Question

Consider the numbers 4^n where n is a natural number.Check whether there is any value of n for which 4^n ends with the digit zero.

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Answer 1

For the number 4" to end with digit zero for any natural number n, it should be divisible by 5. This means that the prime factorisation of 4¹ should contain the prime number 5. But it is not possible because 4" = = (2) 2¹ so 2 is the only prime in the factorisation of 4". Since 5 is not present in the prime factorization, so there is no natural number n for which 4" ends with the digit zero.

Copied because IK the motive behind asking this question...XD

Answer 2

Answer:

4ⁿ's prime factorisation doesn't contain 5 so it cannot end with zero.

Step-by-step explanation:

A number ends with zero only when it contains 5 and 2 as its prime factors.

Let's are some examples :

1. 10 = 2 × 5
2. 30 = 2 × 3 × 5
3. 50 = 2 × 5 × 5

Now coming to 4ⁿ.

Let's prime factorise 4ⁿ

• 4ⁿ = (2 × 2)ⁿ

In the prime factorisation of 4ⁿ we can find only 2 but cannot find 5.

So 4ⁿ cannot end with zero as inorder to end with zero its prime factorisation should contain both 2 as well as 5.

Thanks !