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ooh...i know how to do this one:Originally posted by Constip8edSkunk
but theres this question that i had absolutely no idea: 8889 cards with 5 digit numbers 11111 upto 99999 are randomly arranged 2 form a 44445 digit number. Show that there are no possible combinations where the resulting number is a power of 2003.
Damn, why do i never hear about these things....Originally posted by inasero
Did anyone do it....how #@%#!$ impossible was it!! I hope you all agree with me here...cos if u guys didn't I would feel extremely dumb....HEck!! I feel dumb anyway!!
Thanks for telling me about it Wilson... and man do i feel dumb looking at that one question.Originally posted by inasero
Did anyone do it....how #@%#!$ impossible was it!! I hope you all agree with me here...cos if u guys didn't I would feel extremely dumb....HEck!! I feel dumb anyway!!
Originally posted by spice girl
ooh...i know how to do this one:
if N is a 44445-digit number, then 10^44444 < N < 10^44445
N = 2003^x
10^44444 < 2003^x
log both sides
44444ln10 < xln2003
similarly, xln2003 < 44445ln10
you'll find that x can never be an integer (since you're allowed to use a calculator)
well he got in2 IMO, so yeah....Originally posted by xiao1985
btw, any one got 5/6 q's that person's either a math freak or a math freak...... no offence tho, i adore math freaks..... i want to be one but i can't... too hard...
woa?? gee, wut's his name?? i prolly noe him, knowin he's dat famous dat is.........Originally posted by Constip8edSkunk
well he got in2 IMO, so yeah....