You should make the measurement in the following way.
"[x,...,y] true/false" means radioactivity was found / not found when measuring the balls with number x,...,y. The last two lines of each paragraphs are the tuples wich are possible after the last measurement is true / false. It is easy to process the remaining possibilities
measurement 1 [1,2,3,4,5] true
measurement 2: [4,5,6] true
measurement 3: [1,2,7,8,9,10,11] true
measurement 4: [1,2,7] true
measurement 5: [1,6]
[[1,4],[1,5],[1,6],[2,6]]
[[2,4],[2,5],[4,7],[5,7]]
measurement 1 [1,2,3,4,5] true
measurement 2: [4,5,6] true
measurement 3: [1,2,7,8,9,10,11] true
measurement 4: [1,2,7] false
measurement 5: [4]
[[4,8],[4,9],[4,10],[4,11]]
[[5,8],[5,9],[5,10],[5,11]]
measurement 1: [1,2,3,4,5] true
measurement 2: [4,5,6] true
measurement 3: [1,2,7,8,9,10,11] false
measurement 4: [12,13,14,15]
[[4,12],[4,13],[4,14],[4,15],[5,12],[5,13],[5,14],[5,15]]
[[3,4],[3,5],[3,6],[4,5],[4,6],[5,6]]
measurement 1 [1,2,3,4,5] true
measurement 2: [4,5,6] false
measurement 3: [7,8,9,10,11] true
measurement 4: [1,7] true
measurement 5: [7,8]
[[1,7],[1,8],[2,7],[3,7]]
[[1,9],[1,10],[1,11]]
measurement 1 [1,2,3,4,5] true
measurement 2: [4,5,6] false
measurement 3: [7,8,9,10,11] true
measurement 4: [1,7] false
measurement 5: [2]
[[2,8],[2,9],[2,10],[2,11]]
[[3,8],[3,9],[3,10],[3,11]]
measurement 1 [1,2,3,4,5] true
measurement 2: [4,5,6] false
measurement 3: [7,8,9,10,11] false
measurement 4: [1,12]
[[1,2],[1,3],[1,12],[1,13],[1,14],[1,15],[2,12],[3,12]]
[[2,3],[2,13],[2,14],[2,15],[3,13],[3,14],[3,15]]
measurement 1 [1,2,3,4,5] false
measurement 2: [6,7,8] true
measurement 3: [6]
[[6,7],[6,8],[6,9],[6,10],[6,11],[6,12],[6,13],[6,14],[6,15]]
[[7,8],[7,9],[7,10],[7,11],[7,12],[7,13],[7,14],[7,15],
[8,9],[8,10],[8,11],[8,12], [8,13],[8,14],[8,15]]
measurement 1 [1,2,3,4,5] false
measurement 2: [6,7,8] false
measurement 3: [9,10] true
measurement 4: [9]
[[9,10],[9,11],[9,12],[9,13],[9,14],[9,15]]
[[10,11],[10,12],[10,13],[10,14],[10,15]]
measurement 1 [1,2,3,4,5] false
measurement 2: [6,7,8] false
measurement 3: [9,10] false
measurement 4: [11]
[[11,12],[11,13],[11,14],[11,15]]
[[12,13],[12,14],[12,15],[13,14],[13,15],[14,15]]
Edit:
How can one check this? First of all you should check that these paragraphs are the node of a tree. Then you can check each paragraph by comparing each of the 105 pairs of balls [[1,2],...,[14,15]] with the measurements. Take the first pair [1,2] and the following paragraph
measurement 1 [1,2,3,4,5] true
measurement 2: [4,5,6] false
measurement 3: [7,8,9,10,11] true
measurement 4: [1,7] true
measurement 5: [7,8]
[[1,7],[1,8],[2,7],[3,7]]
[[1,9],[1,10],[1,11]]
Measurement 3 must be true but this is wrong for this pair therefore it can be skipped. Take the pair [1,7]. Measurement 1,3,4,5 is true and measurement 2 is false, therefore it is in the last but one line. For the pair [1,9] again measurement 2 is false and measurement 1,3,4 is true, but measurement 5 is false, therefore it is in the last line. so all node of the tree can be checked. You can do these checks with a simple program (I used one written in maxima).
