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

Let's evaluate each option to determine which expression does not end with 0 for any natural number \( n \).

1. \(4n\) (Assuming \(4 \times n\)):
   • For any natural number \( n \), \( 4n \) may or may not end with 0. It depends on whether \( n \) is a multiple of 5. For example, if \( n = 5 \) or \( n = 10 \), \( 4n \) will end with 0 (e.g., \( 4 \times 5 = 20 \)). Thus, this option can end with 0.
2. \(4^n\):
   • For any natural number \( n \), the expression \( 4^n \) will never end with 0.
   • This is because \( 4^n \) is a power of 4, which is composed only of the factor 2, and lacks the factor 5. A number ends with 0 only when it is divisible by both 2 and 5 (making it a multiple of 10).
3. \(3^n + 1\):
   • For any natural number \( n \), the expression \( 3^n \) will never end with 9, so \( 3^n + 1 \) will never end with 0.
   • The number \( 3^n \) cycles through some units digits (e.g., 3, 9, 7, 1), none of which is 9. Thus, adding 1 will result in numbers ending with 4, 0, 8, or 2.
4. \(10^{n+1}\):
   • This is a power of 10, which always ends with at least one 0. In fact, it ends with \( n+1 \) zeros.
   • Thus, for any natural number \( n \), \( 10^{n+1} \) will certainly end with 0.

Based on the analysis, \(4^n\) is the only option that does not end with 0 for any natural number \( n \).