Question

Show that any number of the form \[{6^n}\], where \[n \in N\] can never end with the digit \[0\].

Answer 1

Hint: Here in this question, we have to show that any number of \[{6^n}\] does not end with the zero. For this, we have to give the values for \[n\] the value should be natural number i.e., \[n \in N\] where, \[n = 1,2,3,....\] then check if the resultant value ends with zero or not and this can also be shown by using a factor.

Complete answer: Consider the question,
Given, \[{6^n}\], where \[n \in N\]
We have to show that any number of \[{6^n}\] can never end with the digit zero.
Let us take the example of a number and it’s factors which ends with the digit \[0\]
So, \[10 = 2 \times 5\]
\[20 = 2 \times 2 \times 5\]
\[50 = 2 \times 5 \times 5\]
\[100 = 2 \times 2 \times 5 \times 5\]
Here, we note that the numbers ending with \[0\] have both 2 and 5 as their prime factors or multiples of 5 and 10.
 Now consider,
\[ \Rightarrow \,\,\,{6^n}\]
Factors of 6 are 2 and 3 i.e., \[6 = 2 \times 3\], whereas
\[ \Rightarrow \,\,\,{6^n} = {\left( {2 \times 3} \right)^n}\]
Here it doesn't have 5 as a prime factor. So, it does not end with zero.
Therefore, any number of \[{6^n}\] , where \[n \in N\] can never end with the digit \[0\].

Note:
By factoring a number means to express it as a product of two numbers or it can also be defined as the division of a given number by some other number such that the remainder is zero. Remember if terms like \[{x^n}\], the value of x must contain zero or zeroes at the end otherwise it must have the factor as 5 or 10 otherwise the term \[{x^n}\] doesn’t contain zero at the end.