Solve math problems?

Answer: 3 times. It only takes three times to weigh, and two counterfeit coins can be found.

For the convenience of description, we number these seven coins as A, B, C, D, E, F and G respectively.

Situation ①: Balance of the first weighing balance

First weighing: Take a group of A, B and C and a group of D, E and F and put them on both sides of the scale. If the balance is balanced, G must be real money, and A, B, C, D, E and F each have a counterfeit money.

Second weighing: put A and B on both sides of the scale. If they are balanced, C is counterfeit money; If it is unbalanced, the worst is counterfeit money (the first counterfeit money is found).

The third weighing: put D and E on both sides of the scale. If they are balanced, F is counterfeit money; If it is unbalanced, the worst is counterfeit money (the second counterfeit money is found).

Situation ②: The first weighing balance is unbalanced.

First weighing: take a group A, B and C and a group D, E and F, and put them on both sides of the scale respectively. If the balance is unbalanced, the heavier party will have one or two counterfeit coins.

Second weighing: If the heavy side is group A, B and C, put A and B on both sides of the scale respectively. If they are balanced, then A and B are both real money or counterfeit money. If it is unbalanced, it is counterfeit money.

The third weighing:

Situation ① Balance A and B for the second time. Take A and C and put them on both sides of Libra. If A is heavy, A and B are counterfeit. If C is heavy, then C and G are counterfeit money (two counterfeit money are found).

Case (2) The second weighing A and B are unbalanced, and the heavier one is called C. If it is balanced, it is all counterfeit money. If it is not balanced, the heavier sum G is counterfeit (two counterfeit coins are found).

In either case, only three times of weighing are needed to ensure that two counterfeit coins are found.