Assigning Variables

Let the number of kills done by Ann, Braden, Caden, and Doug be ${ K }_{ A }$ , ${ K }_{ B }$ , ${ K }_{ C }$ , ${ K }_{ D }$ and the number of times they died be ${ D }_{ A }$ , ${ D }_{ B }$ , ${ D }_{ C }$ , { D }_{ D }} respectively. Also, let their no. of kills to no. of deaths or the ratio be ${ R }_{ A }$ , ${ R }_{ B }$ , ${ R }_{ C }$ and ${ R }_{ D }$ also in the same order.

Given Things

It is given the kills to death ratio is same for Braden and Caden, which mathematically is ${ R }_{ B }$ = ${ R }_{ C }$ . Also, they were the only 4 players, and nobody died due to suicide or world deaths, it means somebody's kills must have added somebody's deaths, i.e the total number of kills and total number of deaths is the same. Mathematically, ${ K }_{ A }$ + ${ K }_{ B }$ + ${ K }_{ C }$ + ${ K }_{ D }$ = ${ D }_{ A }$ + ${ D }_{ B }$ + ${ D }_{ C }$ + ${ D }_{ D }$ .

Reasoning from more given things

It is given that Ann has a greater ratio than both Braden and Caden and that exactly 3 had the same no. of kills and also exactly 3 people have the same no. of deaths. So now suppose, that Braden has a different number of kills and deaths than others, thus Ann, Caden and Doug would have the same no. of deaths and kills, but if Ann and Caden have the same no. of kills and deaths then their ratio would also be equal, however, it is given Ann has a greater ratio, which means Braden and Caden, along with having the same ratio would also have the same no. of deaths and kills to avoid any contradiction. So now out of 3, we have determined 2 people who will have the same no. of deaths, so there could be 3 possible cases now due to the remaining 1 person.

Cases

• 1. Ann has different no. of deaths and kills, which implies then that Doug has the same no. of kills and deaths as Braden and Caden, i.e ${ R }_{ D }$ = ${ R }_{ B }$ = ${ R }_{ C }$ .

• 2 Ann has the same no. of kills but different no. of deaths due to which Doug has the same no. of deaths but different no. of kills i.e ${ K }_{ A }$ = ${ K }_{ B }$ = ${ K }_{ C }$ and ${ D }_{ D } = { D }_{ B } = { D }_{ C }$ .

• 3 The 3rd case is just vice versa of the 2nd case, and thus ${ K }_{ D } = { K }_{ B } = { K }_{ C }$ and ${ D }_{ A }={ D }_{ B }={ D }_{ C }$ .

Equations and Permutations

We know for sure that ${ R }_{ B }$ = ${ R }_{ C }$ , ${ R }_{ B }$ = ${ D }_{ C }$ and ${ D }_{ B }$ = ${ K }_{ C }$ so instead of writing the ratio, kills and deaths of C i.e Caden we will use of Braden's as they are the same. So using the cases 1, 2 and 3 and all the given statements, we can infer that due to the 3 cases and simplifying accordingly, the ratio of D can be either, ${ R }_{ D }={ R }_{ B }$ , or $\large { R }_{ D }=\frac { { R }_{ B }+3{ R }_{ A }(1-{ R }_{ B }) }{ R_{ A } }$ or $\large { R }_{ D }=\frac { R_{ B } }{ 3({ R }_{ B }-1)+{ R }_{ A } }$ .

Now, Doug figures out that the ratio for him could be either $\frac { 3 }{ 4 }$ or $\frac { 9 }{ 11 }$ or $\frac { 6 }{ 5 }$ . So as we found out 3 possible equations, and Doug has 3 possible values, we could have total 6 permutations as if one eq. can have any 1 value out of 3, then the 2nd eq. can have any 1 out of the remaining 2, and the 3rd eq. would have to then select the last remaining value, i.e $3 \times 2 \times 1 = 6$ .

Checking all the 6 permutations, we find only when ${ R }_{ D }={ R }_{ B }=\frac { 3 }{ 4 }$ , or ${ R }_{ D }=\frac { { R }_{ B }+3{ R }_{ A }(1-{ R }_{ B }) }{ R_{ A } } =\frac { 6 }{ 5 }$ and ${ R }_{ D }=\frac { R_{ B } }{ 3({ R }_{ B }-1)+{ R }_{ A } } =\frac { 9 }{ 11 }$ are only equations which are the only consistent equations, otherwise the other permutations have improper or the solutions don't match, and also Doug to know the values he would have to know Ratio of A and B otherwise he couldn't have inferred his possible values. Thus, solving we find, ${ R }_{ A }=\frac { 5 }{ 3 }$ , thus sum of digits is $\boxed{5+3=8}$

Let the number of kills for Ann, Braden, Caden, and Doug be $k_A$ , $k_B$ , $k_C$ , and $k_D$ , let the number of deaths be $d_A$ , $d_B$ , $d_C$ , and $d_D$ , and let the k:d ratios be $A$ , $B$ , $C$ , and $D$ . Then $\frac{k_A}{d_A} = A$ , $\frac{k_B}{d_B} = B$ , $\frac{k_C}{d_C} = C$ , and $\frac{k_D}{d_D} = D$ . Since Braden and Caden had exactly the same k:d, $B = C$ , and since there were no suicides or world deaths, $k_A + k_B + k_C + k_D = d_A + d_B + d_C + d_D$ .

Since exactly three players had the same number of kills, one person had a different number of kills, and since exactly three players had the same number of deaths, one person had a different number of deaths. Since Ann had a different k:d than Braden and Caden, she either had the different number of kills, the different number of deaths, or both. Since Braden and Caden had exactly the same k:d, and Ann had at least one of the different number of kills and/or deaths, Braden and Caden must have had both the same number of kills and the same number of deaths as well. Therefore, $k_B = k_C$ and $d_B = d_C$ , and there are therefore three cases to consider:

Case 1: Ann had the different number of kills and the different number of deaths.

Then Doug had the same number of kills and deaths as Braden and Caden, so Doug’s k:d is the same as Braden’s and Caden’s k:d, so $D = B$ .

Case 2: Ann had the different number of kills and Doug had the different number of deaths.

Then $k_B = k_C = k_D$ and $d_A = d_B = d_C$ , and all the equations simplify to $D = \frac{B}{A - 3 + 3B}$ .

Case 3: Ann had the different number of deaths and Doug had the different number of kills.

Then $k_A = k_B = k_C$ and $d_B = d_C = d_D$ , and all the equations simplify to $D = \frac{B + 3A - 3AB}{A}$ .

Setting the three equations for $D$ , namely $(B, \frac{B}{A - 3 + 3B}, \frac{B + 3A - 3AB}{A})$ , to all possible permutations of Doug’s deductions for his k:d, namely $(\frac{3}{4}, \frac{9}{11}, \frac{6}{5})$ , reveals that only $B = \frac{3}{4}$ , $\frac{B}{A - 3 + 3B} = \frac{9}{11}$ , and $\frac{B + 3A - 3AB}{A} = \frac{6}{5}$ are the only consistent equations and have a solution of $A = \frac{5}{3}$ and $B = \frac{3}{4}$ .

Therefore, Ann’s k:d is $A = \frac{5}{3}$ , so $p = 5$ , $q = 3$ , and $p + q = \boxed{8}$ .