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Topic: Long Division (Read 409 times) |
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ThudnBlunder
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Long Division
« on: Jan 23rd, 2009, 7:35am » |
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Deduce that 6k2 - 37k + 45 evenly divides 12k3 - 5k2 - 251k + 389 for only one integer value of k and find this value.
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balakrishnan
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Re: Long Division
« Reply #1 on: Jan 23rd, 2009, 9:47am » |
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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.
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