Difference between revisions of "1967 IMO Problems/Problem 6"
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medals were awarded. How many days did the contest last, and how many | medals were awarded. How many days did the contest last, and how many | ||
medals were awarded altogether? | medals were awarded altogether? | ||
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==Solution== | ==Solution== | ||
+ | |||
This is not a particularly elegant solution, but if you start from 1 and go all the way in a clever method, by only guessing those that are 1 more than a multiple of 7, you arrive at the answer of 36. | This is not a particularly elegant solution, but if you start from 1 and go all the way in a clever method, by only guessing those that are 1 more than a multiple of 7, you arrive at the answer of 36. | ||
+ | |||
+ | |||
+ | ==Comment (added by pf02, August 2024)== | ||
+ | |||
+ | Indeed, as the author says, the above is not an elegant solution. Also, | ||
+ | it does not give any insight into the uniqueness of the answer to the | ||
+ | problem. I would also comment that choosing to verify the statement | ||
+ | only for multiples of <math>7</math> plus one is not a "clever method". And, note | ||
+ | that when the author says "arrive at the answer of <math>36</math>", they mean | ||
+ | "the contest lasted for <math>6</math> days, and <math>36</math> medals were awarded". | ||
+ | |||
+ | Below, I will give another solution, which is more in the spirit and | ||
+ | style of contemporary problem solving. | ||
+ | |||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Denote <math>m_0 = m</math>. Let <math>m_k</math> be the number of medals left on day <math>k</math> | ||
+ | after the medals for the day have been awarded. The problem says | ||
+ | that <math>m_k = m_{k-1} - k - \frac{m_{k-1} - k}{7}</math> for | ||
+ | <math>k = 1, 2, \dots, (n - 1)</math>, and <math>m_{n - 1} = n</math>, and <math>n > 1</math>. | ||
+ | Simplify the recursive relation and get | ||
+ | <math>m_k = \frac{6}{7}m_{k - 1} - \frac{6}{7}k</math>. Solving for <math>m_{k - 1}</math> | ||
+ | we get <math>m_{k - 1} = \frac{7}{6}m_k + k</math>. | ||
+ | |||
+ | The idea of the solution is to do a "backward" recursive computation. | ||
+ | We will start with <math>m_{n - 1} = n</math>, "compute" <math>m_{k - 1}</math> in terms of | ||
+ | of <math>m_k</math> for <math>k = (n - 1), \dots, 2, 1, 0</math> and impose the condition | ||
+ | that <math>m_0 = m</math>. We will get an equation in <math>m, n</math>, which will yield | ||
+ | the values for <math>m, n</math>. | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | TO BE CONTINUED. SAVING MID WAY, SO I DON'T LOOS WORK DONE SO FAR. | ||
+ | |||
+ | |||
== See Also == {{IMO box|year=1967|num-b=5|after=Last Question}} | == See Also == {{IMO box|year=1967|num-b=5|after=Last Question}} |
Revision as of 12:29, 24 August 2024
In a sports contest, there were medals awarded on
successive days
. On the first day, one medal and
of the remaining
medals
were awarded. On the second day, two medals and
of the now remaining
medals were awarded; and so on. On the n-th and last day, the remaining
medals were awarded. How many days did the contest last, and how many
medals were awarded altogether?
Solution
This is not a particularly elegant solution, but if you start from 1 and go all the way in a clever method, by only guessing those that are 1 more than a multiple of 7, you arrive at the answer of 36.
Comment (added by pf02, August 2024)
Indeed, as the author says, the above is not an elegant solution. Also,
it does not give any insight into the uniqueness of the answer to the
problem. I would also comment that choosing to verify the statement
only for multiples of plus one is not a "clever method". And, note
that when the author says "arrive at the answer of
", they mean
"the contest lasted for
days, and
medals were awarded".
Below, I will give another solution, which is more in the spirit and style of contemporary problem solving.
Solution 2
Denote . Let
be the number of medals left on day
after the medals for the day have been awarded. The problem says
that
for
, and
, and
.
Simplify the recursive relation and get
. Solving for
we get
.
The idea of the solution is to do a "backward" recursive computation.
We will start with , "compute"
in terms of
of
for
and impose the condition
that
. We will get an equation in
, which will yield
the values for
.
TO BE CONTINUED. SAVING MID WAY, SO I DON'T LOOS WORK DONE SO FAR.
See Also
1967 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Question |
All IMO Problems and Solutions |