AMC 10 · 2005 · #17

Grade 11 algebranumber-theory
logarithm-propertiesprime-factorizationexponents convert-to-algebraeasier-related-problem ↑ Prerequisites: logarithm-propertiesprime-factorizationfraction-arithmetic
📏 Long solution 💡 3 insights
Problem
Four rational numbers weight the logarithms of four primes so the total is exactly 2005. Count how many such four-tuples exist.

Pick an answer.

(A)
0
(B)
1
(C)
17
(D)
2004
(E)
infinitely many
How to solve
Strategy Convert to Algebra

A sum of logarithms is hard to reason about directly, so we convert it into algebra about exponents (Tool #13): exponentiating turns the equation into 2^a3^b5^c7^d = 10²⁰⁰⁵. Rational exponents are still awkward — unique factorisation only compares integer exponents — so we introduce a common denominator N (Tool #4) and raise both sides to the N-th power, reducing the question to an easier related problem (Tool #9) about integers only. Then reorganising both sides into prime factorisations (Tool #15) forces the comparison. The step that must not be skipped is the one that looks obvious: "there is no 3 on the right, so b=0" is the whole claim, not a free move, and it is only legitimate once the exponents are integers.

1STEP 1

Move coefficients into exponents

Each coefficient moves into an exponent.

alog₁₀₂ + blog₁₀₃ + clog₁₀₅ + dlog₁₀₇ = log₁₀₂^a + log₁₀₃^b + log₁₀₅^c + log₁₀₇^d
2STEP 2

Collapse to one power equation

The sum collapses into one power equation.

2^a · 3^b · 5^c · 7^d = 10²⁰⁰⁵ = 2²⁰⁰⁵ · 5²⁰⁰⁵
3STEP 3

Clear the denominators

A common denominator makes every exponent an integer.

(2^a3^b5^c7^d)^N = 2^A · 3^B · 5^C · 7^D = 2²005N · 5²005N
4STEP 4

Push everything to one side

Moving everything to one side leaves a product equal to one.

2^A' · 3^B · 5^C' · 7^D = 1, A' = A-2005N, C' = C-2005N
5STEP 5

Match the prime factorisations

Disjoint primes force every exponent to be zero.

A' = B = C' = D = 0
6STEP 6

Read off the tuple

So the whole tuple is forced, with no freedom anywhere.

a = 2005, b = 0, c = 2005, d = 0
7STEP 7

Check it, then count

It checks out, so exactly 1 tuple exists, choice (B).

2005log₁₀₂ + 0 + 2005log₁₀₅ + 0 = 2005(log₁₀₂+log₁₀₅) = 2005log₁₀₁₀ = 2005
Answer
1
Both halves were established separately: the tuple (2005,0,2005,0) satisfies the equation by direct substitution using log₁₀₂+log₁₀₅ = 1, and the prime-factorisation argument shows nothing else can. Both are needed — proving only that one tuple works would leave (E) open, and proving only uniqueness would leave (A) open. The choices 17 and 2004 have no possible source: a single linear equation has a solution set that is empty, a single point, or an infinite family, never a finite set of size 17.
💡Key takeaway

One equation with four unknowns usually has endless solutions, but primes are stubborn: 2, 3, 5 and 7 cannot imitate each other, so only a single tuple survives.

  • Move coefficients into exponents
  • Collapse to one power equation
  • Clear the denominators
  • Push everything to one side
  • Match the prime factorisations
  • Read off the tuple
  • Check it, then count