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#1 Posted 11:55pm 20-04-09 divisible?find the least positive integer m such that 2[p]2000[/p] divides 2003[p]m[/p]-1 |
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#2 Posted 11:43am 11-12-09 Re: divisible?show your working... |
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#3 Posted 10:34pm 11-12-09 Re: divisible?As 2[p]2000[/p] and 2003[p]m[/p] are coprime to each other , so 2003[p]φ(p) [/p] = 1 ( mod 2[p]2000 [/p] ) now φ(p) = 2[p]2000[/p] ( 1 - 1 / 2 ) = 2[p]1999[/p] so m = φ(p) = 2[p]1999[/p]
First is first , and the rest are no-where . Edited on 00:18am 29-12-09 |
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#4 Posted 00:19am 29-12-09 Re: divisible?Err , can anybody confirm what I did ?
First is first , and the rest are no-where . |
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#5 Posted 08:07am 16-01-10 Re: divisible?from a lemma known as "lifting the exponent" m = 1998 |
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