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Sports Medal Muse (Posted on 2022-04-30) Difficulty: 3 of 5
In a sports competition, there were M medals awarded on N successive days, where N is greater than 1.
  • On the first day: 1 Medal and 1/9 of the remaining (M-1) medals were awarded.
  • On the second day: 2 Medals and 1/9 of the now remaining medals were awarded, ...and, so on.
  • On the Nth and the last day: the remaining N medals were awarded.
How many days did the contest last and how many medals were awarded altogether?
Provide valid reasoning for your answers.

See The Solution Submitted by K Sengupta    
Rating: 4.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts Possible Solution | Comment 2 of 6 |
Consider the penultimate day, and let the number of medals remaining on that day be R.
Then on the last day, N=8R/9

Since 8 and 9 have no common factors, R=9 seems like an obvious substitution. Note that if R>9, say 18, then the result on the last day is greater than N, a contradiction, while if R<9, N has a fractional part, so this solution, if valid, is unique. 

If therefore a solution exists, then N=8 and R=9.

To check this, assume that 8 medals were awarded on the Nth (8th) day, with 0 medals remaining.
On the previous (7th) day, (N-1)+1/9(R) medals were awarded, for a total of 8, with 8 medals remaining.
Note that for the 6th day, we can gross up the remaining medals by 9/8 to achieve the number before distribution.
On the 6th day, (N-2)+1/9(9/8*(8+8)) medals were awarded, for a total of 8, with 16 medals remaining.
On the 5th day, (N-3)+1/9(9/8*(8+8+8)) medals were awarded, for a total of 8, with 24 medals remaining.
and so on.
So the same number of medals (8) were distributed each day for 8 days, giving a total of 64.

  Posted by broll on 2022-04-30 23:58:17
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