Question

$5^6 - 1$ is divisible by:

• A. $13$
• B. $31$
• C. $5$
• D. None of these

Hint:

Always check modular arithmetic before relying on factor patterns—it avoids missing hidden factors or misinterpreting divisibility.

Answer 1

The Correct Option is A

Step 1: Apply difference of squares and sum of cubes factorisation $5^6 - 1 = (5^3 - 1)(5^3 + 1)$. Step 2: Factor each term further - $5^3 - 1 = (5 - 1)(25 + 5 + 1) = 4 \times 31$ - $5^3 + 1 = (5 + 1)(25 - 5 + 1) = 6 \times 21 = 2 \times 3 \times 21$ Step 3: Combined factorisation $5^6 - 1 = 4 \times 31 \times 2 \times 3 \times 21$. Step 4: Check divisibility From the factors, $13$ is not immediately visible—but $21 = 3 \times 7$, so no $13$ there. Wait—let's check: Actually, $5^6 - 1 = (5^2 - 1)(5^4 + 5^2 + 1)$ - $5^2 - 1 = 24 = 2^3 \times 3$ - $5^4 + 5^2 + 1 = 625 + 25 + 1 = 651 = 3 \times 217 = 3 \times 7 \times 31$ So indeed, no $13$ factor. The correct check: $5^6 \equiv 1 \ (\text{mod} \ 13)$ since $5^6 \ (\text{mod} \ 13) = (5^2)^3 = (-1)^3 \equiv -1$, contradiction—wait, mistake. Correct computation: $5^2 \equiv -1 \ (\text{mod} \ 13) \Rightarrow 5^4 \equiv 1 \ (\text{mod} \ 13) \Rightarrow 5^6 \equiv 5^2 \equiv -1$, so $5^6 - 1 \equiv -2$ not divisible—so answer actually "None of these".