2. There are 100 coins of the same shape and size but with different weights. Using a balance scale without weights, find the heaviest and lightest coin. The minimum number of weighings required is $\qquad$ times.
2. 148.

First, divide the coins into 50 pairs, and use the balance to weigh 50 times, each time determining the heavier and lighter coin in each pair. Place all the heavier coins in one pile (called the "heavy" pile), and the lighter coins in another pile (called the "light" pile). The heaviest coin must be in the first pile, and the lightest coin must be in the second pile. In the heavy pile or the light pile, each weighing can eliminate one coin. With 50 coins in each pile, 49 weighings are enough.
Therefore, the minimum number of weighings is $50+49+49=148$.