AIME 2023 I · 第 10 题

Solution (Bounds and Decimal Part Analysis, Rigorous)

Define $\left\{ x \right\} = x - \left\lfloor x \right\rfloor$.

First, we bound $U$.

We establish an upper bound of $U$. We have

$$\begin{aligned} U & \leq \sum_{n=1}^{2023} \frac{n^2 - na}{5} \\ & = \frac{1}{5} \sum_{n=1}^{2023} n^2 - \frac{a}{5} \sum_{n=1}^{2023} n \\ & = \frac{1012 \cdot 2023}{5} \left( 1349 - a \right) \\ & \triangleq UB . \end{aligned}$$ We establish a lower bound of $U$. We have

$$\begin{aligned} U & = \sum_{n=1}^{2023} \left( \frac{n^2 - na}{5} - \left\{ \frac{n^2 - na}{5} \right\} \right) \\ & = \sum_{n=1}^{2023} \frac{n^2 - na}{5} - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & = UB - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & \geq UB - \sum_{n=1}^{2023} \mathbf 1 \left\{ \frac{n^2 - na}{5} \notin \Bbb Z \right\} . \end{aligned}$$ We notice that if $5 | n$, then $\frac{n^2 - na}{5} \in \Bbb Z$. Thus,

$$\begin{aligned} U & \geq UB - \sum_{n=1}^{2023} \mathbf 1 \left\{ \frac{n^2 - na}{5} \notin \Bbb Z \right\} \\ & \geq UB - \sum_{n=1}^{2023} \mathbf 1 \left\{ 5 \nmid n \right\} \\ & = UB - \left( 2023 - \left\lfloor \frac{2023}{5} \right\rfloor \right) \\ & = UB - 1619 \\ & \triangleq LB . \end{aligned}$$ Because $U \in \left[ - 1000, 1000 \right]$ and $UB - LB = 1619 < \left( 1000 - \left( - 1000 \right) \right)$, we must have either $UB \in \left[ - 1000, 1000 \right]$ or $LB \in \left[ - 1000, 1000 \right]$.

For $UB \in \left[ - 1000, 1000 \right]$, we get a unique $a = 1349$. For $LB \in \left[ - 1000, 1000 \right]$, there is no feasible $a$.

Therefore, $a = 1349$. Thus $UB = 0$.

Next, we compute $U$.

Let $n = 5 q + r$, where $r = {\rm Rem} \ \left( n, 5 \right)$.

We have

$$\begin{aligned} \left\{ \frac{n^2 - na}{5} \right\} & = \left\{ \frac{\left( 5 q + r \right)^2 - \left( 5 q + r \right)\left( 1350 - 1 \right)}{5} \right\} \\ & = \left\{ 5 q^2 + 2 q r - \left( 5 q + r \right) 270 + q + \frac{r^2 + r}{5} \right\} \\ & = \left\{\frac{r^2 + r}{5} \right\} \\ & = \left\{ \begin{array}{ll} 0 & \text{ if } r = 0, 4 \\ \frac{2}{5} & \text{ if } r = 1, 3 \\ \frac{1}{5} & \text{ if } r = 2 \end{array} \right. . \end{aligned}$$ Therefore,

$$\begin{aligned} U & = \sum_{n=1}^{2023} \left( \frac{n^2 - na}{5} - \left\{ \frac{n^2 - na}{5} \right\} \right) \\ & = UB - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & = - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & = - \sum_{q=0}^{404} \sum_{r=0}^4 \left\{\frac{r^2 + r}{5} \right\} + \left\{ \frac{0^2 - 0 \cdot a}{5} \right\} + \left\{ \frac{2024^2 - 2024a}{5} \right\} \\ & = - \sum_{q=0}^{404} \left( 0 + 0 + \frac{2}{5} + \frac{2}{5} + \frac{1}{5} \right) + 0 + 0 \\ & = - 405 . \end{aligned}$$ Therefore, $a + U = 1349 - 405 = \boxed{\textbf{944}}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

~minor edits by KevinChen_Yay

Solution 2

We define $U' = \sum^{2023}_{n=1} {\frac{n^2-na}{5}}$. Since for any real number $x$, $\lfloor x \rfloor \le x \le \lfloor x \rfloor + 1$, we have $U \le U' \le U + 2023$. Now, since $-1000 \le U \le 1000$, we have $-1000 \le U' \le 3023$.

Now, we can solve for $U'$ in terms of $a$. We have:

$$\begin{aligned} U' &= \sum^{2023}_{n=1} {\frac{n^2-na}{5}} \\ &= \sum^{2023}_{n=1} {\frac{n^2}{5} - \frac{na}{5}} \\ &= \sum^{2023}_{n=1} {\frac{n^2}{5}} - \sum^{2023}_{n=1} {\frac{na}{5}} \\ &= \frac{\sum^{2023}_{n=1} {{n^2}} - \sum^{2023}_{n=1} {na}}{5} \\ &= \frac{\frac{2023(2023+1)(2023 \cdot 2 + 1)}{6} - \frac{a \cdot 2023(2023+1)}{2} }{5} \\ &= \frac{2023(2024)(4047-3a)}{30} \\ \end{aligned}$$ So, we have $U' = \frac{2023(2024)(4047-3a)}{30}$, and $-1000 \le U' \le 3023$, so we have $-1000 \le \frac{2023(2024)(4047-3a)}{30} \le 3023$, or $-30000 \le 2023(2024)(4047-3a) \le 90690$. Now, $2023 \cdot 2024$ is much bigger than $90690$ or $30000$, and since $4047-3a$ is an integer, to satsify the inequalities, we must have $4047 - 3a = 0$, or $a = 1349$, and $U' = 0$.

