Question

The sum of digits of no. 10^4n^2 + 1 ^2 where n is positive enteger

Answer 1

Hii

Be it any number just calculate the total number of zeros that will precede the unit digit in the number

eg 101*101 = 10201
1001 *1001 =1002001
10001*10001 = 100020001

Hence in any case the sum of digigts is always going to be 4

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