Question

If n=123456789*76543211 + (23456789)^2 then sum of the digits of n

Answer 1

Answer:

1

The sum is 10^16

so the sum of digits = 1

Answer 2

Answer:

0

Step-by-step explanation:

N= (123456789) * (76543211) + (23456789) ^2

     by observation

    (100000000 +23456789) * (100000000 -23456789) + (23456789)^2

 by identity (a+b)(a-b) = a^2 - b^2

   100000000^2 -(23456789)^2 +(23456789)^2

after cancellation

 = 100000000^2

 = (10^8)^2

 =10^16

Therefore sum of digits =1