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, 2022

Algebra Difficulty 7.5 National olympiad, round 2 Prove it Balkan Mathematical Olympiad

For every natural number $x$ , let $P(x)$ be the product of the digits of the number $x$ . Is there a natural number $n$ such that the numbers $P(n)$ and $P(n^2)$ are non-zero squares of natural numbers, where the number of digits of the number $n$ is equal to

a. $2021$

b. $2022$

Solution

The answers are affirmative in both cases.

a.
Take $n = \overbrace{33\cdots3}^{2019}68$ . Then $P(n) = (4 \cdot 3^{1010})^2$ . Also,
$\begin{aligned} n^2 &= \left( \frac{10^{2021} - 1}{3} + 35 \right)^2 = \frac{(10^{2021} + 104)^2}{9} \\ &= \frac{10^{4042} + 208 \cdot 10^{2021} + 10816}{9} \\ &= \frac{10^{4042} - 10^{2021}}{9} + 209 \cdot \frac{10^{2021} - 1}{9} + 1225 \\ &= \underbrace{1\cdots1}_{2021} \underbrace{0\cdots0}_{2021} + \underbrace{2\cdots2}_{2021} \underbrace{00}_{2021} + \underbrace{9\cdots9}_{2021} + 1225 \\ &= \underbrace{1\cdots1}_{2019} \underbrace{332\cdots200}_{2019} + 10^{2021} + 1224 \\ &= \underbrace{1\cdots1}_{2019} \underbrace{342\cdots23424}_{2017}. \end{aligned}$
$\text{Thus } P(n^2) = (3 \cdot 2^{2012})^2.$

b.
Take $n = \overbrace{1133\cdots3}^{2020}$ . Then $P(n) = (3^{1010})^2$ . Also,
$\begin{aligned} n^2 &= \frac{(34 \cdot 10^{2020} - 1)^2}{9} = \frac{1156 \cdot 10^{4040} - 68 \cdot 10^{2020} + 1}{9} \\ &= 128 \cdot 10^{4040} + \frac{4(10^{4040} - 10^{2020})}{9} - 7 \cdot 10^{2020} - \frac{10^{2020} - 1}{9} \\ &= \overbrace{1280\cdots0}^{4040} + \overbrace{4\cdots40\cdots0}^{2020} - 7 \cdot 10^{2020} - \underbrace{1\cdots1}_{2020} \\ &= \overbrace{1284\cdots4368\cdots89}^{2018\ 2019}. \end{aligned}$
$\text{Thus } P(n^2) = (9 \cdot 2^{5049})^2.$

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.