Question

The digit in the unit's place of the product $3^{999\times 7^{1000}}$ is \(\underline{\hspace{1cm}}\).

• A. 7
• B. 1
• C. 3
• D. 9

Hint:

Always check repeating cycles of units digits (mod 10). Both $3$ and $7$ have cycles of length 4. Reducing exponents mod 4 gives the correct digit quickly.

Answer 1

The Correct Option is A

Step 1: Units digit cycle of $3^n$.
The units digit of $3^n$ repeats in a cycle of $4$: $3,9,7,1$. Since $999 \bmod 4 = 3$, we have $3^{999}$ ending with the same digit as $3^3=27$. Thus, $3^{999}$ ends with 7.

Step 2: Units digit cycle of $7^n$.
The units digit of $7^n$ repeats in a cycle of $4$: $7,9,3,1$. Since $1000 \bmod 4 = 0$, we have $7^{1000}$ ending with the same digit as $7^4=2401$. Thus, $7^{1000}$ ends with 1.

Step 3: Multiply units digits.
\[ 7 \times 1 = 7 \] So, the final units digit of $3^{999}\times 7^{1000}$ is 7. \[ \boxed{7} \]