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Problem 658

AIME late

Combinatorics Difficulty 5.2 Find the answer

3. Let's call a natural number $n$ squareable if the numbers from 1 to $n$ can be arranged in such an order that each member of the sequence, when added to its position, results in a perfect square. For example, the number 5 is squareable, as the numbers can be arranged as: 32154, in which $3+1=2+2=1+3=4$ and $5+4=4+5=9$ . Determine which of the numbers $7, 9, 11, 15$ are squareable.

Your answer

Official solution

# Answer: 9 and 15.

Solution. The number 7 cannot be squareable, since both numbers 1 and 6 must be in the third position, which is impossible.

The number 9 is squareable, as the numbers from 1 to 9 can be arranged in the following order: $8,2,6$ , $5,4,3,9,1,7$ , thus satisfying the required condition.

The number 11 is not squareable, since both numbers 11 and 4 must be in the fifth position.

The number 15 is squareable, as the numbers from 1 to 15 can be arranged in descending order, and then $1+15=2+14=\ldots=15+1=16$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.