AMC 10 · 2018 · #19

Grade 11 number-theory
Archetype Geometric Series Share └ Sub-archetype Cyclic Participant Share
geometric-series-infiniteprime-factorizationfundamental-counting-principlefraction-arithmetic identify-subproblemspattern-recognitioneasier-related-problem ↑ Prerequisites: geometric-series-infiniteprime-factorization
📏 Medium solution 💡 3 insights
Problem
A set holds every positive integer whose prime factors all come from two, three, and five — numbers like 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, but never 7, 11, or 14. Add the reciprocals of all of them. That infinite sum equals a fraction in lowest terms. Find the sum of its numerator and denominator.

Pick an answer.

(A)
16
(B)
17
(C)
19
(D)
23
(E)
36
How to solve
Strategy Identify Subproblems

Tool #7 (Identify Subproblems): the sum looks like one messy list, but every element of A is built from three independent choices — the power of 2, the power of 3, and the power of 5. That splits the single hard sum into three easy geometric series that get multiplied back together at the end. Tool #5 (Look for a Pattern) is what reveals the 2^a 3^b 5^c shape hiding in the list of denominators. Tool #9 (Solve an Easier Related Problem) supplies the piece each subproblem needs: first add up only the reciprocals of the powers of 2, then reuse that same move for 3 and for 5.

1STEP 1

Describe the set with exponents

Each element is three exponents.

A={ 2^a 3^b 5^c : a,b,c nonnegative integers }
2STEP 2

Add the powers of two first

The powers of two alone form a geometric series.

Σ_a=0^∞1/2^a=1/(1-1/2)=2
3STEP 3

Split the sum into three series

Independence makes the sum a product of three series.

Σ_k ∈ A1/k=(Σ_a=0^∞1/2^a)(Σ_b=0^∞1/3^b)(Σ_c=0^∞1/5^c)
4STEP 4

Evaluate the other two series

The other two series work the same way.

S₃=1/(1-1/3)=3/2, S₅=1/(1-1/5)=5/4
5STEP 5

Recombine and read off m + n

Multiplying and adding gives 19.

2·3/2·5/4=15/4 → m+n=15+4=19
Answer
19
Add the fourteen terms printed in the problem: they come to about 3.10. The answer 15/4=3.75 is larger, as it must be, and only a little larger, which fits a tail whose terms are already below 1/20 and shrinking fast. The total also has to be finite even though the full harmonic series is not, and it is: throwing away every number with a prime factor of 7 or more thins the list enough to converge. A structural check confirms the shape too — the product 2/1·3/2·5/4 has each prime p contributing p/(p-1), and the 2 in the first numerator cancels the 2 in the second denominator, leaving 15/4.
💡Key takeaway

When every number in a list is built from independent choices, add up each choice on its own and multiply the results together.

  • Describe the set with exponents
  • Add the powers of two first
  • Split the sum into three series
  • Evaluate the other two series
  • Recombine and read off m + n

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