2012 AMC 12B Problem 24

Attempt Problem 24 of the 2012 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 12B solutions, or check the answer key.

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24.

Define the function f1f_1 on the positive integers by setting f1(1)=1f_1(1) = 1 and if n=p1e1p2e2⋯pkekn = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k} is the prime factorization of n>1,n \gt 1, then f1(n)=(p1+1)e1−1(p2+1)e2−1⋯(pk+1)ek−1. \begin{aligned} &f_1(n) = (p_1 + 1)^{e_1 - 1}(p_2 + 1)^{e_2 - 1} \\ &\quad \cdots (p_k + 1)^{e_k - 1}. \end{aligned} For every m≥2,m \ge 2, let fm(n)=f1(fm−1(n)).f_m(n) = f_1(f_{m-1}(n)). For how many NN in the range 1≤N≤4001 \le N \le 400 is the sequence (f1(N),f2(N),f3(N),…)(f_1(N), f_2(N), f_3(N), \ldots) unbounded? Note: a sequence of positive numbers is unbounded if for every integer B,B, there is a member of the sequence greater than B.B.
1515
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1919
Answer: D
Concepts:prime factorizationdivisibilityrecursion
Small hint:
If N2N_2 is a multiple of N1N_1 then f1(N2)f_1(N_2) is a multiple of f1(N1),f_1(N_1), so unboundedness is inherited by multiples; find the minimal “essential” NN
Big hint:
An essential NN has every exponent at least 22 and at most two prime factors; analyze powers of 2,32,3 first, then the few remaining prime powers and prime pairs below 400400
Solution:
If N2N_2 is a multiple of N1N_1 then f1(N2)f_1(N_2) is a multiple of f1(N1),f_1(N_1), so if SN1S_{N_1} is unbounded so is SN2.S_{N_2}. Call NN essential if it is unbounded but no proper divisor is. An essential NN must have all exponents at least 2,2, and (p1⋯pk)2≤400(p_1\cdots p_k)^2\le400 forces at most two primes.
For n=2a3b,n=2^a3^b, two iterations send the exponent pair to (2a−4,2b−3).(2a-4,2b-3). Thus the orbit is unbounded exactly when a≥5a\ge5 or b≥4,b\ge4, producing the essential values 25=322^5=32 and 34=81.3^4=81. For a single other prime, the bound leaves only 52,53,72,73,112,132,172,192;5^2,5^3,7^2,7^3,11^2,13^2,17^2,19^2; direct application of f1f_1 leaves only 73=3437^3=343 essential. With two primes, p1p2≤20p_1p_2\le20 leaves the pairs (2,5),(2,7),(3,5)(2,5),(2,7),(3,5) besides (2,3).(2,3). Applying f1f_1 to their possible squareful products leaves only 24⋅52=400.2^4\cdot5^2=400.
Their multiples up to 400400 number ⌊40032⌋=12,\lfloor\frac{400}{32}\rfloor=12, ⌊40081⌋=4,\lfloor\frac{400}{81}\rfloor=4, ⌊400343⌋=1,\lfloor\frac{400}{343}\rfloor=1, and ⌊400400⌋=1,\lfloor\frac{400}{400}\rfloor=1, with no overlaps, for a total of 12+4+1+1=18.12+4+1+1=18.
Thus, the correct answer is D.
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