properties of prime numbers

Properties Of Prime Numbers

Consider the number 8ⁿ, where n is a natural number.

Is there any value of n for which 8ⁿ ends with zero?

It is difficult to answer this question directly. However, we can answer this question by making use of the fundamental theorem of arithmetic. It states that

“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”.

This means that if we are given a composite number, then that number can be written as a product of prime numbers in only one way (except for the order of prime numbers).

For example: the composite number 255 can be written as the product of primes as follows.

255 = 3 × 5 × 17

Also, 255 can be written as 3 × 17 × 5 or 5 × 3 × 17 or 5 × 17 × 3 or 17 × 3 × 5 or 17 × 5 × 3.

Thus, we can see that 255 can be expressed as a product of unique prime numbers 3, 5, and 17 but the order of representation may differ.

Now, by making use of the above theorem, we can answer the question which we were discussing in the beginning. Let us see how.

Suppose the number 8ⁿ ends with zero for some value of n.

Since the number ends with zero, it should be divisible by 10.

Now, 10 = 2 × 5

Thus, this number should be divisible by 2 and 5 also.

Therefore, the prime factorization of 8ⁿ should contain both the prime numbers 2 and 5.

We have, 8ⁿ = (2³)ⁿ = 2³ⁿ

⇒ The only prime in the factorization of 8ⁿ is 2.

Thus, by fundamental theorem of arithmetic, there is no other prime in the factorization of 8ⁿ.

Hence, there is no natural number n for which 8ⁿ ends with the digit zero.

In this way, we can make use of the above theorem.

Let us now look at some more examples to understand this concept better.

Example 1:

Prove that the number 9ⁿ, where n is a natural number, cannot end with a zero.

Solution:

Suppose the number 9ⁿ ends with a zero for some value of n.

Since the number ends with zero, it should be divisible by 10.

Now, 10 = 2 × 5

Thus, this number should be divisible by 2 and 5 also.

Therefore, the prime factorization of 9ⁿ should contain both the prime numbers 2 and 5.

We have, 9ⁿ = (3²)ⁿ = 3²ⁿ

⇒ The only prime in the factorization of 9ⁿ is 3.

Thus, by fundamental theorem of arithmetic, there is no other prime in the factorization of 9ⁿ.

Hence, there is no natural number n for which 9ⁿ ends with the digit zero.

Example 2:

Check whether the numbers 49ⁿ, where n is a natural number, can end with a zero.

Solution:

Suppose the number 49ⁿ ends with a zero for some value of n.

Since the number ends with zero, it should be divisible by 10.

Now, 10 = 2 × 5

Thus, this number should be divisible by 2 and 5 also.

Therefore, the prime factorization of 49ⁿ should contain both the prime numbers 2 and 5.

We have, 49ⁿ = (7²)ⁿ = 7²ⁿ

⇒ The only prime in the factorization of 49ⁿ is 7.

Thus, by fundamental theorem of arithmetic, there is no other prime in the factorization of 49ⁿ.

Thus, there is no natural number n for which 49ⁿ ends with the digit zero.