Impossible Scores

By the Chicken McNugget theorem , for two relatively prime positive integers $m$ and $n$ , the largest value that cannot be formed by a combination of the two is given by $mn - m - n$ . In this case, $3$ and $8$ are relatively prime as $\text{gcd}(8,3) = 1$ . The maximum number that cannot be obtained by using these numbers is $24 - 8 - 3 = 13$ .

We can easily list these numbers till $13$ and there are precisely $7$ of them, obtained from the corollary that there must be $\dfrac{(m-1)(n-1)}{2}$ such numbers.

They are $1 + 2 + 4 + 5 + 7 + 10 + 13 = \boxed{42}$ .

We can apply a process here ...

Let us divide the numbers into 3 groups on the basis of what remainder it leaves when divided by 3.

$3k, 3k+1, 3k+2$

Now, if we get a certain number which falls in one of these groups, we can keep on adding 3s to that number and get all the numbers of that group.

For example, if we can have '4' (3k+1), we can have 7, 10, 13 , etc ...

So, basically, we are looking for the minimum number of each form that we can have

• $3k$ : The minimum number we can have is $3$ itself
• $3k+1$ : The minimum number achievable is $16 = 8 + 8$
• $3k+2$ : The minimum number achievable is $8$

Thus, the scores not achievable are:

• Those of the form $3k+1$ and less than $16$ , vi.z: $1, 4, 7, 10, 13$
• Those of the form $3k+2$ and less than $8$ , viz.: $2, 5$

Sum of these numbers: $1+4+7+10+13+2+5 = \boxed{42}$

PS: We could have tried dividing into groups on the basis of the remainders left by 8 as well. But then there will be 8 different forms. The answer would be same, but this seemed shorter

Given a number n, it is a score if: $n-8i \equiv 0 \mod{3}$ . If $n \equiv 1 \mod{3}$ , then we have that: $1-8i \equiv 0 \Rightarrow -8i \equiv 2 \Rightarrow -2i \equiv 2 \Rightarrow 2i \equiv 3-2 = 1 \mod{3}$ Thus, in this case, the minimum i equals 2. If $n \equiv 2 \mod{3}$ , then: $2 -8i \equiv 0 \Rightarrow -8i \equiv 1 \Rightarrow -2i \equiv 1 \Rightarrow 2i \equiv 3-1=2 \mod{3}$ Thus, in this case, the minimum i equals 1. Since $n+1>8i$ , we have that all the numbers were $n \equiv 1 \mod{3}$ that can't be expressed by a score are: 1, 4, 7, 10 and 13. And the numbers were $n \equiv 2 \mod{3}$ that can't be expressed by a score are 2 and 5. Thus the total is $1+2+4+5+7+10+13=\boxed{42}$

simply you can do this that there are numbers 1,2,4,5,7,10,13 after that all number can be do by summation of 8 and 3... so total is 1+2+4+5+7+10+13=42

lcm of 3 and 8 is 24.

multiples of 3 to 24 include 3,6,9,12,15,18,21, 24

multiples of 8 to 24 include 8.16.24

adding all possible cases we have 3,6,8,9,12,15,16,21,24.

by merging all possible combinations we have 3,6,8,9,11,12,14,15,16,17,18,19,20,21,22,23,24.

we see that the list includes a consecutive row of positive integers starting with 14. so we consider all numbers from 14 to 27 to check if any more basic number combinations (because any positive integer that is 28 or greater can be formed by adding integers in the set [14, 27]). There we find that 25,26 and 27 are possible solutions.

Therefore the only impossible cases are 1+ 2+ 4+ 5 +7 + 10+13 = 42.

The Frobenius number for 3 and 8 is (3*8)- (3+8)= 24-11= 13. That's the largest number that cant be achieved beyond which all numbers are possible. Rest is simple task of finding numbers below 13 that cant be reached.

When do we stop counting numbers?

We start listing the numbers which cannot be written in the form $3x+8y (x,y \in \mathbb{Z_+})$ , and find three consecutive numbers in the form $3x+8y$ . This is because if $n, n+1$ and $n+2$ are three such numbers, then by adding 3 repeatedly to those numbers, we can generate the rest of the set of natural numbers, which will be expressible in the form $3x+8y$ . Those three numbers here are 14, 15 and 16. And so our final number not in the form $3x+8y$ is 13.

For example, if $n=3x_1+8y_1$ , then $n+3=3(x_1+1)+8y_1$ ; if $n+1=3x_2+8y_2$ , then $n+4=3(x_2+1)+8y_2$ ; and if $n+2=3x_3+8y_3$ , then $n+5=3(x_3+1)+8y_3$ and so on.

The technique works if we try to find 8 consecutive numbers of the form $3x+8y$ , too, but why take the trouble.

Knowing that we stop at 13, we add the numbers not in the form $3x+8y$ , $1+2+4+5+7+10+13=\boxed{42}$ .

let z be any score we can get using 3 and 8..........then z can be written as 3x+8y=z......where x,y are the number of times we got 3 and 8 respectively...............we know that "for any three consecutive numbers there exists a 3 multiple among the numbers"...............................this can be further written as in {n,n-1,n-2} there exists a multiple of 3....................by using the same logic we can write {z,z-8,z-16} among these three numbers there also a exists a multiple of 3..............so the value of z cannot be greater than 16 because if it is greater than 16 we get 3x=z or z-8 or z-16...by substituting y=0 or 1 or 2 so among these three numbers one will definitely be a multiple of 3..and we can get the integer value of x....so any number greater than 16 can definitely be formed by using 3 and 8.....so by considering numbers less than 16 we have the numbers which cannot be formed by 3 and 8 are 1,2,4,5,7,10,13 and their sum leads to 42....

Chicken McNugget theorem :D

According to question given only following scores are not possible and rest all scores are possible by combination of 3 and 4. The scores are : 1, 2, 4, 5, 7, 10, 13

Therefore, their sum 1+2+4+5+7+10+13 = 42

I did this using trial and error, starting at number 1 and checking each number if it matches the criteria. By the time I reached # 14 each succeeding number would be possible combinations of 8 and 3. I solved this in less than 2 minutes. lol