AMC 10 · 2019 · #15

Grade 11 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

logarithm-propertiesperfect-squaresbound-inequality-then-enumeratesystematic-enumeration convert-to-algebrabound-inequality-then-enumeratesystematic-enumeration ↑ Prerequisites: logarithm-propertiesperfect-squares

📏 Medium solution 💡 3 insights

Problem

Positive real numbers a and b satisfy one equation: the square root of the base-ten logarithm of a, plus the square root of the logarithm of b, plus the logarithm of the square root of a, plus the logarithm of the square root of b, all add to 100. Each of those four terms is itself a positive integer. Find the product of the two numbers.

Pick an answer.

(A)

$10^{52}$

(B)

$10^{100}$

(C)

$10^{144}$

(D)

$10^{164}$

(E)

$10^{200}$

How to solve

Strategy Introduce a Variable

Tool #4 (Introduce a Variable): the whole equation is built from just two numbers, log a and log b. Name them, and the four scary-looking terms collapse into √(x) + √(y) + x/2 + y/2. Then name the square roots themselves, because those are the quantities the problem promises are integers. Tool #3 (Eliminate Possibilities): the two integrality demands — a whole square root and a whole half — cut the candidates down to a thin family. Tool #2 (Make a Systematic List): what survives is one small equation in two positive integers, so list the possible values and read off the pair that fits.

1STEP 1

Name the two logarithms

Give the two logarithms names.

√(x) + √(y) + x/2 + y/2 = 100

2STEP 2

Square roots must be whole

Whole square roots make them perfect squares.

√(x) = m, x = m², √(y) = n, y = n²

3STEP 3

Halves force even squares

The halving forces those roots to be even.

m = 2p, x = 4p²; n = 2q, y = 4q²

4STEP 4

Rebuild the equation

Rebuilding the equation simplifies it.

2p + 2q + 2p² + 2q² = 100 ⟹ p² + p + q² + q = 50 ⟹ p(p+1) + q(q+1) = 50

5STEP 5

List the products

List the products to find the pair.

p(p+1) ∈ {2, 6, 12, 20, 30, 42}, 20 + 30 = 50 ⟹ {p, q} = {4, 5}

6STEP 6

Back to ab

Converting back gives ten to the one hundred sixty-fourth.

log a = 64, log b = 100 ⟹ ab = 10⁶⁴ · 10¹⁰⁰ = 10¹⁶⁴ → (D)

Answer

10¹⁶⁴

Put the numbers back into the original equation: √(log a) = 8, √(log b) = 10, log √(a) = 64/2 = 32, log √(b) = 100/2 = 50. All four are positive integers, and 8 + 10 + 32 + 50 = 100, exactly as required. A second check pins down the size: the last two terms give (log a + log b)/2 = 100 - (8 + 10) = 82, so log(ab) = 164 — the exponent must be a bit under 200 but well above 100, which rules out 10¹⁰⁰ and 10²⁰⁰ immediately. The near misses fail for the same structural reason: log(ab) = 144 would force m + n = 100 - 144/2 = 28 together with m² + n² = 144, but m + n = 28 already makes m² + n² ≥ 28²/2 = 392. Only 164 keeps the two demands compatible.

💡Key takeaway

Name the two logarithms, let "whole square root" and "whole half" squeeze them down to log a = 64 and log b = 100, and the product is ab = 10¹⁶⁴ — choice (D).

Name the two logarithms
Square roots must be whole
Halves force even squares
Rebuild the equation
List the products
Back to ab