2012 AMC 12B Problem 24

Below is the professionally curated solution for Problem 24 of the 2012 AMC 12B, 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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Concepts:prime factorizationdivisibilityrecursion

Difficulty rating: 2520

24.

Define the function f1f_1 on the positive integers by setting f1(1)=1f_1(1) = 1 and if n=p1e1p2e2pkekn = 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)e11(p2+1)e21(pk+1)ek1.f_1(n) = (p_1 + 1)^{e_1 - 1}(p_2 + 1)^{e_2 - 1} \cdots (p_k + 1)^{e_k - 1}. For every m2,m \ge 2, let fm(n)=f1(fm1(n)).f_m(n) = f_1(f_{m-1}(n)). For how many NN in the range 1N4001 \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

1616

1717

1818

1919

Solution:

If N1N2N_1\mid N_2 then f1(N1)f1(N2),f_1(N_1)\mid f_1(N_2), 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 (p1pk)2400(p_1\cdots p_k)^2\le400 forces at most two primes.

Checking prime powers and prime pairs, the essential values are 25=32,2^5=32, 34=81,3^4=81, 73=343,7^3=343, and 2452=400.2^4\cdot5^2=400.

Their multiples up to 400400 number 400/32=12,\lfloor400/32\rfloor=12, 400/81=4,\lfloor400/81\rfloor=4, 400/343=1,\lfloor400/343\rfloor=1, and 400/400=1,\lfloor400/400\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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