle S(10^{18}) \equiv 178193288596327153 \mod 647464218817951081$.

More Answers: