With m two point shots and n five point shots, you score 2m+ 5n points. You can get X points if there are integers such that 2m+ 5n= X.

2 divides into 5 twice with remainder 1. That is 5- 2(2)= 1 (bet you knew that!) For X a positive integer, multiply both sides of the equation by X: (X)(5)- (2X)2= (X)(5)+ (-2X)(2)= X. So one solution is m= X, n= -2X or "X five point shots and -2X two point shots give a total of X points". Of course, that is NOT a valid solution because you can't have "-2X" shots for X a positive number.

However, m= X- 2k, n= -2X+ 5k, for any integer k, also give 5(X- 2k)+ 2(-2X+ 5k)= 5X- 10k - 4X+ 10k= X since the two 10k parts cancel.

Now, we want to have so and so . That is, k must be a an integerbetween2X/5 and X/2.

Certainly there will be such an integer if the diference between those numbers is greater than or equal to 1: or . For X= 10, for example, X/2= 5 and 2X/5= 4 so 10 points can be got with 10- 2(4)= 2 5 point shots and 0 2 points shots or with -2(10)+ 5(5)= 5 2 point shots. For any X larger than or equal to 10, we can make X points with 2 and 5 point shots. For X< 10, we have to look at the specific numbers: 9= 5+ 2(2), 8= 4(2), 7= 5+ 2, 6= 3(2), 5= 1(5), 4=2(2) so that, 1 and 3 are the only "unattainable" values.

Now, what about 3 and 5 point shots? (I think we just recently had a problem about 3 and 5 cent stamps!)

We can make X shots if there exist integers m and n such that 3m+ 5n= X. 3 divides into 5 once with remainder 2 and 2 divides into 3 once with remainder 1: 3- 2= 1 and then 3- (5- 3)= 2(3)- 5= 1. Multiplying by X, (2X)(3)- X(5)= (2X)(3)+ (-X)(5)= X. Again, -X is not a valid answer, but, again, m= 2X- 5k, n= -X+ 3k give 3(2X- 5k)+ 5(-X+ 3k)= 6X- 15k- 5X+ 15k= X.

To have non-negative solutions we must have so and so . That is, k must be an integer between 2X/5 and X/3. Such an integer will certainly exist if the distance between X/3 and 2X/5 is greater than or equal to 1: or .

Of course, X= 15= 3(5) or 5(3) and now we know any number larger than 15 can be written as a combination of 3 and 5. What about numbers less than 15? The difference between X/3 and 2X/5 will be less than 1 but there still might be an integer between them. We need to look at the numbers individually. 14= 5+ 9= 1(5)+ 3(3). 13= 10+ 3= 2(5)+ 3. 12= 4(3). 11= 5+ 2(3). 10= 2(5). 9= 3(3). 8= 5+ 3. 7 is not a multiple of 5 or 3 and 7= 5+ 2 so 7cannotbe written as a combination of 3 and 5. 7 points cannot be scored with 3 and 5 point shots.