How do I prove the following question
Prove that in any partition of N9 (The first nine natural numbers) into three sets, there will be at least one set whose product of numbers is greater than or equal to 72.
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I would go for a proof by contradiction.
Note that the product of the first nine natural numbers is 9! = 362880. Furthermore, if we multiply the products of the different sets, we should arrive at the same answer.
Now, assume that every product of the sets in the partition is less than 72. I.e the products could be at most 71. Even if all of the three products is the maximum allowed value, the product of all the numbers would be at most 71 * 71 * 71 = 357911.
This falls short of the known value of 362880. So we find a contradiction.
The contradiction occurs because of our assumption, i.e. that all of the sets in the partition have a product less than 72. Therefore, this assumption can not be true. Hence there must be at least one set with a product equal to or bigger than 72.