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a real TOUGH one....

neways m weak in geometry..!!!??:

neways m weak in geometry..!!!??:

@jabes

Posted 30 Sep '06

> 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
hey bro its quite simple...
23!= 1.2.3.4.5.6.....21.22.23
now multiply both numerator n denom by 24
which wil give 23!= 24!/4!
similarly other e...

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

hey bro its quite simple...

23!= 1.2.3.4.5.6.....21.22.23

now multiply both numerator n denom by 24

which wil give 23!= 24!/4!

similarly other egs....the catch was to find those numbers which are factorial results...

@jabes

Posted 30 Sep '06

> 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 ???
hey guys ...i think answer should b 720...
explanation...
5 diff balls distributed to 3 diff ppl...each gettin a...

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 ???

hey guys ...i think answer should b 720...

explanation...

5 diff balls distributed to 3 diff ppl...each gettin atleast 1...

suppose 5 balls are identical...

now these balls can be distributed in 5-1C2 ways....tht is 6 ways...

but the balls are diff so we wil multiply it with 5!....giving answer as 120 x 6 ie 720...

what say.....i learnt this funda in IMS...pls tel if m wrong

@jabes

Posted 30 Sep '06

i wil also go for option D
explanation by some egs.....5!=6!/3! 23!=24!/4! 119!=120/5! 719!=720!/6!
since all other options are contradictin the above egs. so the option 4 is corect

i wil also go for option D

explanation by some egs.....5!=6!/3! 23!=24!/4! 119!=120/5! 719!=720!/6!

since all other options are contradictin the above egs. so the option 4 is corect

explanation by some egs.....5!=6!/3! 23!=24!/4! 119!=120/5! 719!=720!/6!

since all other options are contradictin the above egs. so the option 4 is corect

@jabes

Posted 30 Sep '06

> 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 ...

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

hey thanks a lot dude...good one.....neveer thought in this direction...thanks neways...

@jabes

Posted 30 Sep '06

> *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 player...

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 ;)

hey good job bro...it was a sitter...but by lookin at the question it never appeared

@jabes

Posted 30 Sep '06

> 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) ...

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???

sirjeeee....i think answer should be 20....????

@jabes

Posted 28 Sep '06

> 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 + ... +100q...

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....

Whew...!!!!!

I don't know if i m right or wrong...

Can sm1 ... Warrior ... Ashish... verify this... ? :-|

hey bro.....

could u pls elaborate...ki 33 kaise aaya....

sum of first row comes out to be 39 ....n how come the sum of the same number comes out to be 33...

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