bs0409
February 1, 2013, 10:58am
17546
@anuragu79 said: Please elaborate a bit...
a+b+c+d=90 has 93C3 solutions.
Now we hv to remove solutions where at least one of a,b,c,d is >=46
We can note that only one of a,b,c,d can be more than 46.
So choose it in 4C1 ways.
Let the chosen one be a. So we can say a=a'+46 where a'>=0
The equation becomes, a'+b+c+d=44 which has 47C3 solutions.
ANS will be 93C3-4*47C3
bs0409
February 1, 2013, 11:00am
17547
@Abhinav.kumar said: Pls elaborate..m nt gtng...
999*abc=def132
LHS=(1000-1)abc=abc000 - abc
Now apply simple logic
abc000
000abc
--------
def132
So abc must be 1000-132=868
logrhythm
February 1, 2013, 11:31am
17548
@bs0409 said: a+b+c+d=90 has 93C3 solutions.
mere soln mein kya problem thi BS bhai??
(45-a) + (45-b) + (45-c) + (45-d) = 90
why isn't this working.. :(
bs0409
February 1, 2013, 11:36am
17549
@Logrhythm said: mere soln mein kya problem thi BS bhai??
(45-a) + (45-b) + (45-c) + (45-d) = 90
why isn't this working..
This method only works sometimes and not always........
If the equation is x1+x2+.............xn=K
and xi's
then this method will work if K>=(n-1)a
Here K=90
n=4
a=45
(n-1)a=3*45=135
So condn is not satisfied.
logrhythm
February 1, 2013, 11:42am
17551
@bs0409 said: Solve
a'+b'+c'+d' = 116 -> 119c3
a"+b'+c'+d' = 71 -> 74c3
hence, 119c3 - 4*74c3 ??
ye hoga??
bs0409
February 1, 2013, 11:49am
17554
@Logrhythm said: a'+b'+c'+d' = 116 -> 119c3
a"+b'+c'+d' = 71 -> 74c3
hence, 119c3 - 4*74c3 ??
ye hoga??
Ye 116 kidhar se laye boss???
bs0409
February 1, 2013, 11:51am
17556
@mailtoankit said:
@anuragu79 said:
CLOSE but not correct
Give it a little bit more thought. Dont give answers based on instinct because they often turn out to be wrong......
anuragu79
February 1, 2013, 11:56am
17557
@bs0409 said: CLOSE but not correct
Not able to find the error pls tell me
logrhythm
February 1, 2013, 11:59am
17558
@bs0409 said: Solve
ok let me give it another try...
when any one is 0...
a+b+c = 120
4c1*122c2
a'+b+c = 74
3*77c3
=> 4c1*122c2 - 3*77c3
when any two are 0
a+b = 120
4c2*121c2
a'+b = 74
2*75c1
=> 4c2*121c2 - 2*75c1
when any 3 are 0
4c3
when none is zero
a+b+c+d = 120
123c3
a'+b+c+d = 74
4*77c3
=> 123c3 - 4*77c3
hence, total = (4c1*122c2 - 3*77c3) + (4c2*121c2 - 2*75c1) + (4c3) + (123c3 - 4*77c3)
I have made a mess of it...haven't i
bs0409
February 1, 2013, 12:02pm
17559
@Logrhythm said: haha kuch alag hi kar diya....sikhao bhai kaise hota hai ye....muje nahi aata..
@anuragu79 said: Not able to find the error pls tell me
Try inclusion-exclusion principle.
NOW u shld be able to DO it......
bs0409
February 1, 2013, 12:10pm
17561
@mailtoankit said: sir, jaisa apne pehele post mein bataya,waisa hi kiya meine....ab yahan par konsi extra condition aa gayi. .mujhe samajh nahi aa raha..
OK. let me post it.
Upto 123c3 - 4*77c3 it is correct.
Now see one thing. Two of a,b,c,d can also be more than 45.
By the inclusion-exclusion principle, u hv to add those cases.
So ANS=123c3 - 4*77c3 +4C2*31C3
logrhythm
February 1, 2013, 12:14pm
17562
@bs0409 said: OK. let me post it.
bhai how is my answer incorrect?? meine 0 liya hai case....wo dekho ek bar...
bs0409
February 1, 2013, 1:06pm
17563
solve : N^4+4=p where p is prime and N is a natural number.
logrhythm
February 1, 2013, 1:28pm
17565
@bs0409 said: solve : N^4+4=p where p is prime and N is a natural number.
n^4
1^4 = 1
2^4 = 6
3^4 = 1
4^4 = 6
5^5 = 5
6^4 = 6
7^4 = 1
8^4 = 6
9^4 = 1
0^4 = 0
so only n = 1 is the solution.......rest all wld either be divisible by 2 or 5..
..ab yahan par konsi extra condition aa gayi.