Question

Show that 7n cannot end with the digit 0 for any natural no.n

Answer 1

Hey dear!

Here is yr answer.....


Given, 7^n ( where n is any natural no.)


If any no. is going to end with 0 , then they are divisible by 2 and 5

Therefore, in its prime factorization 2 and 5 should occur!

But, 7n = (7×1)^n


Here, 2 and 5 are not occured.....


Therefore, 7^n never ends with 0



Hope it hlpz..

Answer 2

Solution:-

given by form :- 7^n

》according to question:-

》natural number is n = 0 ,1,2,3,4,5....

》here 7^n

》if n= 0 then 7^0=1

》if n=1 then 7^1= 7

》if n =2 then 7^2 = 49

similarly

》if n=3 then 7^3 = 343

》here we find us that any value of n..

》7^n cannot end with the digit 0

》proved :- 7n cannot end with the digit 0 for any natural no.n

☆i hope its help☆