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Source: 2025 Euler Olympiad, Round 1

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21 posts

#1 h Mar 31, 2025, 6:19 PM

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Find the minimum value of $m + n$ , where $m$ and $n$ are positive integers satisfying:

$2023 \vert m + 2025n$
$2025 \vert m + 2023n$

Proposed by Prudencio Guerrero Fernández

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MR.1

#2 h Mar 31, 2025, 7:20 PM

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what a great problem

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#4 h Mar 31, 2025, 7:55 PM

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solution hopefully

This post has been edited 3 times. Last edited by Soupboy0, Mar 31, 2025, 10:44 PM

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KAME06

#6 h Mar 31, 2025, 10:33 PM

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Soupboy0 wrote:

solution hopefully

disclaimer: if this is wrong can someone please tell me why

Yes, it is. You must prove that there aren't $m, n$ such $m+n$ is less than $5058$

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KAME06

#7 h Mar 31, 2025, 11:17 PM

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Notice that the statement is equivalent to $2023 \mid m+2n$ and $2025 \mid m-2n$ .
Also, $(m+2n)-(m+2n) \equiv 2023a-2025b \equiv -a-b \equiv 0 (\text{ mod }4) \Rightarrow a+b \equiv 0 (\text{ mod }4)$
-If $2023=m+2n$ , notice that $-2023=-(m+2n) < m-2n=2025b < m+2n=2023$ , so the unique possibility is $m-2n=0 \Rightarrow m=2n$ , but $2023=2n+2n=4n$ . Contradiction because $4$ doesn't divide $2023$ .
-If $2023(2)=4046=m+2n$ , notice that $-4046=-(m+2n) < m-2n =2025b < m+2n=4046$ , so $b=-1, 0, 1$ , but $a+b \equiv 2+b \equiv 0 (\text{ mod }4)$ . Contradiction.
-If $2023(3)=6069=m+2n$ , notice that $-6069=-(m+2n) < m-2n =2025b < m+2n=6069$ . We need $b \equiv 1 (\text{ mod }4)$ , so $b=1 \Rightarrow m-2n=2025$ . Solving the system of equations: $(m, n)=4047, 1011 \Rightarrow m+n=5058$ .
-If $2023(4)=8092=m+2n$ , notice that $-2025(4) < -8092=-(m+2n) < m-2n =2025b < m+2n=8092<2025(4)$ . We need $b \equiv 0 (\text{ mod }4)$ , so $b=0 \Rightarrow m=2n \Rightarrow 8092=4n \Rightarrow 2023=n$ so $m+n=4046+2023=6089$ .
-If $2023(5) \le m+2n$ , as $n<m+n$ , then $2023(5) <m+n + (m+n) =2(m+n) \Rightarrow 5058 \le m+n$ .
We conclude that the minimum of $m+n$ is $5058$ .

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#8 h Apr 1, 2025, 12:28 AM • 1 Y

Y by NicoN9

The answer is $\boxed{5058}$ , achieved by $m = 4047, n = 1011$ . Now assume there was a smaller possible value for $m + n$ .

It's clearly equivalent to $2023 \mid m + 2n$ and $2025 \mid m - 2n$ . Let $2023x = m+2n, 2025y = m - 2n$ . Clearly, $2023x \equiv 2025 y \pmod 4$ , so $y \equiv 3x \pmod 4$ . Additionally, note that $2023x + 2025 y = 2m < 2(m+n) < 10116$ . This gives us the bounds $x < 6$ and $y < 5$ .

Now we have $(x,y) \in \{ (1,3), (2,2), (3,1), (4,4), (5, 3)\}$ . The first two possibilities clearly fail as they give $2023x < 2025y$ and the last two clearly fail as they give $2023x + 2025y \ge 10116$ , so we only need to check the third one. However, this yields the solution $(m,n) = (4047, 1011)$ , so we have $m + n \ge 5058$ always.

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