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| A Greatest (Integer) Problem (Posted on 2007-04-16) |
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Analytically determine all possible solutions of the equation :
[p/2]2 + [p/3] = 498, whenever p is a positive integer.
Note: [x] is defined as the greatest integer less than or equal to x.
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Submitted by K Sengupta
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Rating: 5.0000 (1 votes)
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Solution:
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[p/2]^2 + [p/3] = 498
Or, (p/2 -1)^2 + p/3 -1 < 498 < =(p/2)^2 + p/3.......(#)
Or, (p^2)/4 - 2p/3 < 498 < =(p/2)^2 + p/3
Now, for p > 0,
(p^2)/4 + 2p/3 - 498 < 0, gives:
p < (267.8208356+ 8)/6=45.9701393
For p > 0,
(p^2)/4 + p/3 - 498 > = 0, gives:
p > =(267.8208356-4)/6=43.9701393
Accordingly, 43.9701393 < = p < 45.9701393
The only positive integers lying in the above interval occurs whenever p = 44, 45
For p = 45, [p/2]^2 + [p/3] = 499, a contradiction.
For p = 44, we see that, [p/2]^2 + [p/3] = 498
Consequently the only possible solution is p = 44.
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