Question

Show that 21^n cannot end with the digits ,0,2,4,6,8 for any natural number n

Answer 1

Step-by-step explanation:

let P(n):21^n is an odd no.

for n=1, 21^1 is odd.p(1) is true.

assume that P(k) is true, ie.

21^k is odd.

now we have to prove P(k+1) is true whenever P(k)is true.

let 21^k = 2m+1 (m is any natural no.)

21^(k+1)= 21^k×21

=(2m+1)×21

=42m+21

=an even no. +an odd no. = an odd no.

=cannot end with digits 0, 2,4,6,8

therefore P(n) is true for any natural no. n