To find the remainder of the sum $S = 1^{1001} + 2^{1001} + \cdots + 100^{1001}$ when divided by 10100, we need to simplify the problem using modular arithmetic. Let's break it down step by step:

Step 1: Factorize 10100

$10100 = 2^2 \cdot 5^2 \cdot 101$

Step 2: Use Chinese Remainder Theorem (CRT)

To solve the problem modulo 10100, we'll first solve it modulo 4, 25, and 101 separately, and then combine the results using the Chinese Remainder Theorem.

Step 3: Compute $S \mod 4$

For $k = 1, 2, \ldots, 100$:

• $k \mod 4$ cycles through $\{1, 2, 3, 0\}$.
• $k^{1001} \mod 4$:
  • $1^{1001} \mod 4 = 1$
  • $2^{1001} \mod 4 = 0$ (since $2^2 = 4$)
  • $3^{1001} \mod 4 = 3$ (since $3 \mod 4 \equiv -1$, and $(-1)^{1001} = -1$)
  • $4^{1001} \mod 4 = 0$

Each residue class of $k \mod 4$ sums up as follows:

• For $k \mod 4 = 0$, there are 25 terms.
• For $k \mod 4 = 1, 2, 3$, there are 25 terms each.

Thus, the sum modulo 4 is:

$25 \cdot 1 + 25 \cdot 0 + 25 \cdot 3 + 25 \cdot 0 = 25 \cdot 1 + 25 \cdot 3 = 25 + 75 = 100 \equiv 0 \mod 4.$

Step 4: Compute $S \mod 25$

By Euler's Theorem, since $\phi(25) = 20$, for $k$ coprime to 25, $k^{20} \equiv 1 \mod 25$. Therefore, $k^{1001} = (k^{20})^{50} \cdot k \equiv k \mod 25$.

Thus:

$S \mod 25 = \sum_{k=1}^{100} k^{1001} \mod 25$

Since $k^{1001} \mod 25 \equiv k \mod 25$, the sum becomes the sum of the first 100 natural numbers:

$\sum_{k=1}^{100} k = \frac{100 \cdot 101}{2} = 5050$

Now, $5050 \mod 25 = 0$, so:

$S \mod 25 = 0.$

Step 5: Compute $S \mod 101$

Using Fermat's Little Theorem, $k^{100} \equiv 1 \mod 101$ for $k$ coprime to 101. Therefore, $k^{1001} \equiv k^{1} \mod 101$.

Thus:

$S \mod 101 = \sum_{k=1}^{100} k^{1001} \mod 101 \equiv \sum_{k=1}^{100} k \mod 101 = \frac{100 \cdot 101}{2} = 5050$

Calculating $5050 \mod 101$:

$5050 \div 101 = 50, \quad \text{so} \quad 5050 \mod 101 = 0.$

Step 6: Combine Using CRT

We have:

• $S \equiv 0 \mod 4$
• $S \equiv 0 \mod 25$
• $S \equiv 0 \mod 101$

Since 4, 25, and 101 are pairwise coprime, by the Chinese Remainder Theorem, $S \mod 10100$ must be 0.

Conclusion

The remainder when $1^{1001} + 2^{1001} + \cdots + 100^{1001}$ is divided by 10100 is:

$\boxed{0}$.

Do you need any further details or have any questions? Here are some questions you might consider:

1. How does the Chinese Remainder Theorem simplify problems involving moduli?
2. Can we use other theorems to solve similar problems?
3. What is the significance of Euler's Theorem in modular arithmetic?
4. How would the approach change if the exponent was different?
5. Can this method be extended to larger sums or different bases?

Tip: Always check for patterns or use theorems like Euler's or Fermat's to simplify large exponent problems in modular arithmetic.