Can 6n End With the Digit 0 for Any Natural Number N?
To prove that 6n does not end with the digit 0 for a natural number denoted by n, refer to the solution below.
The fundamental theorem of arithmetic states that any number that ends in 0 must be a factor of both 2 and 5.
100 = 2 x 2 x 5 x 5
= 22 x 52
50 = 2 x 5 x 5
= 21 x 52
10 = 2 x 5
= 21 x 51
6n = (2 x 3)n
= 2n x 3n
6n is a factor of 2 but it's not a factor of 5
Therefore, 6n here is not ending with 0 for any natural number.
What are Natural Numbers?
All positive integers from 1 to infinity are included in the number system, which also contains natural numbers, which are also used for counting. There is no zero included (0). In actuality, counting numerals are 1, 2, 3, 5, 6, 7, 8, 9, etc.
Real numbers, which only comprise positive integers such as 1, 2, 3, 4,5, and 6, excluding zero, fractions, decimals, and negative numbers, include natural numbers as a subset.
Examples of Natural Numbers
The positive integers, usually referred to as non-negative integers, are a subset of the natural numbers that are 1, 2, 3, 4, 5, 6, …∞. In other terms, natural numbers are a collection of all whole numbers other than 0. Examples of natural numbers include 23, 56, 78, 999, 100202, etc.
Check Whether 6n Can End With the Digit 0 for Any Natural Number N
It is proved that 6n cannot end with the digit 0 for any natural number n. Natural numbers are the numbers that are positive integers and include numbers from 1 to ∞. The set of natural numbers is indicated using the letter “N”.
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