Question

Show that the given number can never end with unit digit zero; ${9^n}$.

Answer 1

Hint – In this question write 9 as a sum of 1+8, then apply binomial expansion for the term of the form ${\left( {1 + x} \right)^n}$. If we can prove that by putting any values of n the expansion oscillates between numbers other than zero then it means that ${9^n}$ can never end with a unit digit.

Complete step-by-step answer:

We have to show that ${9^n}$ can never end with unit digit zero.

Proof –
According to binomial theorem the expansion of ${\left( {1 + x} \right)^n}$ is
${\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{4!}}{x^4} + .........$

Now ${9^n}$ is written as ${\left( {1 + 8} \right)^n}$ so expand this according to binomial theorem we have,
${\left( {1 + 8} \right)^n} = 1 + 8n + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{8^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{8^3} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{4!}}{8^4} + .........$

Now as we see that in the expansion of ${\left( {1 + 8} \right)^n}$ the first term is 1 and all the remaining terms are positive and multiple of 8.

So if we put n= 0
The sum is 1.

So if we put n = 1
The sum is 9.

If we put n = 2
The sum is 81.

And If we put n = 3
The sum is 729.

And so on……….

So the unit digit always oscillates between 1 and 9.

So we can say that ${\left( {1 + 8} \right)^n}$ = ${9^n}$ can never end with unit digit zero.

Hence proved.

Note – It’s a unique question of its kind, it is always advised to remember the direct binomial expansion of${\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{4!}}{x^4} + .........$. Key point here is we can’t put the value of n as negative, this equation holds true for n belonging to the whole number only that is from 0 to infinity.