Question:
The rightmost non-zero digit of the number \(30^{2720}\) is
The rightmost non-zero digit of the number \(30^{2720}\) is
Show Hint
Write \(30^{2720} = 3^{2720} \times 10^{2720}\); the factor \(10^{2720}\) only adds zeros at the end. Use the cycle of last digits of powers of 3.
Updated On: Nov 4, 2026
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- 9
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The Correct Option is A
Solution and Explanation
Step 1: Split the number:
\(30^{2720} = 3^{2720} \times 10^{2720}\). Multiplying by \(10^{2720}\) just appends 2720 zeros, so the rightmost non-zero digit of \(30^{2720}\) is the last digit of \(3^{2720}\).
Step 2: Cycle of last digits of powers of 3:
\(3^1 \to 3,\ 3^2 \to 9,\ 3^3 \to 7,\ 3^4 \to 1\), then repeats with period 4.
Step 3: Use the exponent:
\(2720 = 4 \times 680\), so the exponent is a multiple of 4 and \(3^{2720}\) ends in 1.
Final Answer:
The correct option is (1): 1
\(30^{2720} = 3^{2720} \times 10^{2720}\). Multiplying by \(10^{2720}\) just appends 2720 zeros, so the rightmost non-zero digit of \(30^{2720}\) is the last digit of \(3^{2720}\).
Step 2: Cycle of last digits of powers of 3:
\(3^1 \to 3,\ 3^2 \to 9,\ 3^3 \to 7,\ 3^4 \to 1\), then repeats with period 4.
Step 3: Use the exponent:
\(2720 = 4 \times 680\), so the exponent is a multiple of 4 and \(3^{2720}\) ends in 1.
Final Answer:
The correct option is (1): 1
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