| The Mathematical olympiad forum? | |
|---|---|
| Tweet Topic Started: Mar 31 2009, 10:01 AM (1,064 Views) | |
| Einstein the 2nd | Mar 31 2009, 10:01 AM Post #1 |
![]()
|
Firstly, I do not think that anyone will come here. Secondly, I am asking help because i really do not know how to do these questions and i am not ashamed to ask questions because, if you never ask you'll never know. ( Hitler don't sprout crap about this thing, i know its wrong) Thirdly,STRICTLY NO DIGRESSING IN THIS FORUM LIKE YOU DO FOR THE OTHERS. i.e. from math olympiad to syazani being short, to "Yes you finally get it!", etc. |
![]() |
|
| Einstein the 2nd | Mar 31 2009, 10:09 AM Post #2 |
![]()
|
Ok, so erm... how do you do this Prove that for every integer n, the number 77n +1 is the product of at least 2n + 3 primes. (the primes do not have to be distinct) Solution: let A(n) = 77n +1. So lets assume its as the question says and has 2n+3 prime divisors. Then we can express A(n+1) as [(A(n)-1)7] +1. Go do your binomial theorem, and you will know why. there is symetry right? 1,7,21,35,35,21,7,1. the there's a (A(n)6) and the rest...which after factoring, we get: (A(n+1) ) = A(n)(A(n))6 - 7 (A(n)-1)( A(n))2 - A(n) + 1)2) so, from this, 7(A(n)-1) = 77n+1 = 72k The 2k just expresses the thingy i said earlier on which also makes the number a square. Perfect squares. So the thingy i said earlier on is also a perfect square! Then with difference of squares property, It is a product of 2 numbers, so + 2 prime factors! so A(n+1)= 2 + 2n + 3 prime factors, hence, proven. QED!!! YAYAYAY!! [/sup][/sup] Edited by Einstein the 2nd, Apr 4 2009, 12:39 PM.
|
![]() |
|
| Williamcxp | Mar 31 2009, 02:54 PM Post #3 |
|
and no one replies.. not even hitler.. sorry i do not understand the question much less know how to do.. but i would be interested to be taught. ^^ |
![]() |
|
| Hitler | Mar 31 2009, 02:55 PM Post #4 |
![]()
|
i am not helping you with this unless you help me with the modulo part. i have not tried yet but i already know you specially chose a question requiring the use of it to tick me off. |
![]() |
|
| Hitler | Mar 31 2009, 02:55 PM Post #5 |
![]()
|
k last post of the day bye now... |
![]() |
|
| Einstein the 2nd | Apr 1 2009, 10:00 AM Post #6 |
![]()
|
I can't even solve it?? |
![]() |
|
| Hitler | Apr 1 2009, 12:21 PM Post #7 |
![]()
|
you do not have to know it requires modulo. you just have to do it until you get stuck at the modulo part. |
![]() |
|
| Einstein the 2nd | Apr 1 2009, 12:36 PM Post #8 |
![]()
|
fine, thanks for reminding me.... ahaha just joking. |
![]() |
|
| Hitler | Apr 1 2009, 12:51 PM Post #9 |
![]()
|
of course you were. if you were not you would not be able to post that reply. |
![]() |
|
| Einstein the 2nd | Apr 1 2009, 12:56 PM Post #10 |
![]()
|
Stupid.
Edited by Einstein the 2nd, Apr 1 2009, 01:02 PM.
|
![]() |
|
| 1 user reading this topic (1 Guest and 0 Anonymous) | |
| Go to Next Page | |
| « Previous Topic · School Discussion · Next Topic » |









7:21 PM Jul 11