Difference between revisions of "1970 IMO Problems/Problem 4"
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The only primes dividing numbers in the set can be 2, 3 or 5, because if any larger prime was a factor, then it would only divide one number in the set and hence only one product. Three of the numbers must be odd. At most one of the odd numbers can be a multiple of 3 and at most one can be a multiple of 5. The other odd number cannot have any prime factors. The only such number is 1, so the set must be <math>\{ 1, 2, 3, 4, 5, 6 \}</math>, but that does not work because only one of the numbers is a multiple of 5. So there are no such sets. | The only primes dividing numbers in the set can be 2, 3 or 5, because if any larger prime was a factor, then it would only divide one number in the set and hence only one product. Three of the numbers must be odd. At most one of the odd numbers can be a multiple of 3 and at most one can be a multiple of 5. The other odd number cannot have any prime factors. The only such number is 1, so the set must be <math>\{ 1, 2, 3, 4, 5, 6 \}</math>, but that does not work because only one of the numbers is a multiple of 5. So there are no such sets. | ||
| + | ==Solution 2== | ||
| + | As in the previous solution, none of the six consecutive numbers can be multiples of <math>7</math>. This means that together, they take on the values <math> \{ 1, 2, 3, 4, 5, 6, \} \mod 7</math>. The product of all the numbers in this set, then, is <math>-1 \mod 7</math>, by [[Wilson's Theorem]]. However, <math>-1</math> is not a [[quadratic residue]] <math>\mod 7</math>, which means that we cannot partition the original set into two sets of equal product. Thus, no such <math>n</math> exist. | ||
| + | ==See Also== | ||
{{IMO box|year=1970|num-b=3|num-a=5}} | {{IMO box|year=1970|num-b=3|num-a=5}} | ||
[[Category:Olympiad Number Theory Problems]] | [[Category:Olympiad Number Theory Problems]] | ||
Latest revision as of 02:31, 2 January 2023
Contents
Problem
Find the set of all positive integers 
with the property that the set 
can be partitioned into two sets such that the product of the numbers in one set equals the product of the numbers in the other set.
Solution
The only primes dividing numbers in the set can be 2, 3 or 5, because if any larger prime was a factor, then it would only divide one number in the set and hence only one product. Three of the numbers must be odd. At most one of the odd numbers can be a multiple of 3 and at most one can be a multiple of 5. The other odd number cannot have any prime factors. The only such number is 1, so the set must be 
, but that does not work because only one of the numbers is a multiple of 5. So there are no such sets.
Solution 2
As in the previous solution, none of the six consecutive numbers can be multiples of 
. This means that together, they take on the values 
. The product of all the numbers in this set, then, is 
, by Wilson's Theorem. However, 
is not a quadratic residue 
, which means that we cannot partition the original set into two sets of equal product. Thus, no such exist.
See Also
| 1970 IMO (Problems) • Resources | ||
| Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
| All IMO Problems and Solutions | ||
