Daily Trivia And Puzzles Thread

ok post another one

 

A NON-STOP TRAIN TRAVELLING AT 40 KM/HR STARTS FROM EASTPUR FOR WESTPUR AND ANOTHER STARTS FROM WESTPUR TO EASTPUR AT 45 KM/HR AT THE SAME TIME.HOW FAR APART ARE THEY ONE HOUR BEFORE THEY MEET?EASTPUR AND WESTPUR ARE 650 KM APART.


simple one for 100 points

 

85 km .,.............................

 

86kmz ??????? shjiiiiiiiiiiiiiiii late by a min :aggre:

 

oops spell mistake 85* kmz
lol typo stupid typo !!!!!!!!

 

since this thread has started .... a little puzzle from my side

problem: you have 100 doors in a row that are all initially closed. you make 100 passes by the doors starting with the first door every time. the first time through you visit every door and toggle the door (if the door is closed, you open it, if its open, you close it). the second time you only visit every 2nd door (door #2, #4, #6). the third time, every 3rd door (door #3, #6, #9), etc, until you only visit the 100th door.
question: what state are the doors in after the last pass? which are open which are closed?


reps for anyone who gets it absolutely correct first

Ats

PS: if anyone knows the solution before hand .. plzz donot tell

 

1st open rest closed :-s ???

 

nopes ... think a bit harder ... its mathematical .. no tricks involved

 

For getting the last state of door number 'N'. Apply the following algo...

1. Represent N in its factors including 1. e.g. 10 = 1*2*5, 21 = 1*3*7
2. Count the number of factors 'X'. e.g. 10's X = 3, 21's X = 3

If X == ODD, Door state = OPEN
if X == EVEN, Door state = CLOSE

 

if 'X' is equal for 10 (even) and 21 (odd) .. how do you conclude that when X is odd door is open and when X is even door is closed ????

your answer is not right but you are on the right track ..... i'll give you a hint ..

number of factors of 10: 1, 2, 5, 10 (he will visit #10 for #10, #20, #30 ..... ) == 4

number of factors of 21: 1, 3, 7, 21 (he will visit #21 for #21, #42, #63 ..... ) == 4

now proceed ... i have told you enough

ats

 

Boss, X represents the NUMBER OF FACTORS so for 10 its 1*2*5 = 3, it means that (s)he will visit the door for numbers #1, #2, #5 and then in last #10, i can not make 10 = 1*2*5*10.

And hence keep altering the doors till you last visit.

 

ok ... i got what u are saying ... still .. its too general an answer .... u have to give me a specific answer as to which doors will be open and which will be closed .... in other words .. for what numbers the X is odd and for what numbers it is even

 

mere main itna dimaag nahi hai :help :annoyed: :think: :help: :

 

will someone plz post an esy way and precise answer lol

 

The solution:

you can figure out that for any given door, say door #42, you will visit it for every divisor it has. so 42 has 1 & 42, 2 & 21, 3 & 14, 6 & 7. so on pass 1 i will open the door, pass 2 i will close it, pass 3 open, pass 6 close, pass 7 open, pass 14 close, pass 21 open, pass 42 close. for every pair of divisors the door will just end up back in its initial state. so you might think that every door will end up closed? well what about door #9. 9 has the divisors 1 & 9, 3 & 3. but 3 is repeated because 9 is a perfect square, so you will only visit door #9, on pass 1, 3, and 9... leaving it open at the end. only perfect square doors will be open at the end.

Cheers

Ats

 

hey i knew that one -perfect squares
had read it in mindsport some time back
ats 100 points for u
and everyone is welcome to post any riddles they might have

 

On each line place a three letter word which can be attached to the end of the word to the left and to the beginning of the word to the right to give two other words.

FORE (...) PAGE

FOND (...) HEM

 

foresee - seepage
fondant - anthem

 

sahi jawab

next one-
why cant micheal schumacher do any good for his motherland?

 

you want a serious answer ? or som kinda fun ????