Question

check whether 20^n can end with the digit 5 for any natural number n

Answer 1

Answer:

Step-by-step explanation:

= 20ⁿ

= ( 5 x 4 )ⁿ

= 5ⁿ x 4ⁿ

To end with the digit 5 any number should be in the form of either 5^m x 3^n or 5^m x 7^n ( Where m and n are whole numbers ).But here it is not so.

Hence 20ⁿ can't end with 0.

Answer 2

Given,

Number = $20^n$

To Find,

Whether $20^n$ can end with the digit 5 for any natural number n =?

Solution,

We have to find whether the one place digit of $20^n$ can be equal to 5 or not.

From factorization of 20 = 2*2*5

We know that, for the last place to be 5, 5 should be multiplied by an odd number, then only the last digit is 5 such as 5*3 = 15, 5*5 = 25.

But in 20 there is no odd number except 5. Whenever 5 is multiplied by 2 or multiples of 2, we get 0 as the last place digit.

$20^2 = 20 * 20 = 400\\20^3 = 20 * 20 * 20= 8000\\20^4 = 20*20*20*20 = 160000$

As we can see, for no value of n  $20^n$ can end with digit 5.

Hence, $20^n$ can never end with the digit 5 for any natural number n.