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

Step 1: => d(1) + d(2) + d(3) + …. + d(n) is odd

Step 2: d(square number) = odd

Step 3: for $n$ = 1

Step 4:  $d$ ( i ) = odd

Step 5: For $n$ = 3

Step 6:  $d$ ( i ) = d(1) + d(2) + d(3)

Step 7: odd + even + even = odd

Step 8: => $n$ in [1,3]

Step 9: For $n$ = 4

Step 10:  $d$ ( i ) = even as d(4) = odd

Step 11: => $n$ in [4, 8] in even

Step 12: => $n$ in [1, 3]  [9, 15]  […

Step 13: [12, 22 – 1]  [32, 42 – 1]  [52, 62 – 1] … [432, 442 – 1]

Step 14: => Number of element.

Step 15: (22 – 1 – 12 + 1) = 3

Step 16: (42 – 1 – 32 + 1) = 7

Step 17: => (22 – 12) + (42 – 32) + … + (442 – 432)

Step 18: => 3 + 7 + 11 + … 22 terms

Step 19: => 2 \times 3 + ( 22 – 1)  4 

Step 20: = 11 × [6 + 84] = 90 × 11

Step 21: => Sum of digits = 18.

Step 22: ❑ ❑ ❑