Math Problem Statement

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).

Solution

To solve 17920+27920++14147920mod231^{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 aa and a prime pp, we have:

ap11modpfor any integer a not divisible by p.a^{p-1} \equiv 1 \mod p \quad \text{for any integer } a \text{ not divisible by } p.

Here, p=23p = 23, so for any integer aa that is not divisible by 23:

a221mod23.a^{22} \equiv 1 \mod 23.

Given this, we can reduce the exponent 79207920 modulo 22:

7920mod22=792022×792022=792022×360=0.7920 \mod 22 = 7920 - 22 \times \left\lfloor \frac{7920}{22} \right\rfloor = 7920 - 22 \times 360 = 0.

Thus, for any a≢0mod23a \not\equiv 0 \mod 23:

a7920=a22k1k=1mod23.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{1,2,,22}a \in \{1, 2, \dots, 22\}, we have:

a79201mod23.a^{7920} \equiv 1 \mod 23.

However, 230mod2323 \equiv 0 \mod 23, and similarly, any multiple of 23 will be congruent to 0 modulo 23. Thus, for a0mod23a \equiv 0 \mod 23, a79200mod23a^{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:

141423=61.\left\lfloor \frac{1414}{23} \right\rfloor = 61.

So, there are 61 terms divisible by 23, and for each of these terms a79200mod23a^{7920} \equiv 0 \mod 23.

Step 4: Compute the total sum

Now, the sum of 17920+27920++14147920mod231^{7920} + 2^{7920} + \dots + 1414^{7920} \mod 23 consists of:

  • 61 terms where a79200mod23a^{7920} \equiv 0 \mod 23, contributing 0.
  • The remaining 141461=13531414 - 61 = 1353 terms where a79201mod23a^{7920} \equiv 1 \mod 23, contributing 1 each.

Thus, the total sum is:

1353×1+61×0=1353.1353 \times 1 + 61 \times 0 = 1353.

Finally, reduce this modulo 23:

1353mod23=135323×135323=135323×58=13531334=19.1353 \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 17920+27920++14147920mod231^{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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Modular Arithmetic
Number Theory
Patterns in Powers Modulo Prime

Formulas

a^(p-1) ≡ 1 (mod p) for p prime (Fermat's Little Theorem)
Sum formula for powers mod prime

Theorems

Fermat's Little Theorem

Suitable Grade Level

Grades 11-12 or Early College