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by shashank.ism » Tue Feb 09, 2010 1:05 pm
How many natural numbers 'k' exist such that (2k + 3) divides (24k^2 + 201)?

4
0
6
7
8
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by harsh.champ » Wed Feb 10, 2010 11:47 pm
shashank.ism wrote:How many natural numbers 'k' exist such that (2k + 3) divides (24k^2 + 201)?

4
0
6
7
8
24k2 + 201 = 6 (2k + 3)(2k - 3) + 255
Since, 2k + 3 divides 24k2 + 201, it should also divide 255.
255 is a multiple of 3.
Also, 255 = 3 × 5 × 17
Therefore, 2k + 3 has to be factor of 255
There are 8 factors of 255
Out of which
(i) 2k + 3 = 1 (not possible)
(ii) 2k + 3 = 3 (not possible)
Hence a total of 6 possible values of k are there.

So,C is the answer.
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by amit28it » Mon Dec 26, 2011 2:51 am
A/c to natural numbers definition it's answer is right as solved above . So just follow it .
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