Question I don't get

[MathStudent1999]

11. In Martian Football you get 7 points for a touchdown and 5 points for a field goal. These are the only way you can score. What is the largest score that is logically impossible to achieve in Martian Football?


First before we get to the largest impossible score, wouldn't there be a infinite amount of unachievable score, because you can only score 7 or 5 points at a time. So 1,2,3,4,6,8,9,10,11,13,16... and so on would be all unachievable scores. Can someone help me on this question? Would the answer be ∞?

 

[tkhunny]

"logically impossible" ??? Is there some other kind of impossible?

This is a wonderful Exploration leads to conjucture sort of thing.

If you add 5s, you get values ending in 0 and 5
If you add 7s, you get values ending in 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
If you have some number of 7's and you swap them for some number of 5's, you can adjust your score. I explain:

10*7 = 70 Can we get 71. By subtracting 7 7's and addin 10 5s, we get 71.
This is a little trickier with 69.
14*5 - 70. By subtracting 10 5s and adding 7 7s, we get 69.

You'll have to keep in mind that there may be multiple solutions to a particular value.

HOWEVER, notice how it too 10 5s and 7 7s to pull that off. This suggests that the effort might be unsuccessful with values under 49.

Explore! Discover!

 

[TchrWill]

Martian Football you get 7 points for a touchdown and 5 points for a field goal. These are the only way you can score. What is the largest score that is logically impossible to achieve in Martian Football?

Here is one solution path with different numbers allowing you to work up your particular version.

There is this weird football league who plays the game a little differently. Here is their scoring system: Each feild goal counts for 5 points. Each touchdown counts for 3 points. The only way they can score points is by getting field goals and touchdowns or a combonation of both. One of the players noticed that not every score is possible to make. For example a score of 1 isn't possiable. This player also knows the highest imposible score.

Figure out what the highest impossible score is and prove it and make up another scoring system and see whether there are impossible scores for those too and find a rule to help to know if there are always impossible scores.


So far I have determined that the higest impossible score is 7 points. But I do not know how to prove it. Can you help me? >>

If the score values of m and n have a common factor, there will always be an infinite number of values that cannot be made. Thus, m and n need be relatively prime.

Consider your score values of m = 3 and n = 5. Create a table starting with 1, writing consecutive numbers in rows with n numbers to a row, stopping with m rows (m < n). All obtainable numbers are now marked off with a / symbol. They consist of 1) the multiples of n (the last column), 2) the remaining (n - 1) multiples of m (3, 6, 9, and 12), and 3) the numbers directly below each marked multiple of m, obtainable by adding to the multiple of m an appropriate multiple of n. We end up with the following, the / following a number meaning it has been crossed out.

.1......2.....3/......4........5/
.6/.....7.....8/......9/.....10/
11/..12/...13/....14/.....15/

for a total of 4 unattainable whole number scores.

If m and n have no common factor, the table contains marks in every column; then all numbers larger than those appearing in the table are obtainable, and the answer to the problem is the quantity of unmarked numbers.

A derived expression for the number of unattainable numbers is with N = (n - 1)(m - 1)/2. For your problem, N = (5 - 1)(3 - 1)/2 = 4(2)/2 = 4.

What if the scoring possibilities were m = 3 and n = 10:

.1.....2.....3/.....4.....5....6/.....7.....8....9/....10/
11...12/..13/...14...15/..16/...17...18/..19/...20/
21/..22/..23...24/...25/..26/..27/. .28/..29/...30/

for a total of 9 unattainable whole number values, verifiable by N = (3 - 1)(10 - 1)/2 = 2x9/2 = 9.

Examining these results, the highest unattainable number is expressable by H = mn - (m + n) = n(m - 1) - m which produces 7 as the highest unattainable number in your problem and 17 in the other example.

I ran through another example with m = 5 and n = 11 resulting in 11 columns and 5 rows. After crossing out the numbers according to the rules, I was left with 20 uncrossed numbers. Using the magic expression, N = (5 - 1)(11 - 1)/2 = 4x10/2 = 20. The highest unattainable number remaining was 39 and H = (11 - 1)5 - 11 = 50 - 11 = 39.

 

[HallsofIvy]

You are essentially dealing with a linear diophantine equation- a linear equation in which the unknowns must be integers.

If you can only score 7 points for a touchdown and 5 points for a field goal, then, by scoring m touchdowns and n field goals, you can score 7m+5n points.

We start by seeking m and n such that 7m+ 5n= 1, using the "Euclidean division algorithm":
5 divides into 7 once, with remainder 2: 2= 7- 5= 7(1)- 5(1).
2 divides into 5 twice with remainder 1: 1= 5- 2(2).

Replace the "2" in the second equation by "7(1)- 5(-1)" so that you get 1= 5- (7(1)- 5(-1))(2)= 7(-2)+ 5- 5(-2)= 7(-2)+ 5(3).

So one solution to "7m+ 5n= 1" is m= -2, n= 3. And, multiplying both sides by x, 7(-2x)+ 5(3x)= x so m= -2x, n= 3x satisfies 7m+ 5n= x. Of course, the "number of touchdowns" cannot be negative! But note that, for any integer k, m= -2x+ 5k, n= 3x- 7k also sastisfies that equastion: 7(-2x+ 5k)+ 5(3x- 7k)= -14x+ 35k+ 15x- 35k= x because the terms involving k cancel.

So for any x, m= -2x+ 5k, n= 3x- 7k will satisify the conditions as long as both -2x+ 5k>= 0 and 3x- 7k>= 0. The first reduces to (3/7)x\le k<= (2/5)x. That means we must have at least one integer between (3/7)x and (2/5)x. What is the largest value of x for which that is not true? Note that (2/5)x- (3/7)x= ((14- 10)/35)x= (4/35)x>= 1 for x>= 35/4< 9 so you only need to check positive integers less than 9.

 

[MathStudent1999]

You are essentially dealing with a linear diophantine equation- a linear equation in which the unknowns must be integers.

If you can only score 7 points for a touchdown and 5 points for a field goal, then, by scoring m touchdowns and n field goals, you can score 7m+5n points.

We start by seeking m and n such that 7m+ 5n= 1, using the "Euclidean division algorithm":
5 divides into 7 once, with remainder 2: 2= 7- 5= 7(1)- 5(1).
2 divides into 5 twice with remainder 1: 1= 5- 2(2).

Replace the "2" in the second equation by "7(1)- 5(-1)" so that you get 1= 5- (7(1)- 5(-1))(2)= 7(-2)+ 5- 5(-2)= 7(-2)+ 5(3).

So one solution to "7m+ 5n= 1" is m= -2, n= 3. And, multiplying both sides by x, 7(-2x)+ 5(3x)= x so m= -2x, n= 3x satisfies 7m+ 5n= x. Of course, the "number of touchdowns" cannot be negative! But note that, for any integer k, m= -2x+ 5k, n= 3x- 7k also sastisfies that equastion: 7(-2x+ 5k)+ 5(3x- 7k)= -14x+ 35k+ 15x- 35k= x because the terms involving k cancel.

So for any x, m= -2x+ 5k, n= 3x- 7k will satisify the conditions as long as both -2x+ 5k>= 0 and 3x- 7k>= 0. The first reduces to (3/7)x\le k<= (2/5)x. That means we must have at least one integer between (3/7)x and (2/5)x. What is the largest value of x for which that is not true? Note that (2/5)x- (3/7)x= ((14- 10)/35)x= (4/35)x>= 1 for x>= 35/4< 9 so you only need to check positive integers less than 9.

Thanks, I now get this question and know how to solve this question!