M28-50

Official Solution:


If \(0 \lt x \lt y\) and \(x\) and \(y\) are consecutive perfect squares, what is the remainder when \(y\) is divided by \(x\)?

Observe that, since \(x\) and \(y\) are consecutive perfect squares, it follows that \(\sqrt{x}\) and \(\sqrt{y}\) are consecutive integers. For example:

4 and 9 are consecutive perfect squares, and their square roots, \(\sqrt{4}=2\) and \(\sqrt{9}=3\), are consecutive integers.

16 and 25 are consecutive perfect squares, and their square roots, \(\sqrt{16}=4\) and \(\sqrt{25}=5\), are consecutive integers.

(1) Both \(x\) and \(y\) have 3 positive factors.

Since only the squares of primes have three factors (for example, the factors of \(p^2\), where \(p\) is a prime, are 1, \(p\), and \(p^2\)), this statement implies that \(x=(prime_1)^2\) and \(y=(prime_2)^2\). Based on the previous observation that the square roots of consecutive perfect squares are consecutive integers, \(\sqrt{x}=prime_1\) and \(\sqrt{y}=prime_2\) must be consecutive integers. The only pair of consecutive integers that are both primes is 2 and 3. Therefore, \(x=(prime_1)^2=4\) and \(y=(prime_2)^2=9\). When dividing 9 by 4, the remainder is 1. Sufficient.

(2) Both \(\sqrt{x}\) and \(\sqrt{y}\) are prime numbers.

Using the same logic as above, since the only pair of consecutive integers that are both primes is 2 and 3, we can conclude that \(\sqrt{x}=2\) and \(\sqrt{y}=3\). Therefore, \(x=4\) and \(y=9\). When dividing 9 by 4, the remainder is 1. Sufficient.


Answer: D