Charity Fair Riddle

Let $S$ stand for the solution amount fund earned on each day; $a,b$ be the odd numbers of men and women on the first day respectively; and $c,d$ be the odd numbers of men and women on the second day respectively with distinct $a,b,c,d > 3$ . We can conclude that $S = 5a + 3b = 5c + 3d$ , and so alternatively, we could then write the matrix system as the following:

$\begin{bmatrix}{a} && {b} \\ {c} && {d} \end{bmatrix} \begin{bmatrix}{5}\\{3}\end{bmatrix} = \begin{bmatrix}{S}\\{S}\end{bmatrix}$

Since $a,b,c,d$ be odd and greater than $3$ as there were men and women in the party, we are looking at the ways the even $S$ can be rewritten in the sum of $2$ composite odd numbers. Then since Holmes found out the numbers of guests immediately after knowing $S$ , that means there are only $2$ ways that $S$ can be rewritten in such summations. That is, $S = 5a + 3b = 5c + 3d$ is the only combination.

In order to find such combinations, let us learn first that $5+3 \equiv 3 \pmod{5} \equiv 5 \pmod{3}$ naturally. Now, if we add $3n$ into it for some non-negative integer $n$ , we will get a new number of $5 + 3 + 3n$ , and to make it even as $S$ , $n$ must also be even. Hence, let us rewrite it as $5+3+3n = 5+3+6m$ for some natural number $m$ . Then there are more ways to break it down the multiples of $3$ , but if $m < 5$ , the only multiple of $5$ remains as $5 + (3+6m)$ because in order to increase the multiple of $5$ and retaining the multiple of $3$ , we must be able to obtain some quotient when dividing $m$ by $5$ . Thus, $m$ must exceed $5$ .

Then let us rewrite it as $5 + 3 + 6m = 5 + 3 + 6(5q + r) = 5 + 3 + 30q + 6r$ for some positive quotient $q$ and non-negative $r$ .

By similar methods, we can altogether rewrite $S = 5+3 + 30q + 6r + 10s$ and, also, $S \geq 5+3 + 30q$ .

Let us take $q=1$ , for example. Then $5+3+30 = 38 = 5+33 = 3+ 35$ . There are no ways to write it as the sum of odd composites, and so we need to add the cominations of $6r$ and $10s$ . The candidate numbers then are:

$5+3+30 +6+10= 54$ ;

$5+3+30 +12+10= 60$ ;

$5+3+30 +6+20= 64$ .

However, Holmes told us that $S$ could not be of men's or women's donations alone. That means $S$ itself is not a multiple of $3$ or $5$ , so only $64 = 25 + 39 = 9 + 55$ remains, but with $S=64$ , $b=3$ is contradicted.

Now for $q = 2$ , we have $5+3+60 = 68 = 33+35$ . This is the only summation of odd composites. By using the same restrictions of $S$ as non-divisible by $3$ or $5$ and $a,b,c,d > 3$ , we will seek out more candidate numbers:

$68 + 18 = 86 = 51 + 35 = 21 + 65$ ;

$68 + 20 = 88 = 33 + 55 = 63 + 25$ ;

$68 + 24 = 92 = 27 + 65 = 57 + 35$ ;

Then for $q = 3$ we will obtain: $98 = 65 + 33 = 35 + 63$ .

Then $98 + 6 =104 = 65 + 39 = 35 + 69 = 9+ 95$ ;

$98 + 20 =118 = 85 + 33 = 55 + 63 = 25 +93$ ;

$98 + 18 =116 = 65 + 51 = 35 + 69 = 95+21$ ;

$98 + 24 =122 = 65 + 57 = 35 + 77 = 95+21$ ;

At this point, if we keep on adding $6r$ or $10s$ , it will cause the number of $S$ summations to be greater than $2$ . Thus, this is our upper limit though $104 = 65 + 39 = 35 + 69$ might be possible, given that we rule out $9+95$ out. We will then investigate our candidate numbers.

In order to visualize $a,b,c,d$ , let us rewrite them in the matrix systems:

$\begin{bmatrix}{13} && {7} \\ {7} && {17} \end{bmatrix} \begin{bmatrix}{5}\\{3}\end{bmatrix} = \begin{bmatrix}{86}\\{86}\end{bmatrix}$

$\begin{bmatrix}{11} && {11} \\ {5} && {21} \end{bmatrix} \begin{bmatrix}{5}\\{3}\end{bmatrix} = \begin{bmatrix}{88}\\{88}\end{bmatrix}$

$\begin{bmatrix}{13} && {9} \\ {7} && {19} \end{bmatrix} \begin{bmatrix}{5}\\{3}\end{bmatrix} = \begin{bmatrix}{92}\\{92}\end{bmatrix}$

$\begin{bmatrix}{13} && {11} \\ {7} && {21} \end{bmatrix} \begin{bmatrix}{5}\\{3}\end{bmatrix} = \begin{bmatrix}{98}\\{98}\end{bmatrix}$

$\begin{bmatrix}{13} && {13} \\ {7} && {23} \end{bmatrix} \begin{bmatrix}{5}\\{3}\end{bmatrix} = \begin{bmatrix}{104}\\{104}\end{bmatrix}$

From Holmes' last clue, even if we know the number of guests in any day, it wouldn't help us with $S$ . At this point, if we know $d=23$ , it will instantly lead to $S = 104$ , for the number stands out distinctly from the other numbers. Similarly, if we know $d = 17$ , then $S = 86$ , because the number, too, stands out uniquely. Moreover, if $c=5$ , then we know $S=88$ , and if $b=9$ , then we know $S=92$ . On the other hand, even if we know $a = 13$ , $b= 11$ , $c = 7$ , or $d=21$ , that wouldn't conclude which $S$ to be because these numbers overlap with other combinations.

As a result, we can conclude that $S = \boxed{98}$ , and of course, $a = 13$ , $b= 11$ , $c = 7$ , or $d=21$ .