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1967 IMO Problems/Problem 6

Revision as of 16:59, 24 August 2024 by Pf02 (talk | contribs)

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 $m_k = \frac{6}{7}m_{k - 1} - \frac{6}{7}k$.

We will now get an explicit formula for $m_k$.

$m_1 = \frac{6}{7}m - \frac{6}{7} \cdot 1$

$m_2 = \frac{6}{7}m_1 - \frac{6}{7} \cdot 2 = \left(\frac{6}{7}\right)^2m - \left(\frac{6}{7}\right)^2 \cdot 1 - \frac{6}{7} \cdot 2$

$m_3 = \frac{6}{7}m_2 - \frac{6}{7} \cdot 3 = \left(\frac{6}{7}\right)^3m - \left(\frac{6}{7}\right)^3 \cdot 1 - \left(\frac{6}{7}\right)^2 \cdot 2 + \frac{6}{7} \cdot 3$

. . . . . . . .

$m_{n - 1} = \frac{6}{7}m_{n - 2} - \frac{6}{7} \cdot (n - 1) = \left(\frac{6}{7}\right)^{n - 1}m - \left(\frac{6}{7}\right)^{n - 1} \cdot 1 - \left(\frac{6}{7}\right)^{n - 2} \cdot 2 - \dots - \left(\frac{6}{7}\right)^2 \cdot (n - 2) - \frac{6}{7} \cdot (n - 1)$

To put it in a shorter way,

$m_{n - 1} = \left(\frac{6}{7}\right)^{n - 1}m - \sum_{k = 1}^{n - 1} \left(\frac{6}{7}\right)^{n - k} \cdot k$

(The reader who is not happy with having obtained this result by having observed the pattern for $m_k$ can easily verify it by induction.)

Note that the sum in the equality above is the sum of the arithmetic-geometric series formed by the geometric series $\{\ \left(\frac{6}{7}\right)^{n - 1}, \left(\frac{6}{7}\right)^{n - 2}, \dots, \left(\frac{6}{7}\right)^2, \frac{6}{7}\ \}$ and the arithmetic series $\{\ 1, 2, \dots, (n - 2), (n - 1)\ \}$. The formula for the sum of the terms of such series is reasonably well known (and not difficult to prove) (see for example Arithmetico-geometric_series or https://en.wikipedia.org/wiki/Arithmetico-geometric_sequence).

Applying the formula for the sum of the arithmetic-geometric series, and using $m_{n - 1} = n$, we get




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
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