74th Romanian Mathematical Olympiad

Question

Consider a positive integer n and the set A_n = 1, 3, 5, , 2n - 1. For each pair (a, b), where a, b A_n we construct the concatenated number m = ab, obtained by joining the numbers a and b. For instance, for 19, 37 A_30, the concatenated number is m = 1937. a) What is the smallest number n N^* for which we get at least a perfect square? b) Find the largest perfect square that can be obtained for n = 50.

Accepted Answer

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

9 9^{2} = 9801

,

9 7^{2} = 9409

,

9 5^{2} = 9025

,

9 3^{2} = 8649

,

9 1^{2} = 8281

. They are not acceptable, as the first two digits form an even number.

The number

8 9^{2} = 7921

is obtained joining

79

and

21

, where

79, 21 \in A_{50}

. In conclusion, the largest perfect square is

7921

.