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Let d(p,n,0) be the multiplicative inverse of n modulo prime p, defined as n × d(p,n,0) = 1 mod p.
Let d(p,n,k) = $\sum_{i=1}^n$d(p,i,k−1) for k ≥ 1.
Let D(a,b,k) = $\sum$(d(p,p-1,k) mod p) for all primes a ≤ p &lt; a + b.
You are given:
D(101,1,10) = 45
D(103,102,102) = 8334
D(106,103,103) = 38162302Find D(109,105,105).

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