No Sum

Probability Level 3

What is the maximum number of numbers can you choose from the first 20 positive integers such that none of the chosen numbers is the sum of exactly 2 other chosen numbers?

Example - You can choose 1,2 and 4 but you can’t choose 1,2 and 3 (since 1+2=3)


The answer is 11.

Chandrashekar Giridharan by Chandrashekar Giridharan , 18, India

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1 solution

Richard Costen
May 17, 2018

10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20 = 11 \boxed{11} numbers. Any two of these numbers add to more than 20. As a side note, one can also choose all the odd numbers but that only gives 10 numbers. If one begins at 1 and moves upwards, one can choose 1, 2, 4, 7, 10, 13, 16, 19 (every 3rd number after 4, because of the 1, 2) but that only gives 8 numbers.

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Thanks...I edited the question

Chandrashekar Giridharan - 3 years ago

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A follow-up question could be to find the maximum such that none of the chosen numbers is the p r o d u c t product of exactly 2 other chosen numbers. The "other" condition means I think we can include 1 in any such subset, in which case I get a maximum of 12, with the elements being 1 , 2 , 3 , 4 , 5 , 7 , 9 , 11 , 13 , 16 , 17 , 19 1,2,3,4,5,7,9,11,13,16,17,19 , i.e., all the primes and perfect squares.

Brian Charlesworth - 3 years ago

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