Messages posted by eragon24

use exp of x.cosecx

since 0/0 form

apply L.hopital

since indeterminancy removed...put x=0

= -1/3

here f is phi.....eulers totient function

ok firstly see that f (100) = 100(1 - 1/2).(1 - 1/5)= 40 .Since 7 and 100 are coprime Hence, by Euler’s Theorem, 7⁴⁰≡1mod 100......Now, f (40) =40( 1- 1/2).(1- 1/5)=16, hence 7¹⁶≡1mod40....Now since 1000= 16 ·62+8. This means that 7¹⁰⁰⁰≡(7¹⁶)⁶².7⁸≡1⁶².7⁸ ≡(7⁴)²≡1²≡1mod40 This means that 7¹⁰⁰⁰= 1+40t for some integer t. now assembling all this we get 7^(7¹⁰⁰⁰)≡7^1+40t≡7.(7⁴⁰)^t≡7 mod 100

this means last two digits are 07

checkitout

checthisout for oder methods and other rmo Qs

using expansion of tan x and sin x

actually this is the state of this site nowadays......jus got degraded completely.........no quality Qs......no quality discussions...........only posting rubbish everytime now or then.....and that too we get replies for such rubbish queries......and has lost serious users drastically........my advice to the admin......either revive it....or shut it down completely .......

checkthisout

and thisalso

yep...other experts who are creating fake id shud take a lesson from it......and for others plz see that post on generating functions.....its worth a thing to be learned.....very useful

just superb

yeah.....it happens only in india...........btw 1st one was very complex :P.....dont know how many times wud u apply kirchoffs law der....rofl

hmm...though i hav not taken bt ts this yr..... but i hav its experience......this toh i wud definately say tht leaving 2-3 full test the standard of the paper is not of jee level tats for sure....sligthly below it....but i wud say give its open tests(tat is first 2-3 full tests) they r jus fabulous......exactly of jee level...not tat tough as of fiitjee's but tricky like jee.....

ten speaker can address in 10! ways.......including S1 S2 S3

now s1 s2 s3 can be arranged in 3! ways=6ways i.e s1s2s3 , s1s3s2,s2s1s3,s2s3s1,s3s1s2,s3s2s1

since we want only the order s1 s2 s3 we will take 1/6 th of the total....

so no of ways such that S₁speaks before S₂ and S₂ speaks before S₃ is (1/6)10!

no of ways in which ten speaker can address is 10! ways

probab=(1/6)10!/10!=1/6

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