Question

Prove that there is no natural number for which ${4^n}$ ends with the digit zero.

Answer 1

Hint – In this question use the concept of binomial expansion. Binomial expansion of any number of the form ${\left( {1 + x} \right)^n}$ is $1 + nx + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{4!}}{x^4} + .........$. So write 4 as (1+3), and use this concept, this will help approaching the problem.

Complete step-by-step answer:
We have to show that ${4^n}$ 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 ${4^n}$ is written as ${\left( {1 + 3} \right)^n}$ so expand this according to binomial theorem we have,
${\left( {1 + 3} \right)^n} = 1 + 3n + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{3^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{3^3} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{4!}}{3^4} + .........$
Now as we see that in the expansion of ${\left( {1 + 3} \right)^n}$ the first term is 1 and all the remaining terms are positive and multiple of 3.
So if we put n= 0
The sum is 1.
So if we put n = 1
The sum is 4.
If we put n = 2
The sum is 16.
If we put n = 3
The sum is 64.
And If we put n = 4
The sum is 256.
And so on……….
So the unit digit always oscillates between 4 and 6.
So we can say that ${\left( {1 + 3} \right)^n}$ = ${4^n}$ can never end with unit digit zero.
Hence proved.

Note – The trick point was that we started putting the values of n from 0, 1, 2, 3 …….. and did not consider negative values of n. The binomial expansion formula for n is applicable for n belonging to the set of whole numbers only. The sum obtained is the value of ${4^n}$, start showing up from 1 then 4 ………. Clearly these are sets of natural numbers. The basic difference between natural number set and whole number set is that natural numbers is a subset of whole numbers and 0 is not included in natural numbers.