Problem Set #3

The question that I have chosen to do this blog on is number 11. Here is the question.

11. Let N = 10^3 + 10^4 + 10^5 +10^6 + 10^7 + 10^8 + 10^9. The sum of the digits of N is
       A. 12    B. 1    C. 6    D. 9    E. 7

At first when i saw this question, i was thinking, oh no, I'm going to have to count so many zeros! i started to find out how many zeros were in each exponent, ( eg. 10^3 = 1000, 10^4 = 10000, etc.) but then i realized that the question wasn't asking me what the answer to the equation was, only for the sum of all the digits. After i realized that, i figured out quite easily that no matter how many zeros there are in the equation, there is only going to be one 1. So the sum of the digits would always be one. So that makes the answer one (:


I liked this question because it made me realize that questions that seem to be hard at first can actually turn out to be really easy. I was so happy when i caught my own mistake and was able to solve the problem easily.


Something that i learned from this question is that i should read the whole question first before i rush into it without really knowing what I'm doing. If i had paid more attention to the question at first, then i could have gotten through this question quickly.


- casie