AIME 2026 I · 第 6 题

Solution 1

Raising both sides to the power of 20, we have

$$x^{\log_{2026}x} = (26x)^{20}$$ Taking log base $x$ of both sides, we obtain

$$\log_{2026}x = 20 \log_{x}26+20$$ Rewrite in log base $e$:

$$\dfrac{\ln x}{\ln 2026} = \dfrac{20 \ln 26}{\ln x} + 20$$ Let $y = \ln x$. Substituting and multiplying both sides by $y \cdot \ln 2026$, we obtain

$$y^2 = 20 \ln 26 \ln 2026 + 20 \ln 2026 \cdot y$$ This becomes

$$y^2 - 20 \ln 2026 \cdot y - 20 \ln 26 \ln 2026 = 0$$ Note that there are $2$ solutions $x_1$ and $x_2$. We wish to find $x_1 \cdot x_2$, or $e^{y_1 + y_2}$. By Vieta's,

$$y_1 + y_2 = 20 \ln 2026$$ Then

$$e^{y_1 + y_2} = e^{20 \ln 2026} = (e^{\ln 2026})^{20} = 2026^{20} = 1013^{20} \cdot 2^{20}$$ Then our answer is

$$21 \cdot 21 = \boxed{441} \text{. } \square$$ ~BrocSoc

Solution 2

Raising both sides to the $20^{\text{th}}$ power:

$$x^{\log_{2026} x}=26^{20}x^{20}$$ Let $x=2026^u$. Now the equation becomes:

$$2026^{u^2}=26^{20}2026^{20u}$$ $$2026^{u^2-20u}=26^{20}$$ The sum of all $u$ is $20$, therefore the product of all $x$ is $2026^{20}=2^{20}\cdot 1013^{20}$ which has $\boxed{441}$ factors.

~ zhenghua

Another version of solution 2 (more in depth) is found in solution 5.

Solution 3 (Cheese)

Just guess $2026^{20}$ or $26^{20}$. Each of these have $21 * 21 = \boxed{441}$ factors.

~Aarav22 (I did not make AIME)

Note to author: it is not intuitively clear why one should guess these numbers, please provide more rationale

Note to note: there are actually not even that many to try. If we only use the numbers $20,26, 2026$, and raise one to the power of another, we see that anything raised to $2026$ will have way too many factors for the AIME answer extraction, and $20^{26}$ also has too many factors for the answer extraction. $2026^{26}$, as it turns out, has $729$ factors, but as long as you don't pull the unlucky d3 (if you choose one of these three options that give an available answer extraction), you actually will get $441$.

~tiguhbabheow (I did make AIME)

Solution 4 (Similar to Solution 2)

Consider the substitution $y=\log_{2026} x$, so $2026^y=x$. Then we have $x^y=2026^{y^2}$, so $2026^{\frac{y^2}{20}}=26\cdot 2026^y$, which implies $2026^{\frac{y^2}{20-y}}=26$. Thus, $\frac{y^2}{20-y}=\log_{2026} 26$, so $y^2-20y-20\log_{2026} 26=0$. We see that this quadratic has real solutions. Let the corresponding solutions for $x$ be $2026^{y_1}$ and $2026^{y_2}$. Then the product of the solutions for $x$ is $2026^{y_1+y_2}$. By Vieta’s formulas, $y_1+y_2=20$, so the product of the solutions for $x$ is $2026^{20}$. Factoring, $2026=2\cdot 1013$, so $2026^{20}=2^{20}\cdot 1013^{20}$, which has $(20+1)(20+1)=441$ divisors.

~pl246631

Solution 5

Take the 20th power on both sides to obtain

$$x^{\log_{2026}x} = (26x)^{20}$$ Given the condition $\log_{2026}(x)$, we find that this equals $a$. Then, $x = 2026^a$. Thus the expression simplifies to

$$2026^{a^2} = (26 \cdot 2026^a)^{20}$$ $$2026^{a^2} = 26^{20} \cdot 2026^{20a}$$ In which we may divide by $2026^{20a}$ on both sides to obtain

$$\frac{2026^{a^2}}{2026^{20a}} = 26^{20}$$ $$2026^{a^2 - 20a} = 26^{20}$$ Taking the logarithm base $2026$ on both sides results in the quadratic

$$a^2 - 20a - \log_{2026}(26^{20}) = 0$$ in simplest form is

$$a^2 - 20a - 20 \log_{2026}(26) = 0.$$ The values of $a$ are determined by

$$a = \frac{20 \pm \sqrt{20^2 + 4 \cdot 20 \log_{2026}(26)}}{2}$$ Notice that the values of $x$ are equivalent to $2026^a$. We require $x \in \mathbb{R}^+$. Thus, any real value $a$, positive or negative, is permitted. Thus two values $x_1 = 2026^{a_1}$ and $x_2 = 2026^{a_2}$ exist. When we take their product, it is equivalent to $2026^{a_1} \cdot 2026^{a_2} = 2026^{a_1+a_2}$, in which we can see that the discriminant of $a$ cancels out. Thus, if we let $b = \sqrt{20^2 + 4 \cdot 20 \log_{2026}(26)}$, then we have $a_1 = \frac{20 + b}{2}$, and $a_2 = \frac{20 - b}{2}$, we see that the product of all $x$ is just $2026^{20}$.

Thus we have $P = 2026^{20}$, whose prime factorization is $P = 1013^{20} \cdot 2^{20}$. The number of factors of a number with prime factorization $p_1^{e_1} \cdot p_2^{e_2} \cdot \ldots \cdot p_n^{e_n}$ is determined by $(e_1+1)(e_2+1)\ldots(e_n+1)$. Thus, the number of factors of $P$ is just $(20+1)(20+1) = 21^2 = \boxed{441}$.

~Pinotation

Solution 6

Let $y=\log_{2026}x$. Raise both sides to the $20$th power and take log base $2026$ of both sides.

$$\log_{2026}(x^{\log_{2026}x}) = \log_{2026}(26x)^{20}$$ By the power rule of logarithms we have

$$\log_{2026}x \cdot \log_{2026}x = 20(\log_{2026}(26 \cdot x))$$ By the product rule of logarithms,

$$\log_{2026}x \cdot \log_{2026}x = 20\log_{2026}26 + 20\log_{2026}x$$ Plugging in $y$ we have

$$y^2 = 20y + 20\log_{2026}26$$ $$y^2 - 20y - 20\log_{2026}26 = 0$$ Let's say the roots of this equation are $y_1$ and $y_2$, then the product of the values of $x$ is

$$2026^{y_1} \cdot 2026^{y_2} = 2026^{y_1 + y_2}$$ By Vieta's we know that $y_1+y_2$ is just 20, so the product of the values of $x$ is $2026^{20}$

$$2026^{20} = 2^{20} \cdot 1013^{20}$$ So the number of factors of $2026^{20} (P)$ is $(20 + 1)(20 + 1) = 21^2 = \boxed{441}$.

~midnightgalaxy

Video Solution (Fast and Easy 🔥🚀)

2026 AIME I #6

PiAcademyUs.org

Video Solution (Easy)

https://www.youtube.com/watch?v=n6Jya5Jq58Y - Continuum Math