Question

The digit in the unit's place of \(3^{999}\times 7^{1000}\) is \underline{\hspace{1.5cm}.}

• A. 7
• B. 1
• C. 3
• D. 9

Hint:

For units digits, use the mod-10 cycles of the base: powers of \(3\) and \(7\) both repeat every 4 steps.

Answer 1

The Correct Option is A

Step 1: Units digit of \(3^{999}\).
Cycle for powers of \(3\): \(3,9,7,1\) (period \(4\)).
\(999 \equiv 3 \pmod{4}\Rightarrow\) units digit \(=7\). \smallskip

Step 2: Units digit of \(7^{1000}\).
Cycle for powers of \(7\): \(7,9,3,1\) (period \(4\)).
\(1000 \equiv 0 \pmod{4}\Rightarrow\) units digit \(=1\). \smallskip

Step 3: Multiply units digits.
\(7\times 1\) has units digit \(=7\).

Final Answer: \(\boxed{7}\)