## Problem II - 5

The natural numbers 22, 23, and 24 have the following property: the exponents of the prime factors in their factorization are all odd:

$$
22=2^{1} \cdot 11^{1} ; \quad 23=23^{1} ; \quad 24=2^{3} \cdot 3^{1}
$$

What is the largest number of consecutive natural numbers that can have this property? Justify your answer.
## Solution

We will prove that it is impossible to find 8 consecutive numbers with this property. Suppose, for the sake of contradiction, that such 8 consecutive numbers exist. One of them, which we will call $n$, is divisible by 8.

Among the 8 numbers, there must be either $n+4$ or $n-4$. Both are divisible by 4, but not by 8, and this is a contradiction because the exponent of 2 in this number must be even.

The example of the numbers

$$
29,30,31,32,33,34,35
$$

finally demonstrates that the maximum sought is 7, and that it is indeed achieved.