AMC 10 · 2004 · #25

Grade 9 algebra
base-conversiontelescoping-productfactorial pattern-recognitioneasier-related-problem ↑ Prerequisites: prime-factorization
📏 Long solution 💡 3 insights
Problem
For each integer base b at least 4, let a_b be the value of the repeating base-b expansion 0.133. Multiply these values for b from 4 to 99; the product can be written as m over n factorial with n as small as possible. Find m.

Pick an answer.

(A)
98
(B)
101
(C)
132
(D)
798
(E)
962
How to solve
Strategy Introduce a Variable

Treat the base as a variable b and find one formula that covers all 96 factors at once. The standard shift-and-subtract trick for repeating expansions turns a_b into a ratio of two polynomials. Then look at the shape of that ratio: if the numerator for base b can be rewritten as the same expression the denominator uses for base b+1, consecutive factors cancel and the whole 96-term product collapses to its two ends. Finally, minimizing n is a separate divisibility question, not a byproduct of the cancellation, so it gets its own argument.

1STEP 1

Turn the repeating expansion into a fraction

Shifting by one period and subtracting turns the expansion into a fraction.

b³ a_b = (b²+3b+3) + a_b ⟹ a_b(b³-1) = b²+3b+3 ⟹ a_b=(b²+3b+3)/(b³-1)
2STEP 2

Recognize the numerator as a cube minus one

The numerator is really the same cube-minus-one shape one step later.

b²+3b+3=((b+1)³-1)/b ⟹ a_b=((b+1)³-1)/(b(b³-1))
3STEP 3

Multiply and telescope

So the long product telescopes, leaving 95238 over 99 factorial.

a₄a₅… a₉₉=(Π_c=5¹⁰⁰(c³-1))/((Π_b=4⁹⁹ b)Π_b=4⁹⁹(b³-1))=(100³-1)/(4³-1)·6/99!=15873·6/99!=95238/99!
4STEP 4

Shrink the factorial

A shared factor cancels, dropping the factorial one step to give 962.

95238/99!=(99 · 962)/(99 · 98!)=962/98!, 962=2 · 13 · 37
5STEP 5

Show 98 cannot be beaten

A divisibility check proves no smaller factorial works, so m is 962, choice (E).

n ≤ 97 → 98 ∣ (98 · 97… (n+1)) ∣ 962, but 7 ∤ 962 → impossible; n=98, m=962
Answer
962
Test the closed form at the smallest base: it predicts a₄ = (16+12+3)/63 = 31/63 ≈ 0.4921, and adding the base-4 digits directly gives (1/4+3/16+3/64)·64/63 = 0.484375 · 64/63 ≈ 0.4921, a match. Test the telescoping on a short product: the same derivation stopped at base 5 predicts a₄a₅ = (6(6³-1))/(63 · 5!) = 1290/7560 = 43/252, and multiplying the two closed forms directly gives 31/63·43/124 = 43/252, the same. Finally every a_b is less than 1, so a product of 96 of them must be astronomically small, and 962/98! certainly is.
💡Key takeaway

Rewrite each factor so its top matches the next factor's bottom, and a 96-term product collapses to just its two ends — then check separately that nothing more can cancel.

  • Turn the repeating expansion into a fraction
  • Recognize the numerator as a cube minus one
  • Multiply and telescope
  • Shrink the factorial
  • Show 98 cannot be beaten