Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?
I'm not really sure how to even approach this question.
Source: Washington's Monthly Math Hour, 2014
Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them
discrete mathematicsintuitionpigeonhole-principle
Related Question
- Pigeonhole Principle – Subset Sum Divisible by n in a Set of Positive Integers
- Given $51$ natural numbers whose sum is $100$, show that it is possible to split them to $2$ sets such that each of them is $50$.
- Show that in any set of $2n$ integers, there is a subset of $n$ integers whose sum is divisible by $n$.
Best Answer
Hint: $2,4,6$ are all divisors of $2+4+6=12$. Similarly $2+4+6+12$ are all divisors of $2+4+6+12=24$
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