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Sum of Prime Factors of f(84) for a Recursive Function | IOQM

JEE Maths question with a full step-by-step solution.

Question
The function ff is defined on the set of integers and satisfies
f(n)={n−3,if n≥1000,f(n+5),if n<1000.f(n)=\begin{cases}n-3, & \text{if } n\ge1000,\\[2pt] f(n+5), & \text{if } n<1000.\end{cases}
 Find the sum of the prime factors of f(84)f(84).
Solution
Answer: 31
Step 1: For any argument below 10001000, the rule f(n)=f(n+5)f(n)=f(n+5) lets us increase the argument by 55 repeatedly without changing the value, until it first reaches 10001000 or more. Starting from 8484,
f(84)=f(89)=f(94)=⋯=f(999)=f(1004).f(84)=f(89)=f(94)=\cdots=f(999)=f(1004).
 Step 2: All these arguments are congruent to 4(mod5)4\pmod5, and the first such value that is at least 10001000 is 10041004. Since 1004≥10001004\ge1000, the first rule applies:
f(1004)=1004−3=1001.f(1004)=1004-3=1001.
 Hence f(84)=1001f(84)=1001. Step 3: Factor 10011001 into primes:
1001=7×11×13.1001=7\times11\times13.
 Step 4: The sum of the prime factors is
7+11+13=31.7+11+13=31.
 Answer: 31
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