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![[Post New]](/templates/default/images/icon_minipost_new.gif) 15 May 2007 03:18:48 IST
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CAUTION : A DANGEROUS PROBLEM AHEAD In a sports contest a total of m medals were awarded over n days. On the first day one medal and 1/7 of the remaining medals were awarded . On the second day two medals and 1/7 of the remaining medals were awarded, and so on. On the last day, the remaining n medals were awarded. How many medals were awarded, and over how many days? COMMENT : THIS Q. WILL TAKE UR LIFE (killer )BUT U WONT B ABLE TO SOLVE IT AWARD : 30 RATING POINTS REQUEST - THIS POST IS NOT FOR PEOPLE WHO HAVE SOLVED THIS Q. BEFORE OR HAVE AN IDEA FROM WHERE TO GET THE SOLUTION ( BY SOME UNKNOWN REFERENCES .............. )
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............tseb eht ma i |
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15 May 2007 07:50:56 IST
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here i go........... so since on da first day.....i medal i medal and 1/7 of da remeanin medals......means 6/7medals which wer remainin......after....the day b4 da first day...... this happens everyday........... on da first da 1+x/7.......medals were given........ on da second day.....2+1/7(x-1-x/7)......here x is da total number of medals......... so we get.......2+1/7(6x-7/7) generalising......we get on the pth day....... m1=x mp+1= 6(mp-p)/p..... using dis formula and sbstituting values..... we get an A.G. series ... x= 1+2(7/6)+3(7/6)2.............n(7/6)n-1 using da formula for summation and upon simplifyin... we get the sum as..... m=36(1-(n+1)(7/6)n+n(7/6)n+1) m=36+(n-6)7n/6n-1 6 does not divide 7.....for dis part to b a whole number..... so we get n=6......to make m an integer.......... since n=6.....m=36......... is da answer....correct.....it involved a lotta tireless......work....... thank u....
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15 May 2007 11:14:12 IST
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no that cant be they give d1= 1 + 1/7(remaining medal ) = 1 + 1/7 (m - 1) DAy two d2 = 2 + 1/7 (m - 2 -d1) Day three d3 = 3 + 1/7 (m - 3 -d2) . . . Day n dn= n + 1/7 [m -n - d(n-1)] Sum of all the medals m = [(1+2+3+.........n) + 1/7{ nm - (1+2+3...+n) - (d1+ d2+ d3+.......d(n-1))} goes on.......
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In the process of learnin..............blunders do happen !!! |
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15 May 2007 11:50:41 IST
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i guess arjun is right............ The answer cant get too complex...
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In the process of learnin..............blunders do happen !!! |
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15 May 2007 12:02:11 IST
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How long NO RESPONSE!!!
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In the process of learnin..............blunders do happen !!! |
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15 May 2007 12:27:04 IST
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Hey everyone I got the solution ........
Arjun is right....
See if we see the trend of giving medals, the remaining medals are always divisible by 7, and for the first case 1 medal is added so the smallest no. like that which is satisfying all the conditions is 36 and that comes in the 6th day.....
So no. of medals is 36 and No. of days is 6.
(This one is my point of view towards the problem NOT COPIED FROM ANYWHERE ELSE)
I think we should rate Arjun........
Cheers.............
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There's Light at the end of every Tunnel, so KEEP MOVING....
Best of luck to all my mates....
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15 May 2007 12:34:10 IST
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thanx....is it correct..........
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15 May 2007 12:35:40 IST
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i worked.........da helll outta dis problem.......
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15 May 2007 12:44:48 IST
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is da proof correct??
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15 May 2007 12:45:41 IST
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I say YeS!!!!
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There's Light at the end of every Tunnel, so KEEP MOVING....
Best of luck to all my mates....
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15 May 2007 14:15:54 IST
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YA CONGRATS ARJUN ,
FABULOUS REPLY................
I WILL GIVE U UR 30 PTS.........
CONGRATS AGAIN....
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............tseb eht ma i |
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15 May 2007 14:51:13 IST
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I WUD B POSTING THE NEXT " KILLER Q. " SHORTLY..........
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............tseb eht ma i |
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15 May 2007 19:14:08 IST
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no probs giv me.man....
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15 May 2007 19:16:18 IST
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i have already given 3rd and 4th .
3rd has been solved also.......
try the 4th............
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............tseb eht ma i |
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