Question

Which of the following numbers will not end with 0 for any natural number \(n\)?

• A. \(4n\) (Note: Assuming \(4 \times n\))
• B. \(4^n\)
• C. \(3^n + 1\)
• D. \(10^{n+1}\)

Hint:

The fundamental theorem of arithmetic ensures that if 5 is not in the prime factorization of the base, no power of that base will ever result in a number ending in 0.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a number to end with the digit 0, its prime factorization must contain both 2 and 5 as factors.
Step 2: Key Formula or Approach:
Check the prime factors of the base of each exponent or the values for \(n\).
Step 3: Detailed Explanation:
1. For \(4^n\): The base is 4. Prime factorization of \(4 = 2 \times 2\). Since there is no factor of 5, \(4^n\) will never end in 0. (It only ends in 4 or 6).
2. For \(4n\): If \(n=5, 10, 15 \dots\), it can end in 0.
3. For \(3^n + 1\): If \(n=2\), \(3^2+1 = 10\). It can end in 0.
4. For \(10^{n+1}\): It always ends in 0 because the base is 10 (\(2 \times 5\)).
Step 4: Final Answer:
\(4^n\) will never end with the digit 0.