9oops sorry nishant sir i dint notice u ve already posted :)
ps. b555 will be angry uve solved this ,he'll be tellin later that he posted it for juniors to solve
, he even banned me from posting clues !
Arithmetic progressions
Consider the following infinite table :
04, 07, 10, 13, 16, 19, ...
07, 12, 17, 22, 27, 32, ...
10, 17, 24, 31, 38, 45, ...
13, 22, 31, 40, 49, 58, ...
16, 27, 38, 49, 60, 71, ...
19, 32, 45, 58, 71, 84, ...
...
The pattern is quite obvious I think.
Each row/column contains an infinite arithmetic progression.
Prove or disprove :
The number k doesn't occur in this table if and only if 2k + 1 is prime.
-
UP 0 DOWN 0 1 8
8 Answers
341good one.
keyword: Simon's Favourite Factorisation Technique :D
62t(m,n) = 1+3m+(2m+1)(n-1)
t(m,n) = 1+3m+2mn+n-2m-1 = m+n+2mn
m+n+2mn=k
2k+1=2m+2n+4mn+1=(2m+1)(2n+1)
Awesome Prophet sir... I am learning a lot of maths from you :)
I dont know if this problem needed the hint of Simmon's Favorite Factorization technique but then I got to know that this was called the same. :)
62btw even i solved on problem by b555 after ages... I dont remember which the last one was :P
9am i eligible to post the full ans b555 ?
or shall i atleast give a clue
a really nice Q from u b555 :)
ps. wow prophet sir u have a name for each and every technique :D
9341This name is more by the way of a joke I think :D
I have messed the name a bit. You can read it at
http://www.artofproblemsolving.com/Wiki/index.php/Simon's_Favorite_Factoring_Trick
39celestine , yes i had banned u from posting the solution (rather even clue) last time .
but i am sorry for that.
i think we should all be having fun with these questions ( u ,kaymant sir , obviously prophet sir, and nowadays even nishant sir seems to be back with this this website) .
and also after aug 1st , u would not be there , i also would not be there.
i mean we may not totally leave this website but we would be obviously busy with college lives and may login only once in a time and solve some questions for juniors rather than check out such questions.
so dude , till july ending lets all really enjoy the true beauty of mathematics.
and by the way are my questions being idiotic or really toughies? if so kindly say , i would change my style of questions. anyone would be wanting to comment on my questions is laways welcome to do so.
but wat i say is till atleast july ending lets all have good fun.
and celestine u can always post answers to all my questions .(which is a very easy task for u).
39just an other way of expressing the solution
We can identify each number in the table by the number of steps we go downwards and the number of steps we have to go to the right,
so all the numbers t in the table must have form 4+3*k+(3+2k)*n where k and n are any 2 natural numbers (includding 0).
This means that for all t in the table 2t+1=9+6k+6n+4kn=(2k+3)(2n+3), so I guess this shows that t isn't in the table iff 2t+1 can't be written as (2k+3)(2n+3) with k,n>=0, which is equivalent to saying that 2t+1 can't be written as the product of 2 factors >1 iff 2t+1 is a prime.