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        riddles >> medium >> Long Division
        
(Message started by: ThudanBlunder on Jan 23^rd, 2009, 7:35am)

Title: Long Division
Post by ThudanBlunder on Jan 23^rd, 2009, 7:35am

Deduce that 6k² - 37k + 45 evenly divides 12k³ - 5k² - 251k + 389 for only one integer value of k and find this value.

Title: Re: Long Division
Post by balakrishnan on Jan 23^rd, 2009, 9:47am

[hide]Write
12k³-5k²-251k+389 as

(3k-5)(4k²+5k-75)+14-k
Now since 6k²-37k+45=(3k-5)(2k-9),
we need that
3k-5 evenly divides (3k-5)(4k²+5k-75)+14-k
or
3k-5 must evenly divide k-14
since 3k-5 does not divide 3, it is equivalent to saying
3k-5 divides (3k-42) or (3k-5)-37
or
3k-5 divides 37
since 37 is a prime, 3k-5 =1 or 37 or k=2 or 14
when k=2 :
12k³-5k²-251k+389 = -37 and
6k²-37k+45 = -5 and so k=2 is not a solution
at k=14, the values are 28823=703*41 and 703.
Hence k=14 is the only solution.[/hide]