Competition problem — Number theory

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Number theory Difficulty 6.8 National olympiad Prove it Romania

Consider a positive integer $n$ and the set $A_n = \{1, 3, 5, \dots, 2n - 1\}$ . For each pair $(a, b)$ , where $a, b \in A_n$ we construct the concatenated number $m = \overline{ab}$ , obtained by joining the numbers $a$ and $b$ . For instance, for $19, 37 \in A_{30}$ , the concatenated number is $m = 1937$ .

a) What is the smallest number $n \in \mathbb{N}^*$ for which we get at least a perfect square?

b) Find the largest perfect square that can be obtained for $n = 50$ .

Solution

a) We cannot obtain a perfect square by concatenating two elements from the set $A_{10} = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\}$ .

The elements $1$ and $21$ from the set $A_{11} = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21\}$ yield the perfect square $121$ . Hence, the answer is $n = 11$ .

b) By concatenating two elements from $A_{50} = \{1, 3, 5, \dots, 97, 99\}$ we can obtain perfect squares with at most four digits.

The largest such perfect squares are $99^2 = 9801$ , $97^2 = 9409$ , $95^2 = 9025$ , $93^2 = 8649$ , $91^2 = 8281$ . They are not acceptable, as the first two digits form an even number.

The number $89^2 = 7921$ is obtained joining $79$ and $21$ , where $79, 21 \in A_{50}$ . In conclusion, the largest perfect square is $7921$ .

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