AGC022B

https://gyazo.com/13f3d39471baeb131d387b0805df3eab

Thoughts.

1: all gcd's are 1

2: One and the other sum, gcd is not 1

3: Element is less than 30000

Hmmm, I can't image it, but from condition 2, 1 can't be a factor.

If there is a 2, then one or more odd numbers exist. And any addition other than 2 is even.

It is easy to show that there is no size 2 solution

Consider the time of 3

If there are two, the other two are odd.

If there are 3, the rest is 3n+1

Odd, 3n+1, and a multiple of 5

25.

Therefore, 2, 3, and 25 are solutions

Think of a time when 4.

One even number in the remaining three.

If there are 3, then the sum of the remaining two is 3n+1

If there is a 5, then the remaining one is 5m.

Hmmm, that's going to be a lot of splitting up.

2+3+25=30=2x3x5.

2 x 3 x 5 x 7 = 210 well distributed...how?

You're thinking too hard about the constraints.

The condition that all gcd's are 1 holds for 2 x 3, 3 x 5, and 5 x 2

But in this instance, condition 2 is impossible.

If condition 1 is not present

Easy, just make them all even numbers or something.

Put two odd numbers to satisfy condition 1.

Condition 2 holds for all but odd numbers.

For condition 2 to hold for this odd number?

Odd prime numbers, both sides must have something in common

Let a, b be two odd numbers and c be the sum of even numbers

Need common divisor for a+c and b, b+c and a

No, because if these two terms match, condition 1 is broken.

What if a=3n, b=5m?

I think I can find it with no problem, but I'd like to code it and check.

c cannot be a multiple of 3 or 5, but it can be an even number, so just adjust as appropriate

If c's remainder at 3 or 5 is determined, so are a's and b's.

Official Explanation

I overlooked the severity of the restrictions.

If you're choosing 2,000 pieces with less than 3,000 elements, you can't take the "mostly even" option.

I assumed that the sum is a multiple of 2, which led me to a method that consists of numbers up to about 4000.

The official explanation method is to assume that this is "the sum is a multiple of 2 and 3".

If the sum is a multiple of 2 and 3, and the components are multiples of 2 or 3, then "2: One and the sum of the others, gcd not 1" is satisfied

If 2 and 3 are in the components, then "1: all gcd's are 1" is also satisfied.

---