The statement P(n)=5^(2n)+1 is divisible by 13 is true, then 5^(99) leaves a remainder, when divided by 13, is

Options

A. 7
B. 6
C. 8
D. none of these

Answer

To solve the problem, we need to determine the remainder of \( 5^{99} \) when divided by 13, given that the statement \( P(n) = 5^{2n} + 1 \) is divisible by 13.

### Step-by-Step Solution:

**Step 1: Understand the given statement**
The statement \( P(n) = 5^{2n} + 1 \) is divisible by 13 means that for any integer \( n \), \( 5^{2n} + 1 \equiv 0 \mod 13 \). This implies that \( 5^{2n} \equiv -1 \mod 13 \).

**Hint for Step 1:** Check the properties of modular arithmetic and how to express divisibility in terms of congruences.

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**Step 2: Find a pattern in powers of 5 modulo 13**
We can calculate the first few powers of 5 modulo 13:
- \( 5^1 \equiv 5 \mod 13 \)
- \( 5^2 \equiv 25 \equiv 12 \mod 13 \)
- \( 5^3 \equiv 5 \cdot 12 \equiv 60 \equiv 8 \mod 13 \)
- \( 5^4 \equiv 5 \cdot 8 \equiv 40 \equiv 1 \mod 13 \)

Notice that \( 5^4 \equiv 1 \mod 13 \). This means that the powers of 5 will repeat every 4 terms.

**Hint for Step 2:** Look for cycles in modular arithmetic to simplify calculations.

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**Step 3: Reduce the exponent modulo 4**
Since \( 5^4 \equiv 1 \mod 13 \), we can reduce the exponent 99 modulo 4:
\[ 99 \mod 4 = 3 \]
Thus, \( 5^{99} \equiv 5^3 \mod 13 \).

**Hint for Step 3:** Use the property of exponents and modular arithmetic to simplify large exponents.

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**Step 4: Calculate \( 5^3 \mod 13 \)**
From our earlier calculations, we found that:
\[ 5^3 \equiv 8 \mod 13 \]

**Hint for Step 4:** Always refer back to your earlier calculations to find the required value.

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**Step 5: Conclusion**
Thus, the remainder when \( 5^{99} \) is divided by 13 is \( 8 \).

**Final Answer:** The remainder is \( 8 \).

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