Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
| |
|
|
You are not logged in. #1 2008-11-21 08:16:04
Daniel's Challenge ThreadRather than me setting exercises for others to do, I'd like people to give me some challenging questions #2 2008-11-21 08:54:08
Re: Daniel's Challenge ThreadHi. The solution uses material you should be familiar with (because I’ve seen you discuss it on the forum). Let me know if you need a hint. #3 2008-11-21 10:41:19
Re: Daniel's Challenge ThreadIt would be helpful if we could multiply by a number n such that 13n is 1 more than a mutliple of 29. The smallest such number is 9. Mutliplying by 9 gives: i.e. Thank you Jane Last edited by Daniel123 (2008-11-21 10:44:17) #4 2008-11-30 00:09:40
Re: Daniel's Challenge ThreadHeres a VERY VERY VERY easy sum: #5 2008-11-30 00:11:49
Re: Daniel's Challenge ThreadWriting the information out algebraically: #6 2008-12-12 12:25:16#7 2008-12-25 12:41:50
Re: Daniel's Challenge ThreadWell the first part comes out from the sum of a geometric series: Induction proves the second pretty quickly: Obviously the first part could also be done (and more quickly) by induction, but I feel induction gets rid of the 'why' behind it. I'm sure you're going to show me a nicer way to do the second part Last edited by Daniel123 (2008-12-25 23:00:29) #8 2008-12-28 14:43:33
Re: Daniel's Challenge ThreadWell, you can do both parts in a nice way, actually. Next one? #9 2009-02-08 23:38:10#10 2010-09-29 06:05:47
Re: Daniel's Challenge Thread
All we hv to do is to prove that 9k^2 - 1 is divisible by 8. |