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Algebra

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21 Jan 2010 00:28:52 IST

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Suppose a real number P is such that [(1/2) p] + [(1/3) p] + [(1/5) p] = p Here, [x] denot

None

Suppose a real number P is such that [(1/2) p] + [(1/3) p] + [(1/5) p] = p Here, [x] denotes t

and no of solutions available are a)28 b)32 c)31 d)30

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ashutosh tripathi

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21 Jan 2010 13:53:33 IST

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p = 6

Hari Shankar

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22 Jan 2010 18:26:45 IST

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Writing the expression as f(p) we note that f(p+30) = 31+f(p).

Evaluating for numbers p lying between 0 and 29 and using the above result you get 28 solutions.

But all this is pretty much brute force. I wonder if there are prettier methods

kiran

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22 Jan 2010 19:17:12 IST

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hari sir can u pls explain how u got f(p+30) = 31+f(p).

The HellBlaizer

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23 Jan 2010 01:45:37 IST

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Well a good question!!

As far as I can see the solution provided by Hari sir is really good and I don't think a more better way can be given actually when we talk about the [ ] function we have to restrict ourselves to looking in closed interval groups !!

Now comming to the solution posted by hari sir:

"Writing the expression as f(p) we note that f(p+30) = 31+f(p)."

The above expression given [(1/2) p] + [(1/3) p] + [(1/5) p] = p can equivalently be written as f(p+30) = 31+f(p) here the f(p+30) denoted the lowest commom multiple of 2,3 and 5 (the product of the following primes) the period after which the generating sequence repeats itself which on further investigation yields the the functional equation of 31+f(p).

I hope you got it!!!

Have a nice day!!

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