show that 9^n cannot end with the digit 2 for any natural number.

for the last digit to be 2,9^n should be divisible by 2.But for whatever real number n it cant be divisible by 2.

  • -12
If 9^n end with 2 it must have 2 in its prime factorization(That is it must be a multiple of 2). 9^n = 3^n * 3^n
We can see prime factorization on 9^n does not contain 2. Hence it can't end with 2.
  • 17
to end with number 2, 9n show divisible by 2 and the prime factorisation of 9n should cannot the number 2 but the prime factorisation  of 9n will be 1 and 3 thus cannot end with digit 2 for any natural number.
  • -10
Because 9^n is not of the form 2^1+4n where n is a non negative integar.
  • -8

  • -9
If odd no. like (3×3)^n is 3^2n so 3 is a odd number and its power is even then remember it will not end with 2 . if you dont trust we do 3^2=9,3^3=27,3^4=81,3^5=243 and so on .. hope it must be helpfull to you
  • 1
expand (10-1)^n binomially the product will be 10*integar +9 or -1 depending on the n
  • 1
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