IIT JEE , AIEEE Forum - Mathematics , Algebra

danny_007

1)the number of ordered triplets (a,b,c) such that LCM of (a,b) = 1000 , LCM (b,c) = 2000 LCM (c,a) = 2000 is?

(answer = 70)

2)the number of solutions of the equation

x⁶-x⁵+x⁴-x³+x²-x+3/4 = 0 is?

chimanshu_007

1000 = 2³ . 5³

2000 = 2⁴.5³

a = 2^x.5^y , b = 2^p.5^r , c = 2^h.5^gso max (x,p) = 3 , max of (x,h) = max (p,h) = 4

max (y,r)=max(r,g)=max(y,g)=3

we can choose y,r,g by

1 with all 3 , 3 times with one 2 , 3 times with one 1 and 3 times with one 0

= 10 ways

C must be 4.......

so we can choose A,B by

1 with both 3,3 , 3 with A = 3 and B=/= 3 and 3 with A =/= 3 & B = 3

= 7 ways....

so total number of ways = 7 x 10 = 70

sandeepramesh

for the second question,

Clearly for x<=0 there's no soln

For x>=1, also there's no soln (this can be got by grouping the terms)

It suffices to prove that there exists no soln for x belonging to (0,1).

Group the terms as (x^4-x^5) + (x^2-x^3) + (x^6-x) + 3/4 = 0

Now clearly the first 2 terms are >0.

To find the minimum of x^6-x:

Differentiate

implies 6x^5=1 implies x^5 = 1/6

implies x = (1/6)^1/5 implies x^6 - x = (1/6)^6/5 - (1/6)^1/5

>-3/4

Hence the result

danny_007

thanx

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