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 Medals (Posted on 2004-01-29)
In a sports contest there were m medals awarded on n successive days (n > 1).
On the first day 1 medal and 1/7 of the remaining (m - 1) medals were awarded.
On the second day 2 medals and 1/7 of the now remaining medals was awarded; and so on.
On the nth and last day, the remaining n medals were awarded.
How many days did the contest last, and how many medals were awarded altogether?

 See The Solution Submitted by Ravi Raja Rating: 3.2000 (5 votes)

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 solution | Comment 1 of 5
m = 36 medals over 6 days

On the 1st day, 1 + (1/7)*35 = 6 medals awarded (30 remain)
On the 2nd day, 2 + (1/7)*28 = 6 medals awarded (24 remain)
On the 3rd day, 3 + (1/7)*21 = 6 medals awarded (18 remain)
On the 4th day, 4 + (1/7)*14 = 6 medals awarded (12 remain)
On the 5th day, 5 + (1/7)*7 = 6 medals awarded (6 remain)
On the 6th day, the 6 remaining medals were awarded.
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I simply set this up on an excel spreadsheet, and realized that because m-1 must be divisible by seven, the original number must be 8, or 15, or 22, etc...

I simply tried the first few numbers and stumbled upon the solution....
`Day           given    rem.  given2   also remain0					361		1	35	5	302		2	28	4	243		3	21	3	184		4	14	2	125		5	7	1	66		6	0	0	0`

 Posted by SilverKnight on 2004-01-29 12:19:34

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