Show that 12^n can never end with the digit 0 or 5 for any natural number n.

The last digits of 1 2 n can only be 2, 4, 6, or 8 based on the last digits of its components. Therefore, it can never end with the digits 0 or 5 for any natural number n . This conclusion is reached by examining the last digits resulting from the prime factors of 12.
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Answered by Anonymous | 2025-07-04

To prove that 1 2 n can never end with the digit 0 or 5 for any natural number n , let's examine the conditions under which a number ends in 0 or 5.

Ending in 0: For a number to end with the digit 0, it must be divisible by 10. This means that in its prime factorization, it must include at least one factor of 2 and one factor of 5.
Looking at 12 , its prime factorization is 2 2 × 3 . Notice it does not have a factor of 5. Therefore, 1 2 n = ( 2 2 × 3 ) n = 2 2 n × 3 n will never include a factor of 5, regardless of the value of n . Hence, 1 2 n can never end with the digit 0.

Ending in 5: For a number to end in the digit 5, it must be divisible by 5. This again implies that the number's prime factorization must include a factor of 5.
As we established earlier, 12's prime factorization is 2 2 × 3 , and it lacks a factor of 5. Therefore, no matter the value of n , 1 2 n will never include a factor of 5, which means it cannot end with the digit 5.


Hence, 1 2 n can never end with the digit 0 or 5 for any natural number n . This explanation uses basic properties of divisibility and prime factorization, essential concepts in number theory and high school mathematics.

Answered by IsabellaRoseDavis | 2025-07-06