9n! + r = r(n!/r + 1 ) so its composite as long as r≥2 and <=n
Show that for any number n, there are consecutive n composites that exist...
This is not an olympiad question.... It is very much from the JEE syllabus.. Just that you have to think a bit.. (hint think of n!)
-
UP 0 DOWN 0 0 39
39 Answers
62@celestine.. You proof decieve me ;)
I thought this is exactly what you were upto :)
N!+N ≠(N+1)!
9
1and hw wld u get 5.....philip?
nd u won't even get 7,11,13,17.............?
9but then honey n is obviously not 1 cos u cant have one consective nos ;)
62that is true honey...
for n=1 you need to take 6 as an example
and there are 2 consecutive composites in 14 and 15
infact if it is true for N=5 then the subsets will give it true for n=1,2 ,3, 4
The proof of what celestine said is that it will have
N! + K will have k as a factor for all k<=N
SO it will be divisible by K :)
hence what celestine said holds :)
1ok so watever i understood
the correct ans is "n!+2 to (n+1)! are composites"
for 3
8 to 24 all nos are composites [11]
plz .....HELP !! [2][2][2]
1ys.............thr wld be terms in between which r nt composite...........wt bout thm?
1@Nishant bhaiya...
One condition must be applied on n... saying dat n≠2k i.e., n must not be an even no.
There cannot be even no. of consecutive composite no.s .....
62no!!!!!
I din see correctly what celestine wrote
the answer is
N!+2... N!+N
these are consecutive N-1 composites
So for N composites we have to take
(N+1)!+2 ...... (N+1)!+N+1
so, 5!+2, 5!+3, 5!+4, 5!+5 will be all composites..
62see ":)"
If i say that 1,2,3,4,5,6 are 6 consecutives then I can also say that
1,2,3,4,5 are 5 consecutives
1uff thanx i was really begining to believe that tomorrow i would see flying foxes and lemurs [1]
this looks much more precise hmm.......
1then the question must have "atleast n-consecutive composites" rather than "consecutive n composites"...
anywayzzz... thnx for d reply...
nyc question... :)
62umm... i am not sure.. but stilll I hope you got the "bigger picture" :)
62it is honey... I din get your point!
5!+2, 5!+3...
are 122, 123, 124, 125, 126 they are all composites!
1bt whn u use n=4 ie 4!+2, 4!+3... or <4 the eq. is nt satisfied? or this eq is only true if u tk 5?????
62No..
see for N consecutive, we have to take N+1
so for 2 consecutives, we have to take
3!+2, 3!+3
both are
8, 9
for 3 consecutive we have to take
4!+2, 4!+3, 4!+4 which is
26, 27, 28...
62among Integers....
1,2,3,4,5,6,7,8,9,10,.....................,c1,c2,c3,c4,....cn,..............
24Sir can u just give an sample question to tell in which format it can come in JEE...................
62eureka... when I say jee syllabus, I mean the knowledge that is needed... and when you give the paper you will realize that every arrow in ur quiver might help at some time... I have given a lot of questions to increase your "interest" and "Thinking process" required in the exam... :)
The simplicity of the answer (with the hint) wiill just bowl you out..
Hint : Think of N! (Already with the question...)
After this, it is just a matter of seconds! (If it clicks)
(Had it been a olympiad question... I wud not have given the hint ;)
1bhaiya can u give something more on that factorial thing
its just kind of tickling my mind as if just in reach...
but i can't get it presently[2]
