- Three hockey teams, the Crabs, the Eels, and the Seals, played each other once with no shutouts ( every team scored at least 1 goal in each game). A team gets 10 points for a win, 5 point for a tie and one point for each goal scored. If the Crabs received 20 points, the Eels 19 points and the Seals 5 points,what was the score of the game between the Crabs and the Eels?

- X is defined as one more than the number of digits in the integer x. For example , 1000 = 5 . If x has 900 digiits , find the value of x

5 to 4

To find the score of the game between the Crabs and the Eels, we must first determine the number of wins, ties, and goals for each team.

Let's assume that the number of wins for the Crabs is denoted by Cw, the number of ties by Ct, and the number of goals scored by the Crabs by Cg. Similarly, for the Eels, we will use Ew for wins, Et for ties, and Eg for goals.

We are given the following information:
- Crabs received 20 points.
- Eels received 19 points.
- No shutouts occurred, meaning each team scored at least one goal in each game.

Using this information, we can set up the following equations:

1. 10Cw + 5Ct + Cg = 20
2. 10Ew + 5Et + Eg = 19

Since the Crabs received 20 points, which is the highest among the three teams, we can assume that they won the most number of games.

Let's consider the possible scenarios:
- If the Crabs won all their games, Cw = 3.
- If the Crabs won two games and tied one, Cw = 2 and Ct = 1.
- If the Crabs won one game, tied one, and lost one, Cw = 1, Ct = 1, and Cg = 1.
- If the Crabs only tied all their games, Ct = 3 and Cg = 0.

Now let's look at the Eels' scenarios:
- If the Eels won all their games, Ew = 3.
- If the Eels won two games and tied one, Ew = 2 and Et = 1.
- If the Eels won one game, tied one, and lost one, Ew = 1, Et = 1, and Eg = 1.
- If the Eels only tied all their games, Et = 3 and Eg = 0.

Since the Crabs received more points than the Eels, let's start by considering the scenarios where the Crabs won more games.

1. If the Crabs won all three games (Cw = 3):
- From equation 1: 10(3) + 5(0) + Cg = 20
- Simplifying: 30 + Cg = 20
- Cg = -10 (which is not possible since goals cannot be negative)
- This scenario is not valid.

2. If the Crabs won two games and tied one (Cw = 2, Ct = 1):
- From equation 1: 10(2) + 5(1) + Cg = 20
- Simplifying: 20 + Cg = 20
- Cg = 0
- In this scenario, the Crabs did not score any goals.

3. If the Crabs won one game, tied one, and lost one (Cw = 1, Ct = 1, Cg = 1):
- From equation 1: 10(1) + 5(1) + 1 = 20
- Simplifying: 10 + 5 + 1 = 20
- This equation is not valid.

4. If the Crabs only tied all their games (Ct = 3):
- From equation 1: 5(3) + Cg = 20
- Simplifying: 15 + Cg = 20
- Cg = 5
- In this scenario, the Crabs scored 5 goals in their three games.

Now, let's plug in the values of Cg in equation 2 to find the value of Eg for the Eels:

For Cg = 0:
- 10Ew + 5Et + 0 = 19
- Simplifying: 10Ew + 5Et = 19

For Cg = 5:
- 10Ew + 5Et + 5 = 19
- Simplifying: 10Ew + 5Et = 14

Now, let's solve both the equations to find the possible scenarios for the Eels:

From equation 1: 10Ew + 5Et + Eg = 19

1. For Cg = 0, the equation becomes: 10Ew + 5Et = 19
- If we plug in Ew = 3 (maximum value), the equation becomes: 10(3) + 5Et = 19
- Simplifying: 30 + 5Et = 19
- This equation is invalid as it does not have a solution.

2. For Cg = 5, the equation becomes: 10Ew + 5Et = 14
- If we plug in Ew = 3 (maximum value), the equation becomes: 10(3) + 5Et = 14
- Simplifying: 30 + 5Et = 14
- This equation is invalid as it does not have a solution.

Since none of the scenarios with different values of Cg and Eg yielded valid equations, there is no valid solution that satisfies the given conditions.

Therefore, we cannot determine the score of the game between the Crabs and the Eels with the given information.