2026 AIME I Problem 6

Question

The product of all positive real numbers

x

satisfying the equation

\sqrt[20]{x^{\log_{2026} x}} = 26x

is an integer

P.

Find the number of positive integer divisors of

P.

Answer

Set

t = \log_{2026} x

and take

\log_{2026}

of both sides to get a quadratic in

t

The two roots satisfy

t_1 + t_2 = 20,

so the product of the solutions

x

is

2026

raised to that power; factor

2026

Let

t = \log_{2026} x.

Taking

\log_{2026}

of both sides of

x^{\frac{\log_{2026} x}{20}} = 26x

gives

\frac{t^2}{20} = \log_{2026} 26 + t,

that is

t^2 - 20t - 20\log_{2026} 26 = 0.

The discriminant

400 + 80\log_{2026} 26

is positive, so there are two real roots

t_1, t_2,

each giving a valid positive solution

x = 2026^{t}.

By Vieta’s formulas

t_1 + t_2 = 20,

so the product of the solutions is

2026^{t_1} \cdot 2026^{t_2} = 2026^{20}.

Since

2026 = 2 \cdot 1013

and

1013

is prime,

P = 2^{20} \cdot 1013^{20}

has

21 \cdot 21 = 441

positive divisors.

Problem 6 in other years