Like all great things, they must come to an end. As such, at 10am PST, this morning – our N900 Push competition came to a close. We had some really awesome answers, some really round about, and of course the obligatory – really bad ones. For those that are just on the EDGE of your seat waiting for the final concluding answer to stop the arguing and fighting – to settle this whole dispute. The answer, and the winners are…After the break.
Just to recap, the question –
You’re working with any one of the team winners in the PUSH N900 competition. Your team needs two like resistors in order to complete the project on time.
Luckily your work drawer is overflowing with resistors, and you know that there are only 4 different values since you pulled them yourself from old junk.
You close your eyes and reach in. What is the minimum amount of resistors you have to pull out to ensure 2 of the same value?
The Correct answer is 5 resistors. [Talin Salway], [Sari Ibrahim], [James R. Smoot], [Jody Halyk], and [Thomas Remmert] all sent in the correct answer, were picked by us, and gracefully pass their tickets to another winner. In the end [Robrecht Noens] was the only one to respond within our 24 hour limit claiming tickets. Congratulations, and hope you have fun at the N900 PUSH Showcase [Robrecht]!
For all those that sadly, did not win. You can still keep up on the party, [Matt] let us know…
We’ll be updating our @PUSHN900 Twitter account regularly using the #PUSHN900 hashtag to help follow what you guys are saying.
We’ll be uploading video using Qik.com/PUSHN900
Keep checking the PUSH N900 Facebook page for updates as well.
And finally an explanation: The entire riddle (as a lot of readers repeated to us in the entries) is based on the Pigeonhole Principle. In the worst case scenario, you reach in and pull out 4 different values. Thus, you must pull out 5 to ensure at least 2 match.