no that cant be

 

 

they give  d1= 1 + 1/7(remaining medal )

                    = 1  +  1/7 (m - 1)

DAy two   d2 = 2  +  1/7 (m - 2  -d1)

Day three d3  = 3  +  1/7 (m - 3 -d2)

.

.

.

Day n       dn= n + 1/7 [m -n - d(n-1)]

 

Sum of all the medals m = [(1+2+3+.........n) + 1/7{ nm - (1+2+3...+n) - (d1+ d2+ d3+.......d(n-1))}

 

goes on.......