Is √p a non-integer? (1) p = r × 10 r, where r is a positive odd in

We can also ask "is \(√p\) an integer?" or "is \(p\) a perfect square?"

Statement 1:

All squares can be broken down to even powers of prime factors, or a product of squares.
Split p into squares, \(r * 10^r = r * 10 * 10^{r-1}\). \(10^{r-1}\) is a square since \(r - 1\) is even. Then the concern is whether the rest of the factors, \(r * 10\) can be a square. \(r\) needs to have a factor of 2 since 10 has a factor of 2, to complete the square. Thus \(r\) has to be even if we want to complete the square, then there is no odd \(r\) that can make \(r * 10\) and \(r * 10^r\) a square. Sufficient.

Statement2:

Again focus on breaking p into smaller squares. 9 is already a square, we are concerned about \(10^s\) now. We only know \(s\) is an integer, any even \(s\) would make \(10^s\) a square but any odd \(s\) wouldn't. So this is insufficient.

Ans: A