Question

Guys plzz help :)

show that 21^n cannot end with the digit 0, 2, 4, 6 and 8 for any natural number n.

Answer 1

21ⁿ cannot end with the digit 0,2,4,6,8
prime factorization of 21 = ( 7×3)ⁿ
Since prime factorization of 21 does not contain 0,2,4,6,8 it cannot end with these digit for any natural n

Answer 2

HELLO DEAR,


WE CAN WRITE 21^n =(3×7)^n

AND WE KNOW THAT ANY POWER OF 3 WE

CAN'T GET EVEN NO. AS ITS END DIGIT ,SAME GOES FOR 7 .

SO,

( 21 )n  ​CANNOT END WITH DIGIT 0 , 2 , 4 , 6 AND 8 FOR ANY NATURAL No. n . 


WE CHECK IT AS BY PUTTING THE VALUES OF n

when n=2


21² =441.

, ( End digit is not as ​0 , 2 , 4 , 6 and 8  )


At n  = 3

( 21 )3 = ​9261                      ( End digit is not as ​0 , 2 , 4 , 6 and 8  )

At  n  =  10 

( 21 )10  = ​16679880978201    ​ ​( End digit is not as ​0 , 2 , 4 , 6 and 8  ) 



SO WE CAN SAY THAT 

( 21 )n ​cannot end with digit 0 , 2 , 4 , 6 and 8 for any natural no. n .                ( Ans )

I HOPE ITS HELP YOU DEAR,
THANKS