Edit: Maxima-Program
f(n,ll_found,ll_notfound):=block(
[
remaining:[],
passed
],
for i:1 thru n do (
for j:i+1 thru n do (
passed:true,
for s in ll_found do
if not(member(i,s) or member(j,s)) then
passed:false,
for s in ll_notfound do
if not(not member(i,s) and not member(j,s)) then
passed:false,
if passed then
remaining:endcons([i,j],remaining)
)
),
return(remaining)
);
block([u,v],
for i:1 thru 14 do (
for j:i thru 15 do (
u:length(f(15,[[1,2,3,4,5],makelist(k,k,i,j)],[])),
v:length(f(15,[[1,2,3,4,5]],[makelist(k,k,i,j)])),
if (u<=32 and v<=32) then print([i,j,u,v])
)
)
);
/* measurement 1: [1,2,3,4,5] */
length(f(15,[[1,2,3,4,5]],[]));
length(f(15,[],[[1,2,3,4,5]]));
/* measurement 1 [1,2,3,4,5] true
measurement 2: [4,5,6] */
length(f(15,[[1,2,3,4,5],[4,5,6]],[]));
length(f(15,[[1,2,3,4,5]],[[4,5,6]]));
/* measurement 1 [1,2,3,4,5] true
measurement 2: [4,5,6] true
measurement 3: [1,2,7,8,9,10,11] */
f(15,[[1,2,3,4,5],[4,5,6],[1,2,7,8,9,10,11]],[])$length(%);
f(15,[[1,2,3,4,5],[4,5,6]],[[1,2,7,8,9,10,11]])$length(%);
/* measurement 1 [1,2,3,4,5] true
measurement 2: [4,5,6] true
measurement 3: [1,2,7,8,9,10,11] true
measurement 4: [1,2,7] */
f(15,[[1,2,3,4,5],[4,5,6],[1,2,7,8,9,10,11],[1,2,7]],[])$length(%);
f(15,[[1,2,3,4,5],[4,5,6],[1,2,7,8,9,10,11]],[[1,2,7]])$length(%);
/* measurement 1 [1,2,3,4,5] true
measurement 2: [4,5,6] true
measurement 3: [1,2,7,8,9,10,11] true
measurement 4: [1,2,7] true
fuenfte Messung: [1,6]
und fertig, da trivial*/
f(15,[[1,2,3,4,5],[4,5,6],[1,2,7,8,9,10,11],[1,2,7],[1,6]],[]);length(%);
f(15,[[1,2,3,4,5],[4,5,6],[1,2,7,8,9,10,11],[1,2,7]],[[1,6]]);length(%);
/* measurement 1 [1,2,3,4,5] true
measurement 2: [4,5,6] true
measurement 3: [1,2,7,8,9,10,11] true
measurement 4: [1,2,7] false
trivial da [4,5] * [8,9,10,11]*/
f(15,[[1,2,3,4,5],[4,5,6],[1,2,7,8,9,10,11]],[[1,2,7]]);
/* measurement 1 [1,2,3,4,5] true
measurement 2: [4,5,6] true
measurement 3: [1,2,7,8,9,10,11] false
measurement 4: [12,13,14,15]
und trivial*/
f(15,[[1,2,3,4,5],[4,5,6],[12,13,14,15]],[[1,2,7,8,9,10,11]]);length(%);
f(15,[[1,2,3,4,5],[4,5,6]],[[1,2,7,8,9,10,11],[12,13,14,15]]);length(%);
/* measurement 1 [1,2,3,4,5] true
measurement 2: [4,5,6] false
measurement 3: [7,8,9,10,11] */
f(15,[[1,2,3,4,5],[7,8,9,10,11]],[[4,5,6]])$length(%);
f(15,[[1,2,3,4,5]],[[4,5,6],[7,8,9,10,11]])$length(%);
/* measurement 1 [1,2,3,4,5] true
measurement 2: [4,5,6] false
measurement 3: [7,8,9,10,11] true
measurement 4: ,[1,7] */
f(15,[[1,2,3,4,5],[7,8,9,10,11],[1,7]],[[4,5,6]])$length(%);
f(15,[[1,2,3,4,5],[7,8,9,10,11]],[[4,5,6],[1,7]])$length(%);
/* measurement 1 [1,2,3,4,5] true
measurement 2: [4,5,6] false
measurement 3: [7,8,9,10,11] true
measurement 4: ,[1,7] wahr
fuenfte Messung: [7,8] arbitrary
trivial*/
f(15,[[1,2,3,4,5],[7,8,9,10,11],[1,7],[7,8]],[[4,5,6]]);length(%);
f(15,[[1,2,3,4,5],[7,8,9,10,11],[1,7]],[[4,5,6],[7,8]]);length(%);
