Please open a donation section too. So, that we can grow as a whole and get quality stuffs more frequently.

The only drawback is their bad plagiarism checker. Ratings of the fair contestants drop because of the massive cheater base and their plagiarism checker failing to detect them. But their problems are noice no doubt.

This is how I did (proof):

I always use a number line for such problems and this is my go-to. In this problem, I always considered c to be 1, and the rest of my solution went up from there.

If we give a look at the number line, then the gcd of any two numbers will never exceed their difference.

Case 1: n is even and say n=82; --> Taking c=1 leaves us with 81 --> 40.5 is what we get if we divide it by 2 (You know why we divide) --> Distribute the 0.5 (40.5+0.5 + 40.5-0.5 gives us 41 + 40) --> In the number line adjacent numbers will always have a gcd of 1. So gcd is 1.

Case 2a: n is odd and say n=81; --> Taking c=1 leaves us with 80 --> 40 is what we get if we divide it by 2 (You know why we divide) --> Move the two numbers to the nearest odd numbers (40-1 + 40+1 gives us 39 and 41) --> In the number line their difference is 2 and no way you can divide those numbers with an odd number which is greater than 1. So gcd is 1. (max diff is 2)

Case 2b: n is odd and say n=83; --> Taking c=1 leaves us with 82 --> 41 is what we get if we divide it by 2 (You know why we divide) --> Move the two numbers to the nearest odd numbers (41-2 + 41+2 gives us 39 + 43) --> In the number line their difference is 4 and no way you can divide those numbers together with an odd number other than 1. So gcd is 1(max diff is 4 and you can't jump from 39 to 43 using 3 steps)

So, it's proved that in all the 3 the cases their gcd will always be 1 for the given constraint while maintaining their summation.