If f=(10^10)-47, what is the sum of all the digit of f?  Choices: A) 40 B) 45 C) 50 D) 55 E) 80
Start small and check for any pattern:   10^2 - 47 = 53 -&gt; number of digits = power of 10; number of 9s = (power of 10) - 2; 53 is the last 2 digits   10^3 - 47 = 953 -&gt; number of digits = power of 10; number of 9s = (power of 10) - 2; 53 is the last 2 digits   ...extrapolate   10^10 - 47 -&gt; number of digits = 10; number of 9s = 8; 53 will be the last 2 digits   f= 9*8 + 5 + 3= 
<gadget id="calculator">9*8 + 5 + 3</gadget>
<output>80</output>
80.   Answer (80).
<result>E</result>
If f=(10^10)-47, what is the sum of all the digit of f?