Finding positive integers with good divisors

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Finding positive integers with good divisors

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Source: Belarusian National Olympiad 2025

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#1 h Mar 28, 2025, 8:32 PM • 1 Y

Y by truongphatt2668

For every positive integer $n$ write all its divisors in increasing order: $1=d_1<d_2<\ldots<d_k=n$ .
Find all $n$ such that $2025 \cdot n=d_{20} \cdot d_{25}$ .

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#3 h Mar 28, 2025, 11:59 PM

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$k \geq 25$ and
$d_{20}d_{k-19}=n=d_{25}d_{k-24}$
$d_{25}=2025d_{k-19}$ and $d_{20}=2025d_{k-24}$ so $n=2025d_{k-24}d_{k-19}$

As all numbers in form $3^a 5^b d_{k-19}|d_{25}|n$ for $a \leq 4,b \leq 2$ then there are at least $15$ such divisors $\leq d_{25}$ also $1=d_1<d_2<...<d_{k-19}<d_{25}$ so $15+(k-20) \leq 25 \to k \leq 30$

So $n=3^a 5^b$ with $25 \leq (a+1)(b+1) \leq 30$ and $a \geq 4, b\geq 2$ or
$n=3^4 5^2 *p$ for some $p \neq 3,5$

Case 1: $n=3^a 5^b$ then $k=(a+1)(b+1)=25,27,28$ or $30$
$d_1=1,d_2=3,d_3=5,d_4=9,d_5=15,d_6=25,d_7=27,d_8=45,d_9=75,d_{10}=81$

$k=25 \to n=3^4*5^4=2025*d_1*d_6$ so $n=3^45^4$ is solution
$k=27 \to n=3^8 5^2 \neq 2025 d_3d_8$
$k=28 \to n=3^6 5^3 \neq 2025 d_4d_9$
$k=30 \to n=3^4 5^5 \neq 2025 d_6d_{11}$ or $n=3^55^4 \neq 2025d_6d_{11}$

Case 2: $n=3^4 5^2*p$ for some $p \neq 3,5$ and $k=30$
but $n \neq 2025 d_6 d_{11}$

So only $n=3^45^4$ is answer

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