User:Cxsmi

About Me

Hi! I'm just another guy who happens to enjoy math. I often pop onto the AOPS wiki and look for problems to solve, and I sometimes even write solutions for them! I've starred ⭐ a few of my favorite solutions below; please feel free to take a look at any of them. Thanks for visiting my user page, and enjoy your stay!

Solutions

AIME

AMC 8

AJHSME

AHSME

AMC 12

2021 AMC 12B Problem 12 Solution 6⭐

Significant Problems

Here are some problems that, to me, have been significant on my math journey. This section is mainly for myself, but please please feel free to look at the problems if you're interested.

Problems

I enjoy writing problems when I see concepts that interest me. I've written a few below; please feel free to solve them! Also, feel free to add solutions or change any mistakes you see. Note: Unless otherwise stated, all questions are in the AIME format -- that is, the answer is an integer between 1 and 999.

Problem 1

Find the least positive integer $n$ that satisfies the following. The notation $\lfloor{x}\rfloor$ represents the greatest integer less than or equal to $x$ .

$\frac{20^{23}}{n} + \frac{24^{23}}{n} = \lfloor{\frac{20^{23}}{n}+\frac{24^{23}}{n}}\rfloor \neq \lfloor{\frac{20^{23}}{n}}\rfloor + \lfloor{\frac{24^{23}}{n}}\rfloor$

Problem 2

Bob has received a test. He must match a list of $2024$ words to a list of $2024$ definitions such that each word matches to exactly one definition and vice versa. He does not know any of these words, so he will guess randomly. How many correct matches can he expect to make?

Problem 3

A sequence is defined recursively on the positive integers as $f(1) = f(2) = f(3) = ... = f(2024) = 1$ and

$f(n) = \sum_{k=1} ^{2024} f(n - k)$

for $n \geq 2025$ . The value of $f(4049)$ can be written in the form $a(2)^b+1$ for positive integers $a$ and $b$ such that $b$ is maximized. What is the remainder when $a + b$ is divided by $1000$ ?

Solutions

These are solutions for the problems above. $\textbf{Scroll down at your own risk!}$

Problem 1 Solutions

Solution 1

We split the condition into two separate conditions, as listed below.

$\frac{20^{23}}{n} + \frac{24^{23}}{n} = \lfloor{\frac{20^{23}}{n}+\frac{24^{23}}{n}}\rfloor$

$\frac{20^{23}}{n} + \frac{24^{23}}{n} \neq \lfloor{\frac{20^{23}}{n}}\rfloor + \lfloor{\frac{24^{23}}{n}}\rfloor$

Rearranging the conditions, we find that

$\frac{20^{23}+24^{23}}{n} - \lfloor \frac{20^{23}+24^{23}}{n}\rfloor = 0$

$(\frac{20^{23}}{n} - \lfloor \frac{20^{23}}{n} \rfloor) + (\frac{24^{23}}{n} - \lfloor \frac{24^{23}}{n} \rfloor) \neq 0$

Recalling that $x - \lfloor {x} \rfloor = frac(x)$ where $frac(x)$ represents the fractional part of $x$ , we rewrite once more.

$frac(\frac{20^{23}+24^{23}}{n}) = 0$ $\textbf{(1)}$

$frac(\frac{20^{23}}{n}) + frac(\frac{24^{23}}{n}) \neq 0$ $\textbf{(2)}$

We now gain some valuable insight. From $\textbf{(1)}$ , we find that $n$ must divide $20^{23} + 24^{23}$ . From $\textbf{(2)}$ , we find that $n$ cannot divide both $20^{23}$ and $24^{23}$ . It is impossible for $n$ to divide only $1$ of $20^{23}$ and $24^{23}$ , as this would make $\textbf{(1)}$ false. It must be that $n$ divides neither $20^{23}$ nor $24^{23}$ . For both this and $\textbf{(1)}$ to be true simultaneously, we must have that if $20^{23} \equiv a \bmod n$ , then $24^{23} \equiv -a \bmod n$ . By inspection, this occurs when $n = 22$ .We now test the factors of $22$ to see if we can find a smaller value. As both $20^{23}$ and $24^{23}$ are congruent to $0$ mod $2$ , $n = 2$ is not a valid solution. However, with $n = 11$ , $20^{23} \equiv (-2)^{23} \bmod 11$ , while $24^{23} \equiv 2^{23} \bmod 11$ . Clearly, $-(-2)^{23} = 2^{23}$ , so our final answer is $\boxed{011}$ .