16) 11
14 Answers
3411)390 2) 22 5)95 11) -1±√2
3) I feel second sentence should read "Which factor, in terms of the factorial of an integer. In which case 1006! is the winning candidate
and i beg of you, please tell me or at least pm me where you get such probs
1@ aditya what about(2,3,6) , (9,4,1) , (18,2,1)..i think one more condition will be required to arrive at a certain conclusion??
2622) simple maths- ans will be 22
5)after some logical guesswork - 95
262@narayan - we have to check the cases where the sum of factors are same . (there is no other set of factors which sum up to 11 other than the one u hav provided)
21Qn -5 , Solution:
Since n has 4 divisors it must be of the form pq or m3, where p,q,m are all primes.
Case 1. When n = m3, the given condition translates to 4(m+ m2)= m3+ 1. not possible since m does not divide 1.
Case 2. When n= pq. The given condition translates to 4(p+q)= pq+1, or (p-4)(q-4)= 15
We obtain systems p- 4 = 1, q- 4 = 15 , and p- 4 = 3 , q-4 = 5
this gives n = pq = 95
edit: din saw prophet sir and aditya's posts
Qn-1 (all real solution) -1
341For the first one, adding the two equations and then adding 1 to both sides produces
(1+a)(1+b)(1+c) = 8044 = 2 \ \times \ 2 \ \times \ 2011
So a = b =1 and c=2010.
You can easily check that the equations are satisfied.
So a+b+c=2012
262section B
2)
taking lcm- S= \sum_{1}^{2011}{\frac{a_{i}+b_{i}}{a_{i}b_{i}}}
S= \sum_{1}^{2011}{\frac{2}{-a-a^{2}}}S= -2\sum_{1}^{2011}{\frac{a+1 -a}{a(a+1)}}
S= -2\sum_{1}^{2011}{\frac{1}{a} -\frac{1}{a+1}}
S= -20111006
11Please tell where do you get such questions?
For 9. if we take any particular point the region comes out to be a circle. Cant we simply find the area of circle?
But i also noticed the 'at least' part.
But i dont know what it signifies.
For 1. one real solution is -1 for sure. but i dont know if there are any other solutions too.
