Difference between revisions of "2020 IMO Problems/Problem 5"
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Contradiction: Let us assume this is not true and for a certain n, there are k distinct positive integers which can be written in ascending order as follows : | Contradiction: Let us assume this is not true and for a certain n, there are k distinct positive integers which can be written in ascending order as follows : | ||
| − | z<sub>k > z<sub>k-1 | + | z<sub>k</sub> > z<sub>k-1</sub> > z<sub>k-2</sub> > … > z<sub>1</sub> |
| + | |||
| + | Since z<sub>k</sub> is the largest of the numbers, it has to be greater than 1. This implies that there will be a prime p<sub>1</sub> that divides z<sub>k</sub>. | ||
| + | |||
| + | Now we know that the arithmetic mean of z<sub>k</sub> and z<sub>k-1</sub> is more than z<sub>k-1</sub>, thus the geometric mean which it is equivalent to must include the term z<sub>k</sub>. Since we know the arithmetic mean is a rational number, this means the root of the products of the n numbers of the geometric mean must be positive, and this will also divide p. | ||
| + | |||
| + | We can use similar argument for the sum of z<sub>k-1</sub> and z<sub>k-2</sub> will be greater than z<sub>k-2</sub> and thus the geometric mean is once again divisible by p. We can repeat this all the way to z<sub>1</sub>, which would mean that by dividing each term by p, we can reach another set of n cards which fulfil the conditions. We can keep applying this until z<sub>k</sub>. However this is a contradiction. | ||
| + | |||
| + | Thus, for all n > 1, all the terms on the card will be equal. | ||
| + | |||
| + | QED | ||
== Video solution == | == Video solution == | ||
Latest revision as of 23:14, 12 August 2025
Contents
Problem
A deck of 
cards is given. A positive integer is written on each card. The deck has the property that the arithmetic mean of the numbers on each pair of cards is also the geometric mean of the numbers on some collection of one or more cards.
For which 
does it follow that the numbers on the cards are all equal?
Solution 1
Claim : For all n > 1, all numbers must be equal
Contradiction: Let us assume this is not true and for a certain n, there are k distinct positive integers which can be written in ascending order as follows :
zk > zk-1 > zk-2 > … > z1
Since zk is the largest of the numbers, it has to be greater than 1. This implies that there will be a prime p1 that divides zk.
Now we know that the arithmetic mean of zk and zk-1 is more than zk-1, thus the geometric mean which it is equivalent to must include the term zk. Since we know the arithmetic mean is a rational number, this means the root of the products of the n numbers of the geometric mean must be positive, and this will also divide p.
We can use similar argument for the sum of zk-1 and zk-2 will be greater than zk-2 and thus the geometric mean is once again divisible by p. We can repeat this all the way to z1, which would mean that by dividing each term by p, we can reach another set of n cards which fulfil the conditions. We can keep applying this until zk. However this is a contradiction.
Thus, for all n > 1, all the terms on the card will be equal.
QED
Video solution
https://www.youtube.com/watch?v=dTqwOoSfaAA [video covers all day 2 problems]
See Also
| 2020 IMO (Problems) • Resources | ||
| Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
| All IMO Problems and Solutions | ||
