@iLoveTorres said:it is 60 only.
fir question samajh nahi aaya........
help me out plzzzz....
help me out plzzzz....
@iLoveTorres said:When playing the lottery there are 6 different balls ranging between 0 and 60 that can be chosen from. How many different possibilities are there when picking lottery numbers?
61c6
@Dexian said:fir question samajh nahi aaya........help me out plzzzz....
@iLoveTorres said:When playing the lottery there are 6 different balls ranging between 0 and 60 that can be chosen from. How many different possibilities are there when picking lottery numbers?
This "15 ,20,21,50,55,59" is one possibility , similarly total how many possibilities are there ,thats the question.
**Hope my understanding of the question is correct otherwise please correct me immediately.
@iLoveTorres said:When playing the lottery there are 6 different balls ranging between 0 and 60 that can be chosen from. How many different possibilities are there when picking lottery numbers?
OA 61C6 :P
@Subhashdec2 said:E(1001)=7203^281 mod 10011001=7*11*133^281 mod 7=3^5 mod 7 = -1*9 mod 7=53^281 mod 11 =33^281 mod 13=3^5 mod 13=97a+5=11b+377c+47=13d+91001n+971971??
what is this E(1001)=720 ???
bhai thoda explain karoge please?
bhai thoda explain karoge please?
@harshit713 said:what is this E(1001)=720 ???bhai thoda explain karoge please?
1001=7*11*13
1001(1-1/7)(1-10/11)(1-12/13)=720
In how many ways ram can invite one or more of five friends and seat them at a circular table?
@maroof10 said:In how many ways ram can invite one or more of five friends and seat them at a circular table?
5c1+5c2+5c3*2!+5c4*3!+5c5*4!=5+10+20+30+24=89?
@maroof10 said:In how many ways ram can invite one or more of five friends and seat them at a circular table?
5c1*1+5c2*1+5c3*2+5c4*6+5c5*24 = 89 ways?? 
If six people are selected out of 10,in how many ways will a particular person be found among those six?
@maroof10 said:If six people are selected out of 10,in how many ways will a particular person be found among those six?
9c5??
Can anyone try to solve this cryptarithmetic sum .......
UMA
SPY
--------
MAMM
AYTL*
AXUY**
-----------
ASYLUM
@The_Loser said:For how many natural numbers, N the expression N ˛ + N + 1 results in a perfect square?
If N is a natural number N^2 N ˛ + N + 1
Alternate solution: If n^2+n+1=k^2 =>4k^2 - (2n+1)^2=3 =>(2k-2n-1)(2k+2n+1)=1*3=>2k-2n-1 =1 and 2k+2n+1=3 (since n has to be positive) =>k=1,n=0. This is true only for n=0 and hence no natural number solution exists for n
Team BV -Vineet
@saurav205 said:Sir,Could you explain a bit in detail, please....
Please see the solution i wrote above :)
Team BV - Vineet
@bodhi_vriksha said:Please see the solution i wrote above Team BV - Vineet
Vineet sir,
I am not talking about the alternate approach.....was talking about tbhe first approach..after tbe inequality what logic did you use..???
@saurav205 said:Vineet sir,I am not talking about the alternate approach.....was talking about tbhe first approach..after tbe inequality what logic did you use..???
Please look at the inequality again. If N were a natural number then the fact that
N^2
Note: For some weird reason PG Editor was not showing up the middle part (the one with N^2+N+1) of the inequality.
Team BV - Vineet
@bodhi_vriksha said:Please look at the inequality again. If N were a natural number then the fact that N^2Note: For some weird reason PG Editor was not showing up the middle part (the one with N^2+N+1) of the inequality.Team BV - Vineet
Thank you sir...got it now...
Yes the middle section was missing...so was unable to understand...