20) The sum of the digits of the square of 999999999999999995 is:
(A) 150
(B) 160
(C) 170
(D) 180
(E) 190.
20) The answer is $(\mathbf{B})$. A number that ends in 5 can be written in the form $10 \cdot a+5$ and its square is $100 a(a+1)+25$. In the given problem, $a=99999999999999999$ and $a+1=10^{17}$. The required square is therefore composed of the digit 9 repeated 17 times, followed by 17 zeros, 2, and 5. The sum of the digits will thus be $17 \cdot 9+2+5=160$.