Wild About Math!

MMM #21: How many divisors?

Blinkdagger has announced the winner for MMM #20. Congratulations, Diego Vila Cid!

Now, onto MMM #21.

I have a Rubik’s Revolution, courtesy of Techno Source (or $10 Amazon.com gift certificate, if you prefer, or $10 in USD via PayPal to non-US folks) to give to the winner.

Here’s the problem:

10 is said to have 4 divisors because 4 whole positive integers (1, 2, 5, and 10) divide it. How many divisors does 10! have, where 10! = 1×2x3…x8×9x10? Show how you calculate your answer. While you might write a computer program to verify your answer by brute force, computer solutions are not accepted.

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, December 16, 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, December 19, 2008.
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.

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