Wild About Math! Making Math fun and accessible


MMM #31: Not very complex

The solution to MMM #30 is up at Blinkdagger. Here we go with #31.

Hint: The solution to the problem is a real number. Show your work.

Here are the rules for the contest:

1. Email your answers with solutions to mondaymathmadness at gmail dot com.
2. Only one entry per person.
3. Each person may only win one prize per 12 month period. But, do submit your solutions even if you are not eligible.
4. Your answer must be explained. You must show your work! Wild About Math! and Blinkdagger will be the final judges on whether an answer was properly explained or not.
5. The deadline to submit answers is Tuesday, May 5, 12:01AM, Pacific Time. (That’s Tuesday morning, not Tuesday night.) Do a Google search for “time California” to know what the current Pacific Time is.)
6. The winner will be chosen randomly from all timely well-explained and correct submissions, using a random number generator.
7. The winner will be announced Friday, May 8, 2009.
8. The winner (or winners) will receive a Rubik’s Revolution or a $10 gift certificate to Amazon.com or $10 USD via PayPal. For those of you who don’t want a prize I’ll donate $10 to your favorite charity.
9. Comments for this post should only be used to clarify the problem. Please do not discuss ANY potential solutions.
10. I may post names and website/blog links for people submitting timely correct well-explained solutions. I’m more likely to post your name if your solution is unique.

Comments (5) Trackbacks (0)
  1. Hi,

    Is the 2nd term – i^(-i) or -i^i ? I can’t quite see if there is a minus sign in the power.


  2. Tom, the second term is i^i.

  3. Yeah, i think the negative sign you see is just the dot of the base i. Easy mistake to make though.

  4. The answer is 0:i^i=(1/i)^(1/i)

  5. the answer is zero

    (1/i) = -i and (1/i)^(1/i) = -i^(-i)=(1/-i)^i
    = i^i

    hence (1/i)^(1/i) – i^i = i^i – i^i = 0

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