Rugby Impossibilities

We first see that it is clear that we can not achieve a score of $1$ and $2$ . Then, we see that we can not get a score of $4$ . Now, we will prove that we can get every other score:

First, we see that in each scoring play, we can get either $3$ , $5$ , or $7$ points.

Now, let $n$ be any number of points.

If $n \equiv 0 \pmod{3}$ we can get the score by scoring many penalties.

Then, let $n \equiv 1 \pmod{3}$ . We can get this score by scoring various tries and some penalties, and we will prove this:

Let $5D$ be the number of points scored with tries. Since $5 \equiv 2 \pmod{3}$ , $5D \equiv 1 \pmod{3}$ , if $D \equiv 2 \pmod{3}$ . Since this $D$ can always be found, from here we can just score enough penalties to reach the desired score. As an example, take $n = 22$ . We can score $5\times 2$ points with tries and $3\times 4$ points with penalties.

For $n \equiv 2 \pmod{3}$ we use the exact same analysis but instead of $5$ we use $7$ points.

Hence the only scores we can not get are $1,2,4$ . Then, since $1 + 2 + 4 = 7$ , $7$ is the answer.

Using the chicken nugget theorem (which states that, given two relatively prime integers A and B, we can add together positive multiples of A and B to get any positive integer above AB-A-B), we get that we can get any score above 7 by applying the theorem to penalties and scored, non-converted tries. From here it is easy to test all numbers below 7, and get that the only impossible scores are 1, 2, and 4.

The three possible increases in score a team can gain are $3$ , $5$ , and $7$ . Upon analysis notice that in the series $1$ to $9$ only scores $1$ , $2$ , and $4$ are not possible. $10$ , $11$ , and $12$ are possible, and all positive numbers beyond that can be achieved by adding multiple penalty shots to one of those three scores. Therefore, the answer must be $1+2+4=\boxed{7}$ .

First, consider the case where the score n is only comprised of 3s and 5s. Thus we can write down: $n=5a+3b \Rightarrow n-5a=3b \Rightarrow n-2i \equiv 0 \mod{3}$ We now consider the three cases for n:

1. $n\equiv0 \mod{3}$

We have that $0-2i \equiv 0 \mod{3} \Rightarrow 2i \equiv 0 \mod{3} \Rightarrow i_{min}=0$ Thus there are no positive integers where $n\equiv0 \mod{3}$ that can't be written as $5a+3b$ .

1. $n\equiv 1 \mod{3}$

We have that $1-2i \equiv 0 \mod{3} \Rightarrow 2i \equiv 1 \mod{3} \Rightarrow i_{min}=2$ Thus the positive integers where $n\equiv0 \mod{3}$ that can't be written as $5a+3b$ are 1, 4 and 7 (since $n+1>5i_{min}$ ).

3. $n\equiv 1 \mod{3}$

We have that $2-2i \equiv 0 \mod{3} \Rightarrow 2i \equiv 2 \mod{3} \Rightarrow i_{min}=1$ Thus the only positive integer where $n\equiv0 \mod{3}$ that can't be written as $5a+3b$ is 2.

Notice that 7 is possible in the actual game: a 5 followed by a 2. This implies that the sum of all integers that can't occur as a score is $1+2+4=\boxed{7}$

simple.. since the points start from 3 onwards, no chance of scoring 1 & 2.. even in case of goal (where you get 2 points) you would have already scored 5 points.. and since no scope of gaining 1 point, min value above 3 points is 5 (3 for penalty and 2 extra for goal).. so no chance of 4..

there fore, total = 1+2+4 = 7..

Starting with the first impossibility i.e 1 since single point cannot be scored,we have two more improbable scores.they are 2,4.As we can score two points only if we can first kick a goal which gives five points.similarly with 4 as it follows the previous idea.so the sum of the improbable scores 1+2+4=7

it is obvious that a team can not score 1, 2, 4 ..now they can score 5, 6, 7, 8, 9, 10 : the remaining digits from 10 can be written as a sum of these numbers from 5 to 10 : example 11 = 5+6 , 12 = 6+6 , 13 = 6+7 ...so every number of points after 10 can be achieved ...so answer to this question = 1+2 +4

Using the chicken mcnugget theorem, that is using two prime nos. and b ( a=5 and b= 3 in here) we can get all nos. above (ab-a-b)=7. Using the given score format can get all scores under 7 ( including 7) except 1,2 and 4. So, the answer is (1+2+4)= 7. Hurray!

For a TRY one can get 5 points. and they get 1 chance at a conversion for which 2 points are also added. which makes 5+2=7.

all numbers can be represented as 3k 3k+1 3k+2 now we have 3k = 3m, 3k+2 = 5n, 3(k+1)+1 = 7p; for positive even k's and 3k+1=5n', 3(k+1)+1 = 7p' for positive odd k's. continuing the logic, we find that we can represent 3,5 and all numbers above 7. thus the answer is 7(1+2+4).

starting with the first impossibility i.e 1 since single point cannot be scored,we have two more improbable scores.they are 2,4.As we can score two points only if we can first kick a goal which gives five points.similarly with 4 as it follows the previous idea.so the sum of the improbable scores 1+2+4=7

Only 1,2,4 can't be the possible scores.

So, the sum of all scores which is impossible for a team to achieve in a game of rugby = 1+2+4 = 7

If we look the numbers of the form 3k, 3k+1 and 3k+2

Any score of the form 3k is possible, because we have a score of 3 achievable

Any score of the form 3k+1 is possible greater than equal to 7. Because that is the min score of the form 3k+1 possible (using 5+2)

Any score of the form 3k+2 is possible greater than equal to 5.

So, the scores which are not possible are:

1. Scores of the form 3k+1, less than 7, viz.: 1 and 4
2. Scores of the form 3k+2, less than 5, viz.: 2

Their sum: 1 + 2 + 4 = $\boxed{7}$