Number TheorymediumFree

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
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Number Theory · hard
How many of the numbers 12,22,32,…,199921^2, 2^2, 3^2, \ldots, 1999^2 have an odd digit in the tens place?
Number Theory · hard
The numbers in the sequence 101, 104, 109, 116, …101,\,104,\,109,\,116,\,\ldots are of the form an=100+n2a_n=100+n^2, where n=1,2,3,…n=1,2,3,\ldots. For each nn, let dnd_n be the greatest common divisor of ana_n and an+1a_{n+1}. Let kk be the maximum value of dnd_n as nn ranges over the positive integers. Find the value of k−120\dfrac{k-1}{20}.
Number Theory · medium
Let N=22225555+55552222N=2222^{5555}+5555^{2222}. Let rr be the remainder when NN is divided by 462462. Find the sum of the distinct prime factors of rr.
Number Theory · medium
If x1,x2,x3,x4,x5x_1,x_2,x_3,x_4,x_5 are five consecutive natural numbers with x1+x2+x3+x4+x5=n3andx2+x3+x4=m2,x_1+x_2+x_3+x_4+x_5 = n^3 \qquad\text{and}\qquad x_2+x_3+x_4 = m^2 , where mm and nn are natural numbers, then the least possible value of n3+m22\dfrac{n^3+m^2}2

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath