Check whether \(6^n\) can end with the digit 0 for any natural number \(n\).
\(6^n\) cannot end with 0 because it always ends with 6 for all natural numbers \(n\).
We are asked to check whether the number \(6^n\) can ever end with the digit 0, where \(n\) is any natural number.
Step 1: Understand what it means for a number to end with 0
If a number ends with 0, it must be divisible by 10.
Being divisible by 10 means the number must contain both prime factors:
\(2\) and \(5\)
This is because:
\(10 = 2 \times 5\)
So, for \(6^n\) to end in 0, the number must include the factor 5.
Step 2: Look at the structure of \(6^n\)
We know:
\(6 = 2 \times 3\)
So,
\(6^n = (2 \times 3)^n\)
Using exponents, this becomes:
\(6^n = 2^n \times 3^n\)
Step 3: Check whether any power of 6 contains the factor 5
The prime factors in \(6^n\) are only \(2\) and \(3\).
There is no factor of \(5\) in \(6^n\) for any natural number \(n\).
Since a factor 5 never appears, \(6^n\) can never be divisible by 10.
Step 4: Observe the last digit pattern of powers of 6
Let us check the first few powers:
\(6^1 = 6\)
\(6^2 = 36\)
\(6^3 = 216\)
\(6^4 = 1296\)
In each case, the last digit is 6.
This pattern continues for every higher power because multiplying a number ending in 6 by 6 again always gives a number ending in 6.
Step 5: Conclude the result
Since no power of 6 contains the factor 5, \(6^n\) can never be divisible by 10 and therefore cannot end with 0.
Thus:
\(6^n\) never ends with 0 for any natural number \(n\).