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Source: 2025 Taiwan TST Round 1 Independent Study 2-N

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CSJL

#1 h Mar 6, 2025, 8:21 AM

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Let $k$ be a positive integer. A positive integer $n$ is called a $k$ -good number if it satisfies
the following two conditions:

1. $n$ has exactly $2k$ digits in decimal representation (it cannot have leading zeros).

2. If the first $k$ digits and the last $k$ digits of $n$ are considered as two separate $k$ -digit
numbers (which may have leading zeros), the square of their sum is equal to $n$ .

For example, $2025$ is a $2$ -good number because
$(20 + 25)^2 = 2025.$ Find all $3$ -good numbers.

This post has been edited 3 times. Last edited by CSJL, Mar 6, 2025, 8:27 AM
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#3 h Mar 6, 2025, 12:43 PM • 1 Y

Y by shanelin-sigma

solution

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5532 posts

#4 h Mar 6, 2025, 5:36 PM • 1 Y

Y by imagien_bad

The only solutions are $494209$ and $998001$ , which can be checked to work. Now we show they are the only ones.

Let $a$ be the number formed by the first $k$ digits of $n$ and $y$ be the number formed by the last $k$ digits of $n$ . Note that $100 \le a < 1000$ .

Note that $1000a + k = (a+k)^2$ . Let $b = a + k$ . The equation becomes $999a + b = b^2 \leftrightarrow 999a = b(b-1)$
This means that $100 \le \frac{b(b-1)}{999}< 1000$ . Since $b$ is positive, this implies $300 < b < 1000$ .

Also, $999 \mid b(b-1)$ .

Case 1: $999$ divides one of $b, b - 1$ .
Since $1 < b < 1000$ , $999$ cannot divide $b - 1$ . Thus, $999\mid b$ , and since $b < 2 \cdot 999$ , $b = 999$ . In this case, we get $a = \frac{999 \cdot 998}{999} = 998$ and $y = b - a = 1$ , so the number is $998001$ , as desired.

Case 2: $999$ divides neither of $b, b - 1$ .
Then since $999 = 3^3 \cdot 37$ and $\gcd(b-1, b) =1$ , $27$ divides one of $b, b - 1$ and $37$ divides the other.

If $27 \mid b - 1$ and $37 \mid b$ , then $b \equiv 1 \pmod{27} \text{ and } b \equiv 0 \pmod{37}$ . CRT shows that $b \equiv 703 \pmod{999}$ , so (since $b < 1000$ ), $b = 703 \implies a = \frac{703 \cdot 702}{999} = 494 \implies y = 703 - 494 = 209,$ meaning the number is $494209$ , as desired.

If $27 \mid b$ and $37 \mid b - 1$ , then $b \equiv 0\pmod{27}$ and $b \equiv 1\pmod{37}$ . CRT shows that $b \equiv 297 \pmod{999}$ . Since $b < 1000$ , $b = 297$ , which is absurd since $b > 300$ .

Thus, the only solutions are the ones described in the beginning.

This post has been edited 2 times. Last edited by megarnie, Mar 6, 2025, 5:39 PM

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#5 h Mar 6, 2025, 8:34 PM • 1 Y

Y by megarnie

Essentially HMMT February 2025 Guts #21.

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Korean_fish_Kaohsiung

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Korean_fish_Kaohsiung

#6 h Mar 17, 2025, 7:13 AM

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According to A238237 on OEIS
done

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