yaar how did u go for 23! and 119! or 719! those are too far fetched values.
i could conjure just the 1st one.enlighten me lord
question. 5 distinct balls need to be distributed among 3 people such that each person has atleast one ball. In how many ways can we do this?
is the answer 540 ???
take nos as 100a+10b+c
we can remove 100a as we need lats two only
so (10b+c)^3 = 1000b^3 +300b^2c+30bc^2+c^3 (ignnore first two of its terms)
also c has to be 6
216+1080b should end in 56
ie 8b+1 should give ne 5 possible when b=3 and 8
a can take any value from 1 to 9 and c=6
so 9*2 =18
Several B-schools took part in a tournament. Each player played one match against each player from a different B-school and did not play anyone from the same B-school. The total number of men taking part differed from the total number of women by 1. The total number of matches with both players of the same gender differed by at most one from the total number of matches with players of opposite gender. What is the largest number of B-schools that could have sent an odd number of players to the tournament?
(a) 3 (b) 5 (c) 6 (d) 11
My answer is option (a) 3 .... I took simplest case ...3teams of one each 1 m ,1 m, 1 w ... 0 teams with even players. this satisfies ...
similar thing with 5 /6/11 does not satisfy ... I guess this is like a sitter that appears tough ... well whether its a sitter or not shall be decided on whether my answer is correct ;)
Binomial theorem has been used here
((15 + 2)^ 19) /25 = (2^19 + 19(2^18 )(15))/25 (rest all will have 25 as a multiple)
((15 - 2)^ 19) /25 = (-2^19 + 19(2^18 )(15))/25
Add both, you'll get the part in bold
Formula for Euler's no E(n), if a no n= p*q*r (where p,q,r are prime nos)
then E(n) = (p-1)(q-1)(r-1)
E(101) = (101-1)= 100
Rani, what were you doin at 4.45 in the morning? :o???
Well well well... i think i m missin smthng here...
The above soln is wrong...
Correct soln shud be like:
(1^1+2^2+3^6 + 4^4k1 + 5^k2 + 6^4k3 + 7^4k4 + 8^4k5 + 9^2k6 + 10Q1) +
(11^4m1 + 12^4m2 + ...... + 19^4m9 + 20q2) +
(21^4p1 + ... + 30q3)
(91^4x1 + ... +100q10)
10Z + 39 + 9*33
Here 39 come's from the first row...
and 9*33 is the product of the sum of last digits of the 9 rows after the 1st row...
So, the expr = 10Z + 39 + 297 = 10Z + 336
So last digit = 6....
I don't know if i m right or wrong...
Can sm1 ... Warrior ... Ashish... verify this... ? :-|