Find the least residue of 1^7920 + 2^7920 + ... + 1414^7920 modulo 23. Hint: there's a pattern here! Find the pattern for small values modulo 23 (but dont change 23 or the exponent to different numbers).

To solve \( 1^{7920} + 2^{7920} + \dots + 1414^{7920} \mod 23 \), we need to identify patterns in powers of integers modulo 23. The large exponent (7920) suggests that we might benefit from properties of modular arithmetic and number theory, such as **Fermat's Little Theorem**.

### Step 1: Use Fermat's Little Theorem
Fermat's Little Theorem tells us that for any integer \(a\) and a prime \(p\), we have:

\[
a^{p-1} \equiv 1 \mod p \quad \text{for any integer } a \text{ not divisible by } p.
\]

Here, \(p = 23\), so for any integer \(a\) that is not divisible by 23:

\[
a^{22} \equiv 1 \mod 23.
\]

Given this, we can reduce the exponent \(7920\) modulo 22:

\[
7920 \mod 22 = 7920 - 22 \times \left\lfloor \frac{7920}{22} \right\rfloor = 7920 - 22 \times 360 = 0.
\]

Thus, for any \(a \not\equiv 0 \mod 23\):

\[
a^{7920} = a^{22k} \equiv 1^k = 1 \mod 23.
\]

### Step 2: Apply this to all terms from 1 to 1414
By Fermat's Little Theorem, for any \(a \in \{1, 2, \dots, 22\}\), we have:

\[
a^{7920} \equiv 1 \mod 23.
\]

However, \(23 \equiv 0 \mod 23\), and similarly, any multiple of 23 will be congruent to 0 modulo 23. Thus, for \(a \equiv 0 \mod 23\), \(a^{7920} \equiv 0 \mod 23\).

### Step 3: Count the multiples of 23
To find how many terms between 1 and 1414 are divisible by 23, we divide 1414 by 23:

\[
\left\lfloor \frac{1414}{23} \right\rfloor = 61.
\]

So, there are 61 terms divisible by 23, and for each of these terms \(a^{7920} \equiv 0 \mod 23\).

### Step 4: Compute the total sum
Now, the sum of \(1^{7920} + 2^{7920} + \dots + 1414^{7920} \mod 23\) consists of:

- 61 terms where \(a^{7920} \equiv 0 \mod 23\), contributing 0.
- The remaining \(1414 - 61 = 1353\) terms where \(a^{7920} \equiv 1 \mod 23\), contributing 1 each.

Thus, the total sum is:

\[
1353 \times 1 + 61 \times 0 = 1353.
\]

Finally, reduce this modulo 23:

\[
1353 \mod 23 = 1353 - 23 \times \left\lfloor \frac{1353}{23} \right\rfloor = 1353 - 23 \times 58 = 1353 - 1334 = 19.
\]

### Final Answer:
The least residue of \( 1^{7920} + 2^{7920} + \dots + 1414^{7920} \mod 23 \) is **19**.

---

Would you like further details on Fermat's Little Theorem or any step in this process?

Here are some related questions:
1. Can Fermat’s Little Theorem be applied to other primes in a similar way?
2. How do we handle exponents larger than the modulus in modular arithmetic?
3. How does modular reduction work for large sums like this?
4. What happens if 23 is not prime? How would the approach change?
5. What are other useful number-theoretic results like Fermat’s Little Theorem?

**Tip**: When dealing with large exponents and modular arithmetic, always check for patterns using number-theoretic results like Fermat's Little Theorem to simplify the problem.

Find the least residue of 1^7920 + 2^7920 + ... + 1414^7920 modulo 23. Hint: there's a pattern here! Find the pattern for small values modulo 23 (but don't change 23 or the exponent to different numbers).

Solve the challenging number theory problem involving the sum of powers modulo 23. This problem highlights the use of Fermat's Little Theorem and modular arithmetic to simplify large exponents. Suitable for advanced high school or early college students.

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