Question

Check whether \[{6^n}\] can end with a digit zero for any natural number n.

Answer 1

Hint: We will use prime numbers to denote the product of the number given. We also know that a number with end digit zero can be expressed with the help of two basic prime numbers 2 and 5. So we will express the given number in the form of the product and will check if it contains 2 and 5 or not.

Complete step-by-step answer:
We are not given the fixed power of 6. Like $6^1$ is 6 only.
$6^2$ is the square of 6 that is 36.
This will keep increasing as we add 1 more power.
But we need to check whether the number in the unit's place or the number that ends with zero is there or not.
So for that we will use the 6 in the form of a product.
We can write the numbers that end with zero as
\[10 = 2 \times 5\]
\[20 = 2 \times 2 \times 5\]
\[100 = 2 \times 2 \times 5 \times 5\]
In general we can say that the product contain 2 and 5 compulsorily
Now let’s check for 6.
\[6 = 2 \times 3\]
But 6 cannot be written in the form of product of 2 and 5.
So it is clear that it cannot be expressed to any power n that gives the end digit as zero.
So, the correct answer is “Option B”.

Note: Here note that we can also check this by taking the different powers of 6 like
\[{6^1} = 6,{6^2} = 36,{6^3} = 216...\]
Always the end digit is 6 and in the next power we will multiply the number again with 6. Such that the resulting product will be 6 at end digit again.
Thus we can simply conclude that no power of 6 can lead to zero at the end.
Also note that the tenth power of 6 won't mean it is 0 at the end ; it only means multiplying 6 ten times.