AMC 10 · 2014 · #20

Grade 11 algebra
Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search
logarithm-propertiesdomain-restrictionquadratic-equationscompleting-the-squareconsecutive-integer-count identify-subproblemsconvert-to-algebra ↑ Prerequisites: logarithm-propertiesquadratic-equations
📏 Long solution 💡 4 insights
Problem
A sum of two logarithms must stay below a bound. Count the whole numbers that qualify.

Pick an answer.

(A)
10
(B)
18
(C)
19
(D)
20
(E)
infinitely many
How to solve
Strategy Identify Subproblems

The set being counted is an intersection of two separate conditions, so Tool #7 (Identify Subproblems) splits the work: first "where do these logarithms exist at all?", then "where is their sum below 2?". Skipping the first subproblem is what makes choice (E) look possible. Tool #13 (Convert to Algebra) handles the second subproblem by merging the two logarithms and using the fact that log₁₀ is increasing to trade the log inequality for a plain product inequality — an equivalence, so the ＜ survives intact. Tool #4 (Introduce a Variable) sets y=x-50 to center the product, turning it into 100-y² and exposing why the inequality collapses to y² ＞ 0. Tool #14 (Extreme Principle) then does the part an equation-solver would miss: the product reaches its maximum of exactly 100 at one point, and because the inequality is strict that single boundary point must be thrown out — and it happens to be an integer. Tool #2 (Make a Systematic List) finishes by counting the integers that survive both conditions.

1STEP 1

Ask where the logarithms exist

Existence alone bounds the range.

x-40 ＞ 0 and 60-x ＞ 0 ⇔ 40 ＜ x ＜ 60
2STEP 2

Merge two logs into one

The two logs merge into one.

log₁₀[(x-40)(60-x)] ＜ 2 ⇔ (x-40)(60-x) ＜ 10²=100
3STEP 3

Center the product at 50

Centring turns it into a plain square.

y=x-50: (x-40)(60-x)=(10+y)(10-y)=100-y², so 100-y² ＜ 100 ⇔ y² ＞ 0
4STEP 4

Solve y² ＞ 0, not y²=0

The condition rules out exactly one value.

y² ＞ 0 ⇔ y ≠ 0 ⇔ x ≠ 50; at x=50: log₁₀₁₀+log₁₀₁₀=2 ≮ 2
5STEP 5

Count what survives both conditions

The count is 18, choice (B).

59-41+1=19, then 19-1=18 → (B)
Answer
18
Spot-check the edges of the claim. At x=41 the product is 1 · 19=19 and log₁₀₁₉≈ 1.28 ＜ 2, so it passes. At x=49 the product is 9 · 11=99 and log₁₀₉₉≈ 1.9956 ＜ 2, so it barely passes — exactly what the model 100-y² predicts, since y=-1 makes the product 99. At x=50 the product is 100 and the left side is exactly 2, which fails the strict ＜ . At x=51 the product is 11 · 9=99 again, so it passes. A second count confirms 18 without redoing any algebra: the substitution x↦ 100-x swaps x-40 with 60-x, so it leaves the whole condition unchanged and pairs the domain up as {41,59},{42,58},…,{49,51}, which is 9 pairs, plus the fixed point 50. Both members of each pair pass and the fixed point fails, giving 2 · 9=18. Choice (E) is impossible because the domain is a bounded interval, and 18 sitting just under the 19 integers of that interval is exactly the size to expect when only a single boundary value is lost, so (B) is correct.
💡Key takeaway

Logarithms only exist where their insides are positive, so pin down 41 ≤ x ≤ 59 first; the product (x-40)(60-x) equals 100-(x-50)², so it hits exactly 100 only at x=50, and a strict ＜ throws that one value away, leaving 18.

  • Ask where the logarithms exist
  • Merge two logs into one
  • Center the product at 50
  • Solve y² ＞ 0, not y²=0
  • Count what survives both conditions

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