o3-mini

Assume for contradiction that sqrt(2) is rational, meaning sqrt(2) = p/q where p and q are integers with no common factors. Squaring both sides gives 2 = p²/q², so p² = 2q². This means p² is even, which implies p is even—write p = 2k. Substituting gives 4k² = 2q², so q² = 2k², meaning q is also even. But if both p and q are even, they share a common factor of 2, contradicting our assumption. Therefore sqrt(2) must be irrational.

Pick a fruit from the box labeled 'Mixed'. Since all labels are wrong, this box contains only one type of fruit—either all apples or all oranges. Say you draw an apple: the 'Mixed' box is actually the Apples box. Now consider the remaining two boxes labeled 'Apples' and 'Oranges'. The box labeled 'Apples' cannot be apples (wrong label) and cannot be mixed (we found that), so it must be Oranges. The box labeled 'Oranges' is therefore the Mixed box. One draw is all you need.