A heptagon with all interior angles less than 180 degrees has at least ( ) obtuse angles.
(A) 6
(B) 5
(C) 4
(D) 3
Solution: The following solution tries to use elementary school level thinking as much as possible:

The sum of the interior angles of a heptagon is $180 \times(7-2)=900^{\circ}$
Try to have as many angles as possible be $90^{\circ}$,
1) If 6 angles are $90^{\circ}$, then the last 1 angle is greater than $180^{\circ}$
2) If 5 angles are $90^{\circ}$, then the sum of the other 2 angles is $900-5 \times 90=450^{\circ}$, $450^{\circ}>2 \times 180^{\circ}$, there must be an angle greater than $180^{\circ}$
3) If 4 angles are $90^{\circ}$, then the sum of the other 3 angles is $900-4 \times 90=540^{\circ}$, $540^{\circ}=3 \times 180^{\circ}$, there must be an angle greater than or equal to $180^{\circ}$
4) If 3 angles are $90^{\circ}$, then the sum of the other 4 angles is $900-3 \times 90=630^{\circ}$, $640<4 \times 180^{\circ}$, so at least 4 angles are obtuse