/* measurement 1 [1,2,3,4,5] true
measurement 2: [4,5,6] false
measurement 3: [7,8,9,10,11] true
measurement 4: ,[1,7] false
trivial, da {2,3} * {8,9,10,11}*/
f(15,[[1,2,3,4,5],[7,8,9,10,11]],[[4,5,6],[1,7]]);length(%);
/* measurement 1 [1,2,3,4,5] true
measurement 2: [4,5,6] false
measurement 3: [7,8,9,10,11] false
measurement 4: [1,12] arbitrary
trivial */
f(15,[[1,2,3,4,5],[1,12]],[[4,5,6],[7,8,9,10,11]]);length(%);
f(15,[[1,2,3,4,5]],[[4,5,6],[7,8,9,10,11],[1,12]]);length(%);
/* measurement 1 [1,2,3,4,5] false
measurement 2: [6,7,8] */
f(15,[[6,7,8]],[[1,2,3,4,5]])$length(%);
f(15,[],[[1,2,3,4,5],[6,7,8]])$length(%);
/* measurement 1 [1,2,3,4,5] false
measurement 2: [6,7,8] true
measurement 3: [6] arbitrary
trivial*/
f(15,[[6,7,8],[6]],[[1,2,3,4,5]]);length(%);
f(15,[[6,7,8]],[[1,2,3,4,5],[6]]);length(%);
/* measurement 1 [1,2,3,4,5] false
measurement 2: [6,7,8] false
measurement 3: [9,10] */
f(15,[[9,10]],[[1,2,3,4,5],[6,7,8]]);length(%);
f(15,[],[[1,2,3,4,5],[6,7,8],[9,10]]);length(%);
/* measurement 1 [1,2,3,4,5] false
measurement 2: [6,7,8] false
measurement 3: [9,10] true
measurement 4: [9] arbitrary
trivial */
f(15,[[9,10],[9]],[[1,2,3,4,5],[6,7,8]]);length(%);
f(15,[[9,10]],[[1,2,3,4,5],[6,7,8],[9]]);length(%);
/* measurement 1 [1,2,3,4,5] false
measurement 2: [6,7,8] false
measurement 3: [9,10] false
measurement 4: [11] arbitrary
trivial */
f(15,[[11]],[[1,2,3,4,5],[6,7,8],[9,10]]);length(%);
f(15,[],[[1,2,3,4,5],[6,7,8],[9,10],[11]]);length(%);
Edit:
How did I derive the result?
With trial and error. The following Lemma guides our trials
Lemma 1 (without proof): In every step of the algorithm the following is true: Let $S$ be the list of the remaining possible solution and $M$ be the list of balls to be measured and $n$ be the number of allowed measurements. Let $ T(S,M)$ be the list of all pairs of $S$ where at least one element of the pair is in $M$ and let $F(S,M)$ the list of all pairs of $S$ where no element of the pair is in $M$. Then the following is a necessary condition for an algorithm to solve all instances of the problem.
$$\begin{align*}
|S|&\leq 2^n\\
|T(S,M)|&\leq 2^{n-1}\\
|F(S,M)|&\leq 2^{n-1}
\end{align*}$$
After this step the new S is $T(S,M)$ or $F(S,M)$ depending of the result of the measurement $M$. The new $n$ is $n-1$.
Finding two balls with at most k measuremants from n balls we call an (n,k)-problem. We want to find an algorithm for the (15,7)-problem.
The (15,7)-problem therefore $\binom{15}{2}=105$ possible solution pairs and therefore an algorithm to find such a pair must make at least 7 measurements in
the worstcase because $2^6<105<2^7$. We want to investigate the first measurement. Let's arange the list of all possible ball combination in the following
triangle manner
[1,2] [1,3] [1,4] ... [1,14] [1,15] 14 pairs row 1
[2,3] [2,4] ... [2,14] [1,15] 13 pairs row 2
. . . .
