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swethav12
11-18-2004, 10:55 PM
Hi
can any solve this problem with explaination

Four bells begin to toll together and then each one at intervals of 6sec, 7sec, 8sec, 9sec respectively. The number of times they will toll together in the next 2hors is..?

thebullfighter
11-18-2004, 10:59 PM
see Q3 here--- (solution, post #10)
http://www.urch.com/forums/showthread.php?t=15393

swethav12
11-18-2004, 11:12 PM
hi bullfighter
can u plz explain me the answer

krakow
11-19-2004, 10:26 AM
Re: 4 Problems.[/b]]Q3 answer is 14 times.

6*7*8*9 = (2*3)(7)(2*2*2)(3*3) = 2^4 * 3^3 * 7
in order for a number to be a multiple of 6, it must have factors of 2, 3
in order for a number to be a multiple of 7, it must have factor of 7
in order for a number to be a multiple of 8, it must have factors of 2, 2, 2
in order for a number to be a multiple of 9, it must have factors of 3, 3

since the 2*3 factor for the 6 overlaps with one of the 2s in the 8, and one of the 3s in the 9, we can take them out. that leaves us with 2^3 * 3^2 * 7 = 504

so the bells will ring together every 504 seconds. in 60 secs/min * 60 min/hr * 2hr = 7200 secs, that will happen 14 times

first bell tolls at: 6th, 12th, 18th second, etc..
fourth bell tolls at: 9th, 18th sec., etc..
so those toll together at 18th second..and also every 18 seconds,
and coincidentially ;) 18 is a LCM of 6 and 9..

the same if we consider 3, or 4 bells.. analogy

LCM seconds is when they all toll together

and the rest of calculations is in above quotation..

swethav12
11-20-2004, 12:41 AM
Thanks krakow for the explaination
now I understood both the question and the answer clearly
I also read some of your answers in some posts u gave they r simple and short
and Bull u know many shortcuts i guess:hmm:
good job guys:tup:
All The Best