Problem
Pigeonhole applied to digit-card numbers
Cards 1, 3 and 6 are rearranged into a three-digit number. Every student builds one. Part (1) has 20 students. Find the guaranteed match count, and the class size that forces six.
Reasoning · Grade 5-2 The Pigeonhole Principle
Your answer
Watch out The biggest possible pile is tempting, but a guarantee comes from the evenest sharing.
How to solve
Strategy Make a Systematic List — Everything depends on how many different numbers the cards can make, and with only three cards I can list them all in order and be sure none is missing or repeated. After that, drawing one box per number and dropping each student into their box turns both questions into questions about how full a box can be. For each part I do not stop at the arithmetic: I write down the actual spread of students that only just fails, which is what pins the answer down exactly instead of leaving it one out.
1STEP 1
List every number the cards can make
The cards make 6 numbers.
136, 163, 316, 361, 613, 631 → 6 numbers
Six items is few enough to write out completely, so there is no chance of missing an arrangement or counting one twice.
2.NBT.A.1Make A Systematic List2STEP 2
Turn students into a box diagram
Share the students into those six boxes.
Once each student sits in a box, the question stops being about numbers and becomes a question about how full a box is, which is something you can count.
4.OA.A.3Draw A Diagram3STEP 3
Part (1): share 20 students out as evenly as possible
20 into six gives 3 remainder 2.
20 ÷ 6 = 3 remainder 2
Dividing 20 by 6 is exactly "share 20 things fairly among 6 places", and the remainder is what tells you the sharing cannot come out level.
4.NBT.B.6Make A Systematic ListSharing 20 students as evenly as possible over the boxes shows what the fullest box is forced to hold.
Why?
If every box held fewer than that, the boxes together could not account for all 20 students.
🧱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?
Splitting the students into whole equal shares plus a leftover is what names both the even share and the extra one box must take.
🧱Division with remainderAny number splits into a whole count of equal groups plus a leftover, and the leftover is always smaller than the group size.
4STEP 4
Part (1): show 3 students in a box is impossible
All at three would be only 18, so it fails.
6 × 3 = 18 ＜ 20
Instead of chasing which number gets crowded, count how many students would fit if none did - six boxes of three only hold eighteen, so two students are left with nowhere safe to go.
4.OA.A.3Change Focus Count The Complement5STEP 5
Part (1): show 4 is the most that can be promised
So 4 students is what can be promised.
4 + 4 + 3 + 3 + 3 + 3 = 20
Writing down one spread where 5 fails is what shows 4 is the true promise and not an underestimate - the guarantee is only as strong as the worst spread you can build.
4.OA.A.3Make A Systematic List6STEP 6
Part (2): build the biggest class that still fails
Avoiding six is possible up to 30.
6 × 5 = 30
Six boxes with five students each is six groups of five, and 6 x 5 = 30 - the fullest the class can get while still dodging a group of six.
4.NBT.B.5Make A Systematic List7STEP 7
Part (2): add one more student
One more makes 31.
6 × 5 + 1 = 31
The 31st student has to join one of the six boxes, and every box is already full at five, so that box becomes a group of six.
4.OA.A.3Change Focus Count The ComplementAnswer
4, 31 students
6 × 5 + 1 = 31
Both answers are counts of students, so they must be whole numbers, and 4 and 31 are. Part (1) makes sense in size: 20 students spread over 6 numbers averages a bit more than 3 per number, so the fullest number should be just above 3, and 4 is exactly that; it must also be no more than 20, which it is. Part (2) is checked from both sides - a class of 30 arranged 5 per number really does dodge a group of 6, while 31 cannot - so the answer is neither one too big nor one too small. The two parts also agree with each other: the same box picture, run forwards for part (1) and backwards for part (2), gives 6 x 3 + 2 = 20 and 6 x 5 + 1 = 31.
Takeaway
Count the boxes first - only six numbers can be made - then fill them as evenly as you can and add one more student, because that student has nowhere new to go.
- List every number the cards can make
- Turn students into a box diagram
- Part (1): share 20 students out as evenly as possible
- Part (1): show 3 students in a box is impossible
- Part (1): show 4 is the most that can be promised
- Part (2): build the biggest class that still fails
- Part (2): add one more student
You might solve next