Let $N=\overline{abcd}$ be a positive integer with four digits. We name plátano power to the smallest positive integer $p(N)=\overline{\alpha_1\alpha_2\ldots\alpha_k}$ that can be inserted between the numbers $\overline{ab}$ and $\overline{cd}$ in such a way the new number $\overline{ab\alpha_1\alpha_2\ldots\alpha_kcd}$ is divisible by $N$. Determine the value of $p(2025)$.
Problem
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Tags: number theory
RudraRockstar
18.06.2019 21:08
Your answer is 45
Solar Plexsus
19.06.2019 10:04
Let $n$ be the numbers of digits of $p(2025)=a$. Hence we have $(1) \;\; 10^{n-1} \leq a < 10^n$. According to the given information $(2) \;\; 2025 \mid 20 \cdot 10^{n+2} + 100a + 25$. Set $M = 20 \cdot 10^{n+2} + 100a + 25$. Clearly $25 | M$ by (2). Consequently, since $2025 = 25 \cdot 81$, $(3) \;\; 81 \mid 20 \cdot 10^{n+2} + 100a + 25$ by (2). Hence $9|M$, yielding $9|a$ by (3), i.e. $a=9b$, which inserted in (3) result in $(4) \;\; 81 \mid 20 \cdot 10^{n+2} + 900b + 25$. Moreover by (1) $(5) \;\; 10^{n-1} \leq 9b \leq 10^n$. If $n=1$, then $b=1$ by (5), yielding $M = 20 \cdot 10^3 + 900 \cdot 1 + 25 = 20925$, contradicting (4). Thus we have $n>1$. Assume $n=2$. According to (4) $81 | 200025 + 900b$, which means $81 | 9b - 45$, i.e. $(6) \;\; 9 \mid b - 5$. Consequently $p(2025)_{\min} = 9 \cdot 5 = 45$.
Bluesoul
26.04.2022 02:06
Very easy So $p(2025)$ must have 6 digits at least, we call it $20ab25$ Using division, $\overline{ab}-25=20$, $\overline {ab}=45$