Question

check whether 6n can end with
the digit o for any natural
number n
​

Answer 1

$\large\underline{\sf{Solution-}}$

Before Prove the statement,

↝ Let us consider an example of a natural number which ends with 0 or having 0 i its unit place.

↝ Let us consider 10 whose prime factorization is

• ↝ 10 = 2 × 5

↝ Let us consider another example 20 whose prime factorization is

• ↝ 20 = 2 × 2 × 5

↝Let us consider one more example 30 whose prime factorization is

• ↝ 30 = 2 × 3 × 5

↝We observe that, for a number to be end with 0,its prime factorization must have factor of 2 and 5 together.

Now,

Consider,

$\rm :\longmapsto\: {6}^{n} \: where \: n \in \: natural \: number$

can be factorized as

$\rm :\longmapsto\: {6}^{n} = {(2 \times 3)}^{n}$

Since, it doesn't contain 2 and 5 as a factor together,

$\bf\implies \: {6}^{n} \: can \: never \: ends \: with \: 0$

Additional Information :-

The fundamental theorem of arithmetic -

• "Every composite number can be factorized as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur".

Euclid's Division Algorithm

• It is a technique to compute the Highest Common Factor (HCF) of two given positive integers. HCF of two positive integers a and b is the largest positive integer d that divides both a and b.

Euclid's Division Lemma

• If we have two positive integers a and b, then there would be whole numbers q and r that satisfy the equation: a = bq + r, where 0 ≤ r < b. a is the dividend. q is the quotient and r is the remainder.