Difference between revisions of "1967 IMO Problems/Problem 6"
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it does not give any insight into the uniqueness of the answer to the | 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 | 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 | + | only for multiples of <math>7</math> plus one is not a "clever method", it is |
− | that when the author says "arrive at the answer of <math>36</math>", they mean | + | quite an obvious thing to do. And, note that when the author says |
− | "the contest lasted for <math>6</math> days, and <math>36</math> medals were awarded". | + | "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 | Below, I will give another solution, which is more in the spirit and | ||
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<math>k = 1, 2, \dots, (n - 1)</math>, and <math>m_{n - 1} = n</math>, and <math>n > 1</math>. | <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 | Simplify the recursive relation and get | ||
− | <math>m_k = \frac{6}{7}m_{k - 1} - \frac{6}{7} | + | <math>m_k = \frac{6}{7}m_{k - 1} - \frac{6}{7}k</math>. |
− | |||
− | + | We will now get an explicit formula for <math>m_k</math>. | |
− | |||
− | |||
− | |||
− | |||
+ | <math>m_1 = \frac{6}{7}m - \frac{6}{7} \cdot 1</math> | ||
+ | <math>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</math> | ||
+ | <math>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</math> | ||
+ | . . . . . . . . | ||
− | + | <math>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)</math> | ||
+ | |||
+ | To put it in a shorter way, | ||
+ | |||
+ | <math>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</math> | ||
+ | |||
+ | (The reader who is not happy with having obtained this result by having | ||
+ | observed the pattern for <math>m_k</math> can easily verify it by induction.) | ||
+ | |||
+ | Note that the sum in the equality above is the sum of the arithmetic-geometric | ||
+ | sequence formed by the geometric progression | ||
+ | <math>\{\ \left(\frac{6}{7}\right)^{n - 1}, \left(\frac{6}{7}\right)^{n - 2}, \dots, | ||
+ | \left(\frac{6}{7}\right)^2, \frac{6}{7}\ \}</math> | ||
+ | (with common ratio <math>\frac{7}{6}</math> (note that the progression starts from a | ||
+ | number smaller than <math>1</math> and is increasing)), and the arithmetic progression | ||
+ | <math>\{\ 1, 2, \dots, (n - 2), (n - 1)\ \}</math> (with common difference <math>1</math>). The | ||
+ | formula for the sum of the terms of such sequences 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 sequence, and | ||
+ | using the fact that <math>m_{n - 1} = n</math>, we get | ||
+ | |||
+ | <math>n = \left(\frac{6}{7}\right)^{n - 1}m - | ||
+ | \frac{\left(\frac{6}{7}\right)^{n - 1} \cdot 1 - 1 \cdot n}{1 - \frac{7}{6}} | ||
+ | -\frac{\frac{7}{6}}{\left(1 - \frac{7}{6}\right)^2} \left(\left(\frac{6}{7}\right)^{n - 1} - 1 \right)</math> | ||
+ | |||
+ | Now do all the simplifications, and rearrange terms. We have | ||
+ | |||
+ | <math>m - 36 = 7(n - 6) \left(\frac{7}{6}\right)^{n - 1}</math> | ||
+ | |||
+ | In order for <math>m</math> to be an integer, we have to have that | ||
+ | <math>\frac{n - 6}{6^{n - 1}}</math> is an integer. This is clearly | ||
+ | not the case when <math>n = 2, 3, 4, 5</math>. It is an integer | ||
+ | when <math>n = 6</math>, in which case <math>m = 36</math>. The fraction is | ||
+ | not an integer for <math>n = 7</math>, and even less so for <math>n > 7</math> | ||
+ | since the exponential at the denominator increases much | ||
+ | faster than the linear function at the numerator. | ||
+ | |||
+ | Thus, <math>m = 36, n = 6</math> is the only solution to the problem. | ||
+ | |||
+ | (Solution by pf02, August 2024) | ||
== See Also == {{IMO box|year=1967|num-b=5|after=Last Question}} | == See Also == {{IMO box|year=1967|num-b=5|after=Last Question}} |
Latest revision as of 20:35, 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", it is
quite an obvious thing to do. 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
.
We will now get an explicit formula for .
. . . . . . . .
To put it in a shorter way,
(The reader who is not happy with having obtained this result by having
observed the pattern for can easily verify it by induction.)
Note that the sum in the equality above is the sum of the arithmetic-geometric
sequence formed by the geometric progression
(with common ratio
(note that the progression starts from a
number smaller than
and is increasing)), and the arithmetic progression
(with common difference
). The
formula for the sum of the terms of such sequences 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 sequence, and
using the fact that , we get
Now do all the simplifications, and rearrange terms. We have
In order for to be an integer, we have to have that
is an integer. This is clearly
not the case when
. It is an integer
when
, in which case
. The fraction is
not an integer for
, and even less so for
since the exponential at the denominator increases much
faster than the linear function at the numerator.
Thus, is the only solution to the problem.
(Solution by pf02, August 2024)
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 |