Let d(m) denote the number of positive integer divisors of a positive integer m . If r is the number of integers n 2023 for which _ i=1 ^n d(i) is odd, find the sum of the digits of r .

Step-by-step solution

_ i=1 ^n d(i) is odd d(1)+d(2)+d(3)+ .+d(n) is odd d (square number) = odd for n=1 _ i=1 ^n d(i)= odd For n=3 aligned & _ i=1 ^n d(i)=d(1)+d(2)+d(3) & odd + even + even = odd & n [1,3] aligned Similarly, For n=4 aligned & _ i=1 ^n d(i)= even as d(4)= odd & n [4,8] even & n [1,3] [9,15] [ & [1^2, 2^2-1 ] [3^2, 4^2-1 ] [5^2, 6^2-1 ] [43^2, 44^2-1 ] aligned Number of element. aligned & (2^2-1-1^2+1 )=3 & (4^2-1-3^2+1 )=7 aligned aligned & (2^2-1^2 )+ (4^2-3^2 )+ + (44^2-43^2 ) & 3+7+11+ 22 terms & 22 2 [2 3+(22-1) 4]