Math Problem Statement
Use the theory of congruence to show that 7|5^2n + 3.2^(5n-2)
Solution
To show that , 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
First, let’s analyze the powers of 5 modulo 7. We calculate the first few powers of 5 modulo 7: Thus, . This means that the powers of 5 repeat every 6 terms in modulo 7 arithmetic.
We can now reduce . Since , we only need to determine to reduce the exponent:
- If , then .
- If , then .
- If , then .
Step 2: Find
Next, we analyze the powers of 2 modulo 7. We calculate the first few powers of 2 modulo 7: So, . This means that the powers of 2 repeat every 3 terms in modulo 7 arithmetic.
We now reduce . Since , we only need to determine to reduce the exponent:
- If , then .
- If , then .
- If , then .
Step 3: Combine the results
We now combine the results of and .
We are looking at the expression . Based on the cases analyzed above, for all values of , we find that the expression simplifies to , meaning that it is divisible by 7.
Thus, we have shown that: for all values of .
Would you like further details on any step, or have any questions?
Here are five related questions to further explore the concept:
- How do modular arithmetic properties help in proving divisibility?
- What is Fermat's Little Theorem, and how could it be applied here?
- Can you show similar proofs using different bases, such as ?
- How does the periodicity of powers in modulo systems work?
- 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