AMC 10 · 2013 · #20

Grade 6 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint
prime-factorizationfactorialprime-numbers extremal-constructionidentify-subproblems ↑ Prerequisites: prime-factorizationfactorial
📏 Medium solution 💡 3 insights
Problem
Write 2013 as a fraction whose top and bottom are each products of factorials, 2013 = (a₁! a₂! ... a_m!)/(b₁! b₂! ... b_n!), where the a's are non-increasing positive integers and the b's are non-increasing positive integers. Here a₁ is the largest number on top and b₁ the largest on the bottom. Among all such ways, pick one where a₁ + b₁ is as small as possible. Report |a₁ - b₁|.

Pick an answer.

(A)
1
(B)
2
(C)
3
(D)
4
(E)
5

AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Extreme Principle

The phrase 'as small as possible' points straight at the Extreme Principle: chase the smallest a₁ and the smallest b₁ that can still work. The lever is prime factors. A factorial n! contains a prime p only if n reaches p, so the largest prime inside 2013 forces a₁, and the largest unwanted prime dragged along forces b₁. Break the problem into 'how small can a₁ be' and 'how small can b₁ be', then build one explicit expression to prove those minimums are reachable.

1STEP 1

Factor 2013 into primes

2013 = 3 x 671 and 671 = 11 x 61, so 2013 = 3 x 11 x 61 — the largest prime factor is 61.

2013 = 3 × 11 × 61
2STEP 2

The prime 61 forces a₁ = 61

A factorial n! holds the prime 61 only once n reaches 61, and the top must, so the smallest possible a₁ is 61.

61 ∣ 2013 → a₁ ≥ 61, min a₁ = 61
3STEP 3

61! drags in 59, forcing b₁ = 59

61! also carries the prime 59, which 2013 does not want, so the bottom must cancel it: b₁ = 59, and a₁ + b₁ = 120 is the floor.

61! contains 59, 59 ∤ 2013 → b₁ ≥ 59, a₁ + b₁ = 120
4STEP 4

Build an expression that hits 120

That floor is reached: (61! 11! 3!)/(59! 10! 5!) = 3660 x 11/20 = 2013, and both lists are in order.

61! 11! 3!/59! 10! 5! = (60 · 61) · 11 · 1/20 = 3660·11/20 = 2013
5STEP 5

Take the difference

The minimum pins a₁ = 61 and b₁ = 59, so |a₁ - b₁| = |61 - 59| = 2, choice (B).

|a₁ - b₁| = |61 - 59| = 2
Answer
2
The two anchors come from opposite ends: 61 is the largest prime inside 2013 (so it must appear on top), and 59 is the largest prime that 61! unavoidably drags along but 2013 does not want (so it must appear on the bottom). Both bounds a₁ ＞ = 61 and b₁ ＞ = 59 are tight, and the explicit expression (61! 11! 3!)/(59! 10! 5!) = 2013 confirms both are reached, so |61 - 59| = 2 is solid.
💡Key takeaway

The biggest prime inside a number decides the top factorial, and the biggest unwanted prime it drags along decides the bottom one.

  • Factor 2013 into primes
  • The prime 61 forces a₁ = 61
  • 61! drags in 59, forcing b₁ = 59
  • Build an expression that hits 120
  • Take the difference

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