101010
Michelle Malkin reported the results of Lockheed Martin's "MATHCOUNTS" competition yesterday. The contest was won by 6th grader Darryl Wu.
The final question he answered (in 45 seconds) to win was:
A set of distinct positive integers has a total of 11 digits, and all the digits are ones. What is the smallest possible sum of the integers in the set?
I'll put the answer in the comments in case you want to try on your own. I'm proud to say that I was able to figure it out myself, so at least I'm as smart as a sixth-grader.
In the comments to Michelle's post, someone wisecracked that the answer was 42 (the answer to everything).
However, as another commenter pointed out, the "distinct positive integers" used in solving the above problem, if summed together in binary, do in fact add up to the binary equivalent of the decimal number 42.
Positively geektacular!
posted by Dan S. @ 3:32 PM
5 comments
5 Comments:
The answer to the question is (1 + 11 + 111 + 11,111 =) 11,234.
In binary, 1 + 11 + 111 + 11111 = 101010, which is binary for 42.
Pretty cool, eh?
Geek.
Thank you.
I'm an English major. Ask me what a gerund is and I'll tell you that.
Just no dangling participles, ok? I try to keep this site PG13.
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