Question

Show that 6^n can never end with digit 0 for any natural number n.​

Answer 1

Answer:

If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5

Prime factorisation of 6n = (2 ×3)n

It can be observed that 5 is not in the prime factorisation of 6n.

Hence, for any value of n, 6n will not be divisible by 5.

Therefore, 6n cannot end with the digit 0 for any natural number n.

Answer 2

$\underline{\underline{\sf{\maltese\:\:To\: prove}}}$

• ${6}^{n}$ can never end with digit 0 for any natural number n.

$\underline{\underline{\sf{\maltese\:\:Proof}}}$

Any number which ends in zero must have at least 2 and 5 as its prime factor.

Since,

6 = 2 × 3

Therefore,

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

$\longrightarrow \: {6}^{n} = {2}^{n} \times {3}^{n}$

Hence, prime factors of 6 are 2 and 3.

Since, ${6}^{n}$ does not contain 5 as a prime factor.

Hence, ${6}^{n}$ can never end in zero.

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LEARN MORE :-

➣ The Fundamental Theorem of Arithmetic

Every composite number can be expressed as a product of primes and this factorisation is unique, apart from the order in which the prime factors occur. Fundamental theorem of arithmetic is also called a Unique Factorisation Theorem.