2023 AIME I Problems/Problem 10

Problem 10

There exists a unique positive integer $a$ for which the sum $U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor$ is an integer strictly between $-1000$ and $1000$ . For that value $a$ , find $a+U$ .

(Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$ .)

Solution (Bounds and Decimal Part Analysis)

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

First, we bound $U$ .

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

We establish a lower bound of $U$ . We have $\begin{align*} U & = \sum_{n=1}^{2023} \left( \frac{n^2 - na}{5} - \left\{ \frac{n^2 - na}{5} \right\} \right) \\ & = \sum_{n=1}^{2023} \frac{n^2 - na}{5} - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & = UB - \sum_{n=1}^{2023} \left\{ \frac{n^2 - na}{5} \right\} \\ & \geq UB - \sum_{n=1}^{2023} \mathbf 1 \left\{ \frac{n^2 - na}{5} \notin \Bbb Z \right\} . \end{align*}$

We notice that if $5 | n$ , then $\frac{n^2 - na}{5} \in \Bbb Z$ . Thus, $\begin{align*} U & \geq UB - \sum_{n=1}^{2023} \mathbf 1 \left\{ \frac{n^2 - na}{5} \notin \Bbb Z \right\} \\ & \geq UB - \sum_{n=1}^{2023} \mathbf 1 \left\{ 5 \nmid n \right\} \\ & = UB - \left( 2023 - \left\lfloor \frac{2023}{5} \right\rfloor \right) \\ & = UB - 1619 \\ & \triangleq LB . \end{align*}$

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

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

Therefor, $a + U = 1349 - 405 = \boxed{\textbf{(944) }}$ .

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

Video Solution by Punxsutawney Phil

https://youtu.be/jQVbNJr0tX8

Video Solution

https://youtu.be/fxsPmL6wuW4

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

2023 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

2023 AIME I Problems/Problem 10

Contents

Problem 10

Solution (Bounds and Decimal Part Analysis)

Video Solution by Punxsutawney Phil

Video Solution

See also