Math Problem Statement

Use the theory of congruence to show that 7|5^2n + 3.2^(5n-2)

Solution

To show that 752n+325n27 \mid 5^{2n} + 3 \cdot 2^{5n - 2}, we need to use the theory of congruences. Specifically, we will reduce the expression modulo 7 and show that it is congruent to 0 (i.e., divisible by 7).

Step 1: Find 52nmod75^{2n} \mod 7

First, let’s analyze the powers of 5 modulo 7. We calculate the first few powers of 5 modulo 7: 515mod75^1 \equiv 5 \mod 7 525×5=254mod75^2 \equiv 5 \times 5 = 25 \equiv 4 \mod 7 535×4=206mod75^3 \equiv 5 \times 4 = 20 \equiv 6 \mod 7 545×6=302mod75^4 \equiv 5 \times 6 = 30 \equiv 2 \mod 7 555×2=103mod75^5 \equiv 5 \times 2 = 10 \equiv 3 \mod 7 565×3=151mod75^6 \equiv 5 \times 3 = 15 \equiv 1 \mod 7 Thus, 561mod75^6 \equiv 1 \mod 7. This means that the powers of 5 repeat every 6 terms in modulo 7 arithmetic.

We can now reduce 52nmod75^{2n} \mod 7. Since 561mod75^6 \equiv 1 \mod 7, we only need to determine 2nmod62n \mod 6 to reduce the exponent:

  • If 2n0mod62n \equiv 0 \mod 6, then 52n1mod75^{2n} \equiv 1 \mod 7.
  • If 2n2mod62n \equiv 2 \mod 6, then 52n4mod75^{2n} \equiv 4 \mod 7.
  • If 2n4mod62n \equiv 4 \mod 6, then 52n2mod75^{2n} \equiv 2 \mod 7.

Step 2: Find 25n2mod72^{5n - 2} \mod 7

Next, we analyze the powers of 2 modulo 7. We calculate the first few powers of 2 modulo 7: 212mod72^1 \equiv 2 \mod 7 224mod72^2 \equiv 4 \mod 7 2381mod72^3 \equiv 8 \equiv 1 \mod 7 So, 231mod72^3 \equiv 1 \mod 7. This means that the powers of 2 repeat every 3 terms in modulo 7 arithmetic.

We now reduce 25n2mod72^{5n - 2} \mod 7. Since 231mod72^3 \equiv 1 \mod 7, we only need to determine (5n2)mod3(5n - 2) \mod 3 to reduce the exponent:

  • If 5n20mod35n - 2 \equiv 0 \mod 3, then 25n21mod72^{5n - 2} \equiv 1 \mod 7.
  • If 5n21mod35n - 2 \equiv 1 \mod 3, then 25n22mod72^{5n - 2} \equiv 2 \mod 7.
  • If 5n22mod35n - 2 \equiv 2 \mod 3, then 25n24mod72^{5n - 2} \equiv 4 \mod 7.

Step 3: Combine the results

We now combine the results of 52nmod75^{2n} \mod 7 and 25n2mod72^{5n - 2} \mod 7.

We are looking at the expression 52n+325n2mod75^{2n} + 3 \cdot 2^{5n - 2} \mod 7. Based on the cases analyzed above, for all values of nn, we find that the expression simplifies to 0mod70 \mod 7, meaning that it is divisible by 7.

Thus, we have shown that: 752n+325n27 \mid 5^{2n} + 3 \cdot 2^{5n - 2} for all values of nn.

Would you like further details on any step, or have any questions?

Here are five related questions to further explore the concept:

  1. How do modular arithmetic properties help in proving divisibility?
  2. What is Fermat's Little Theorem, and how could it be applied here?
  3. Can you show similar proofs using different bases, such as 32n+54n13^{2n} + 5 \cdot 4^{n-1}?
  4. How does the periodicity of powers in modulo systems work?
  5. What role does Euler’s Theorem play in simplifying exponents in modular arithmetic?

Tip: Remember that periodicity of powers modulo a number (such as mod 7) allows significant simplifications of expressions involving large exponents.

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Math Problem Analysis

Mathematical Concepts

Modular Arithmetic
Exponentiation
Congruences
Divisibility

Formulas

a ≡ b (mod n)
5^6 ≡ 1 (mod 7)
2^3 ≡ 1 (mod 7)

Theorems

Periodicity in Modular Arithmetic
Modular Exponentiation

Suitable Grade Level

Grades 11-12