Question 52

Sum of digits of $$(10^{4n^2 + 8} + 1)^2$$. where n is a positive integer, is

Solution

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The given expression is $$(10^{4n^2 + 8} + 1)^2$$.

The expression will be of the form $$(10000....1)^2$$

Any number of the form 1000(x times)1 will be of the form 1000(x times)20000(x times)1.

For example $$1001^2$$=1002001

So, the sum of the digits will always we 1+2+1=4

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