Problem
Fewest groups with no conflicting pair
Five kinds of fish are shared among some tanks. Two fish may not share a tank if one eats the other. All five kinds must be housed. Find the smallest number of tanks needed.
Reasoning · Grade 6-1 Deductive Logic (1)
Your answer
Watch out One arrangement is not proof: you must also show fewer is impossible.
How to solve
Strategy Draw a Diagram — Six separate 'eats' sentences are hard to hold in the head at once, so I first change what I am looking at: for the purpose of sharing a tank, 'B eats C' and 'C eats B' would mean the same thing, so I throw the arrows away and keep only the unordered conflicting pairs. That turns the text into a small picture — five dots, one per kind of fish, with a line joining any two that must be kept apart. Then it is easier to hunt for the two safe pairs (the missing lines) than for the many dangerous ones, so I count the complement. The picture answers both halves of the question: a group of dots that are all joined to each other forces a lower bound on the number of tanks, and a way of colouring the dots gives an arrangement that reaches it.
1STEP 1
Turn every 'eats' into a conflicting pair
Write every 'eats' as a conflicting pair.
A--B, A--C, A--D, A--E, B--C, B--D, C--D, D--E
Sorting the eight statements into 'this pair may never share' is pure classifying and counting; dropping the arrows is allowed because the rule about tanks treats eater and eaten exactly alike.
K.MD.B.3Make A Systematic List2STEP 2
Count the complement: which pairs are safe?
Only two pairs are safe.
10 - 8 = 2 safe pairs: B--E and C--E
There are far fewer safe pairs than dangerous ones, so listing what is allowed is much shorter work than listing what is forbidden — and it is the allowed pairs that will let tanks be shared.
4.OA.A.3Change Focus Count The Complement3STEP 3
Draw the picture
Draw it as dots and lines.
Drawing points and joining them with segments is Grade 4 work, and the drawing does the remembering for you: the number of lines leaving a dot is instantly visible, whereas the same fact buried in six sentences is not.
4.G.A.1Draw A Diagram4STEP 4
Lower bound: four fish that all conflict with one another
There are four fish that all conflict.
pairs inside {A,B,C,D} = 6, all 6 conflict → at least 4 tanks
This is the half of the answer that guessing can never supply. However many arrangements you try, one exhibited group of four mutually hostile fish rules out 3 tanks forever, because those four alone already fill four tanks.
4.OA.A.3Eliminate PossibilitiesFour fish that all conflict with one another force at least four tanks, whatever the arrangement.
Why?
No two of those four may share a tank, so three tanks could not hold four of them one apiece.
🧱More items than boxes forces a shared boxn boxes can hold at most n items one apiece, so item number n plus one has nowhere new to go and must share.
Why?
Exhibiting that one group of four is enough to refute any claim that three tanks would do, without checking every arrangement.
🧱One counterexample refutes an always-claimA claim about every case dies the moment you exhibit one case where it fails, and exhibiting is easier than arguing.
5STEP 5
Achievability: build an arrangement that uses exactly 4 tanks
So fewer than 4 tanks is impossible.
{B, E} ∣ {A} ∣ {C} ∣ {D}
Sorting five objects into four labelled boxes and then checking each box is exactly the classify-and-count skill from Kindergarten; the only box that needs checking is the one holding more than one fish.
K.MD.B.3Make A Systematic List6STEP 6
Put the two halves together, and note the other arrangement
Four tanks really works, so the answer is 4.
at least 4 and 4 is enough → minimum = 4
A smallest-number answer always needs both halves — a working arrangement and a reason nothing smaller can work — and here the two halves meet at the same number, so the answer is settled.
4.OA.A.3Eliminate PossibilitiesAnswer
4 tanks
10 − 8 = 2
The answer has to lie between 1 and 5, since 5 tanks (one fish each) always works and 1 tank obviously does not, and 4 sits sensibly in that range. It cannot be 5, because the exhibited arrangement genuinely puts B and E together. It cannot be 3, because A, B, C and D conflict in all 6 of their pairs and so occupy 4 tanks on their own. Re-checking the winning arrangement against all eight conflicting pairs: A-B, A-C, A-D, A-E, B-C, B-D, C-D and D-E each have their two fish in different tanks, so no fish gets eaten. A quick sanity note on the shape of the answer too — E is the only fish with two safe partners, which is exactly why E is the only fish that ever gets to share.
Takeaway
Four fish that all fight with each other need four separate tanks — so show one arrangement that works and one group that proves nothing smaller can.
- Turn every 'eats' into a conflicting pair
- Count the complement: which pairs are safe?
- Draw the picture
- Lower bound: four fish that all conflict with one another
- Achievability: build an arrangement that uses exactly 4 tanks
- Put the two halves together, and note the other arrangement
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