In the football game on Saturday, the home team scored points by making touchdowns and extra points only (worth 7 points) and field goals (worth 3 points). Some home team scores are impossible, such as 5. What is the largest number of pints that the team could not have scored?
To determine the largest number of points that the home team could not have scored, we need to use a concept called the Frobenius coin problem. This problem deals with finding the largest number that cannot be represented by a combination of given coin denominations.
In this case, the coin denominations are touchdowns + extra points (7 points) and field goals (3 points).
To approach this problem, we need to calculate the largest possible score that the home team can achieve with these coin denominations. We can do this by finding the greatest common divisor (GCD) of the two coin denominations and subtracting it from their product.
In this case, the GCD of 7 and 3 is 1, so the largest impossible score the home team cannot achieve is:
7 * 3 - 7 - 3 = 21 - 7 - 3 = 11 points.
Therefore, the largest number of points that the home team could not have scored is 11.
To find the largest number of points that the home team could not have scored, we can use the concept of the "Frobenius coin problem" (also known as the "coin-exchange problem").
In this case, we have two possible coin values: 7 (touchdowns) and 3 (field goals). To determine the largest number of points that can't be scored, we'll need to find the largest number that cannot be expressed as a non-negative combination of these two values.
To solve the coin-exchange problem and find the largest number that can't be scored, we can use a formula:
largest_number_not_scored = (coin1 * coin2) - (coin1 + coin2)
In this case, with coin1 = 7 and coin2 = 3:
largest_number_not_scored = (7 * 3) - (7 + 3)
largest_number_not_scored = (21) - (10)
largest_number_not_scored = 11
Therefore, the largest number of points that the home team could not have scored is 11.