Now, we can find $U - U'$. We have:

$$\begin{aligned} U - U' &= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor} - \sum^{2023}_{n=1} {\frac{n^2-1349n}{5}} \\ &= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5}} \end{aligned}.$$ Now, if $n^2-1349n \equiv 0 \text{ (mod 5)}$, then $\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5} = 0$, and if $n^2-1349n \equiv 1 \text{ (mod 5)}$, then $\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5} = -\frac{1}{5}$, and so on. Testing with $n \equiv 0,1,2,3,4, \text{ (mod 5)}$, we get $n^2-1349n \equiv 0,2,1,2,0 \text{ (mod 5)}$ respectively. From 1 to 2023, there are 405 numbers congruent to 1 mod 5, 405 numbers congruent to 2 mod 5, 405 numbers congruent to 3 mod 5, 404 numbers congruent to 4 mod 5, and 404 numbers congruent to 0 mod 5. So, solving for $U - U'$, we get:

$$\begin{aligned} U - U' &= \sum^{2023}_{n=1} {\lfloor \frac{n^2-1349n}{5} \rfloor - \frac{n^2-1349n}{5}} \\ &= 404 \cdot 0 - 405 \cdot \frac{2}{5} - 405 \cdot \frac{1}{5} - 405 \cdot \frac{2}{5} - 404 \cdot 0 \\ &= -405(\frac{2}{5}+\frac{1}{5}+\frac{2}{5}) \\ &= -405 \end{aligned}$$ Since $U' = 0$, this gives $U = -405$, and we have $a + U = 1349-405 = \boxed{944}$.

~ genius_007

Solution 3 (Quick)

We can view the floor function in this problem as simply subtracting the remainder of $n^2 - na$ (mod $5$) from the numerator of $\frac{n^2-na}{5}$. For example, $\left\lfloor \frac{7}{5} \right\rfloor = \frac{7-2}{5} = 1$.

Note that the congruence of $n^2 - na$ (mod $5$) loops every time $n$ increases by 5. Also, note that the congruence of $a$ (mod $5$) determines the set of congruences of $n^2 - na$ for each congruence of $n$ (mod $5$).

For example, if $a \equiv 1$ (mod $5$), the set of remainders is $(0, 2, 1, 2, 0)$ for $n \equiv 1,2,3,4,0$ (mod $5$). Let the sum of these elements be $s$. Note that for each “loop” of the numerator (mod $5$), each element of the set will be subtracted exactly once, meaning $s$ is subtracted once for each loop. The value of the numerator will loop $404$ times (mod $5$) throughout the sum, as $5 \cdot 404=2020$. Then

$U \approx \frac {\left( \frac {n(n+1)(2n+1)}{6} - \frac{(a)(n)(n+1)}{2} -404s \right)}{5}$

Where $n=2023$. Note that since $5 \cdot 404=2020$, this is an approximation for $U$ because the equation disregards the remainder (mod $5$) when $n=2021, 2022$, and $2023$ so we must subtract the first 3 terms of our set of congruences one extra time to get the exact value of $U$ (*). However, we will find that this is a negligible error when it comes to the inequality $-1000, so we can proceed with this approximation to solve for$a$.

Factoring our approximation gives $U \approx \frac {\frac{(n)(n+1)(2n+1 - 3a)}{6}-404s}{5}$

We set $a= \frac{(2n+1)}{3} = 1349$ to make $\frac{(n)(n+1)(2n+1 - 3a)}{6}=0$, accordingly minimizing $|U|$, yielding $U \approx \frac{-404s}{5}$

If $a$ increases or decreases by $1$, then $U$ changes by $\frac {(n)(n+1)}{2 \cdot 5} = \frac {2023 \cdot 2024}{10}$ which clearly breaks the inequality on $U$. Therefore $a=1349 \equiv 4$ (mod $5$) giving the set of remainders $(2,1,2,0,0)$, so $s=5$ and our approximation yields $U \approx -404$. However, we must subtract 2, 1, and 2 (*) giving us $U = - 404 - \frac{(2+1+2)}{5} = - 405$, giving an answer of $1349-405= \boxed{944}$

~Spencer Danese

Solution 4 (Calculus)

Consider the integral

$$\int_{0}^{2023} \dfrac{n^2-na}{5} \, dn.$$ We hope this will give a good enough appoximation of $U$ to find $a.$ However, this integral can be easily evaluated to be

$$\dfrac{1}{15}2023^3-\dfrac{a}{10}2023^2=2023^2\left(\dfrac{2023}{15}-\dfrac{a}{10}\right).$$ Because we want this to be as close to $0$ as possible, we find that $a$ should equal $1349.$ Then, evaluating the sum becomes trivial. Set

$$U'=\sum_{n=1}^{2023}\dfrac{n^2-1349n}{5}$$ and

$$U''=\sum_{n=1}^{2023}\{\dfrac{n^2-1349n}{5}\}.$$ Then $U=U'-U''.$ We can evaluate $U'$ to be $0$ and $U''$ to be $-405$ (using some basic number theory). Thus, $U=-405$ and the answer is

$$1349+(-405)=\boxed{944}.$$ ~BS2012

Video Solution

2023 AIME I #10

MathProblemSolvingSkills.com

Video Solution by Punxsutawney Phil

https://youtu.be/jQVbNJr0tX8

Video Solution by Steven Chen

https://youtu.be/fxsPmL6wuW4

Video Solution by Mathematical Dexterity

https://www.youtube.com/watch?v=49XIe2jx9zg