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 .

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Quantrex Academy · Accepted Answer

Correct answer: 18

∑_ 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] & =11 ×[6+84]=90 × 11 & =990 aligned Sum of digits =18 .

Let d(m) denote the number of positive integer divisors of a...

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