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20 May 2007 00:38:58 IST
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10th KILLER QUES. OVER
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OK , a gud 10th one........
Consider the following equation........
Summation ( r = 1 to 120 ) [ r / ( x - r ) ] >= 3
The values of x which satisfy the above eqn . are
x belongs to ( a, b ) union ( c, d ) union ( e , f ) ........................( consider all
the sets of x for which the above holds true )
FIND :: ( b + d + f + ...........) - ( a + c + e + ..... )
Comments (26)
S k
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Joined: 12 May 2007
Posts: 574
20 May 2007 23:17:37 IST
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21 May 2007 16:52:51 IST
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consider,
1/(x+1) >= 3
implies
(3x - 4)/(x-1) <=0
or x lies in (1,4/3] (upper bound(4/3) is the root of numerator and lower bound(1) is root of denominator)
similarly for
1/(x-1) + 2/(x-2) >=3 implies
(3x2 - 11x +9)/(x-1)(x-2) <=0
or x lies in (1, (11 -
13)/6 )
(2, (11 +
13)/6 )
notice that again lower bounds and upper bounds form roots of numerator and denominator respectively.
you will find a similar case for
1/(x-1) + 2/(x-2) + 3/(x-3) >=3
where lower and upper bounds form the roots.
therefore , for 1/(x-1) + 2/(x-2) + 3/(x-3) ....... 120/(x-120) >=3
the roots of numerator and denominator of the simplified expression will give upper and lower bounds.
What is asked is the sum of the roots of the numerator - sum of roots of denominator.
which you can easily find to be 2420.
That solves the problem!
the roots of the denominator will o
1/(x+1) >= 3
implies
(3x - 4)/(x-1) <=0
or x lies in (1,4/3] (upper bound(4/3) is the root of numerator and lower bound(1) is root of denominator)
similarly for
1/(x-1) + 2/(x-2) >=3 implies
(3x2 - 11x +9)/(x-1)(x-2) <=0
or x lies in (1, (11 -
13)/6 )(2, (11 + 13)/6 ) notice that again lower bounds and upper bounds form roots of numerator and denominator respectively.
you will find a similar case for
1/(x-1) + 2/(x-2) + 3/(x-3) >=3
where lower and upper bounds form the roots.
therefore , for 1/(x-1) + 2/(x-2) + 3/(x-3) ....... 120/(x-120) >=3
the roots of numerator and denominator of the simplified expression will give upper and lower bounds.
What is asked is the sum of the roots of the numerator - sum of roots of denominator.
which you can easily find to be 2420.
That solves the problem!
the roots of the denominator will o
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