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 Positive Integer Choice (Posted on 2010-04-22)
Prove that it is possible to choose only up to a set of 100 integers (and no more) from the first 200 positive integers - that is 1, 2,....,200; so that no integer in the set divides any other.

 No Solution Yet Submitted by K Sengupta Rating: 1.0000 (2 votes)

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 Approach. | Comment 1 of 2

Giving an example would suffice as a proof.

The primes up to 200 are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199.

These primes have no divisors (by definition) except one and itself.

Next, upon removal of the smallest prime, the biggest number up of a power to that prime (limited up to 200) can be included in the required list, and repeat for increasing primes. This also allows products of those numbers to be added into the required list.

By working out all the numbers, this would complete the problem.

 Posted by Vee-Liem Veefessional on 2010-04-22 13:33:57

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