. . . .
. . . .
[13,14] [13,15] 2 pairs row 13
[14,15] 1 pair row 14
if the first measurement is made with the list $[1,\cdots,m]$ then $14+13+(15-m)<=2^6$ and $(15-(m+1))+...+2+1<=2^6$ must hold. $2^6=64$ therefore $m$ can
be $4$ or $5$. if $m=3$ and therefore $M=[1,2,3]$ then $|F(S,M)|=1+2+\cdots + 11=66>2^6=64$, if $k=6$ then $M=[1,2,3,4,5,6]$ and $|T(S,M)|=14+13+\cdots + 9
= 69 > 2^6=64$.
if for the first measurement $m=4$ and therefore $M=[15,14,13,12]$ and the result of the measurement is false then one must find with at most $6$
measurement the radioactive pair in the remaining $11$ balls. so we must save the (11,6)-problem.
Lemma 2: There is no algorithm for the (6,4)-problem.
Proof: If $M=[1]$ then $|F(S,M)|=4+3+2+1=10>8=2^3$. If $M=[1,2]$ then $|T(S,M)|=5+4=9>8=2^3$.
Lemmas 3: There is no algorithm for the (8,5) problem.
Proof: for the first measurement: $M=[1]$ or $M=[1,2,3]$ are not possible using the same arguments as in the (6,4) case. Therefore $M$ must be $[1,2]$. If the result of the measurement is false, the algorithm must solve the (6,4)-problem which is impossible by Lemma 2.
Lemma 4: There is no algorithm for the (11,6)-problem.
Proof: if $M=[1,2]$ then $|F(S,M)|=36>32=2^5$, if $M=[1,2,3,4]$ then $T(S,M)=34>32=2^5$. therefore $M$ must be $[1,2,3]$. Suppose that the first measurement
is false 8 balls remains and our algorith must solve the (8,5)-problem wich is impossible by lemma 3.
From Lemma 4 an the reasoning before llemma 2 the following follows:
Lemma 5:
If there is an algorithm for the (15,7)-problem its first measurement must measure a list of 5 balls.
We can assume that the first measurement is on the list [1,2,3,4,5]. The second measurement is on a list [x,x+1,...,y] that has zero, one or more elements
in common with [1,2,3,4,5].
First I tried [6,7,8,9,10,11] but had problems to proceed. Therefore I used my program to check each of the 105 N=[x,...,y] if $T(S,N)<=2^5=32$ and $F(S,N)
<32$.
the following lists were found by the program
[4,5,6]
[5,6,7,8,9]
[6,7,8,9,10,11]
the program also found the lists
[7,...,12]
[8,...,13]
[9,...,14]
[10,...,15]
but these lists are basically the same as [6,...,11] after renaming the balls.
I continued with the list [4,5,6].
The remaining measurements I found by trial an error using my function and lemma 1. If one searches the third measuring after [1,2,3,4,5] and [4,5,6] one can use the fact that the numbers [1,2,3], [4,5], [7,8,9,10,11,12,13,14,15] are equivalent relative to [1,2,3,4,5] and [4,5,6]. this narrows down the cases to investigate.
Best Answer
To solve this without (much) algebra:
Note that green bags must hold $\frac 14$ of the balls (as we divided them, by color, in the ratio 3:1). Thus there are $14$ balls in the green bags, hence $42$ in the red bags.
Now we ask: how many might be in each bag? Well, whatever that number is it must divide both $14$ and $42$, hence it is a common factor of those two numbers. But we can list those common factors: $\{1,2,7,14\}$. Each of these gives us a solution!
I. 1 ball in each of 14 green and 42 red bags.
II. 2 balls in each of 7 green and 21 red bags.
III. 7 balls in each of 2 green and 6 red bags.
IV. 14 balls in each of 1 green and 3 red bags. (Unless you interpret the problem as requiring plural green bags.)
Unless we are told more information about the size of the party we really can't choose amongst these (though at a guess, we are after case II....I is too many and III and IV are too few).
To be clear: I would not assign this to a fifth grader unless I mentioned that there might be several acceptable answers (and probably not even then).