Question 504798
That number is equivalent to

*[tex \LARGE (\frac{10^{12} + 2}{3})^2], or

*[tex \LARGE \frac{10^{24} + 4*10^{12} + 4}{9}], or

(100...00400..004)/9. We can rewrite the fraction as

*[tex \LARGE \frac{1000...0}{9} + \frac{4000...0}{9} + \frac{4}{9}]

*[tex \LARGE = 111...111 + 444...444 + 1] (note that we have to account for the fractional part of the number by adding 1).

Hence, this is simply a 24-digit number consisting of twelve 1's, eleven 5's, and one 6. The digit sum is 12(1) + 11(5) + 6 = 73.