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SPOILERS!!!SPOILERS!!! Do not look without solving the problem!
let (xi,yi) are coordinates of n points. Following sets {xi} {yi} {xi+yi} {xi-yi} are some permutations of {1,2,...,n} by module n.
if n%2==0 Then Sum(xi)+Sum(yi)=Sum(xi+yi) (mod n) it is not true since, n*(n+1)!=(n*(n+1))/2 (mod n) (since n -even)
if n%3==0 Then Sum(xi*xi)+Sum(yi*yi)=Sum( (xi+yi)*(xi+yi) )+Sum( (xi-yi)*(xi-yi) ) it is not true since, ( n*(n+1)*(2n+1) )/3!=( 2*n*(n+1)*(2n+1) )/3 (mod n) (since n%3==0)
Here, I used the fact that if (xi) is permutation of {1,2,...,n} by module n then sum of them is n*(n+1)/2 (mod n). Same with summing (xi*xi). help Some tests:
n - answer
15 - No
99 - No
23 - Yes
97 - Yes
16 - No thanks if you have WA20 try this tests :)
25 YES
145 YES
9991 YES
9995 YES Is it right, that on even numbers answer will always "No"? |
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