Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1967 IMO Problems/Problem 6"

(→‎Solution)
Line 5: Line 5:
 
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?
 +
  
 
==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 $m$ medals awarded on $n$ successive days $(n > 1)$. On the first day, one medal and $\frac{1}{7}$ of the remaining $m - 1$ medals were awarded. On the second day, two medals and $\frac{1}{7}$ of the now remaining medals were awarded; and so on. On the n-th and last day, the remaining $n$ 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 $7$ plus one is not a "clever method". And, note that when the author says "arrive at the answer of $36$", they mean "the contest lasted for $6$ days, and $36$ medals were awarded".

Below, I will give another solution, which is more in the spirit and style of contemporary problem solving.


Solution 2

Denote $m_0 = m$. Let $m_k$ be the number of medals left on day $k$ after the medals for the day have been awarded. The problem says that $m_k = m_{k-1} - k - \frac{m_{k-1} - k}{7}$ for $k = 1, 2, \dots, (n - 1)$, and $m_{n - 1} = n$, and $n > 1$. Simplify the recursive relation and get $m_k = \frac{6}{7}m_{k - 1} - \frac{6}{7}k$. Solving for $m_{k - 1}$ we get $m_{k - 1} = \frac{7}{6}m_k + k$.

The idea of the solution is to do a "backward" recursive computation. We will start with $m_{n - 1} = n$, "compute" $m_{k - 1}$ in terms of of $m_k$ for $k = (n - 1), \dots, 2, 1, 0$ and impose the condition that $m_0 = m$. We will get an equation in $m, n$, which will yield the values for $m, n$.



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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1967_IMO_Problems/Problem_6&oldid=225675"