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Amara, Rowan and CKSprout at the CIC demonstration day
Pigeonhole principle
How can you be certain before you know what everyone will choose?
The team gets the hall ready for CIC.
Amara has finished the welcome signs. Rowan is testing the lights. CKSprout joins them as their problem-solving coach, with a question from the organisers.
Amara “The organisers need to plan for crowds, but visitors will choose their sessions.”
CKSprout “We don’t have the bookings yet. Can we still know that some teams will share a session?”
They cannot predict every visitor’s choice. But perhaps they can prove that something must happen.
CKSTEM Thinking CompassStuck? Find your next move.
You don’t need the whole route yet. Start with one useful move, then notice what it tells you.
What do you know for certain? What are you trying to find?
After checking your steps against every clue, choose your own check:
Ready to explore.
The direction reflects your self-check. Use the lesson’s answer and reasoning to check further when you are ready.
The morning demonstrations
Seven teams. Six sessions. A promise?
Seven teams will each book exactly one of six sessions. Every team is free to choose any session.
Must at least one session have two or more teams?
CKSTEM Thinking CompassPlan and check your reasoning.
Must at least one session have two or more teams?
The morning demonstrations
Seven teams. Six sessions. A promise?
Seven teams will each book exactly one of six sessions. Every team is free to choose any session.
Every team chooses exactly one session
1234567
Session 1
?
Bookings not known yet
Session 2
?
Bookings not known yet
Session 3
?
Bookings not known yet
Session 4
?
Bookings not known yet
Session 5
?
Bookings not known yet
Session 6
?
Bookings not known yet
The seven teams are known. Their session choices are still unknown.
Rowan “What if they spread out as much as possible?”
CKSprout “How could you test your idea?”
You could try this with seven counters and six cups. Can you put each counter in a different cup?
You don’t need the whole route yet. Start with one useful move, then notice what it tells you.
What do you know for certain? What are you trying to find?
After checking your steps against every clue, choose your own check:
Ready to explore.
The direction reflects your self-check. Use the lesson’s answer and reasoning to check further when you are ready.
Every team chooses exactly one session
1234567
Session 1
?
Bookings not known yet
Session 2
?
Bookings not known yet
Session 3
?
Bookings not known yet
Session 4
?
Bookings not known yet
Session 5
?
Bookings not known yet
Session 6
?
Bookings not known yet
The seven teams are known. Their session choices are still unknown.
Rowan “What if they spread out as much as possible?”
CKSprout “How could you test your idea?”
You could try this with seven counters and six cups. Can you put each counter in a different cup?
Answer and reasoning: must a session be shared?
Answer:Yes. Some session must have at least two teams.
CKSprout “What would the most spread-out arrangement look like?”
Amara “One team in each session. Then I can check whether that uses every team.”
Try to avoid any shared session
Session 1
1
1 team
Session 2
2
1 team
Session 3
3
1 team
Session 4
4
1 team
Session 5
5
1 team
Session 6
6
1 team
7No unused session remains.
This is the largest arrangement with at most one team per session. The seventh team must share somewhere, even though we cannot name the session.
CKSprout “How far can the one-team-per-session arrangement go?”
Amara “One team in each session uses only six teams. The seventh team has to join a session that already has a team.”
\[6\times1=6<7\]
So some session must contain at least two teams. This covers every possible booking arrangement. If several teams choose the same session earlier, the promised repeat has already happened.
Check the limit. With only six teams, one team could choose each session. A shared session would be possible, but it would no longer be certain.
Pigeonhole principle
Too many objects for separate groups? Some must share.
The name comes from putting pigeons into pigeonholes. Our “pigeons” are teams; our “holes” are sessions. The categories do not have to be physical boxes.
What makes it useful?
You want to know what must happen. Try to make it fail. How many objects can fit before it has to happen?
What must you check?
Give every object one group. Include all the groups. Check that two objects in the same group really do what the question asks.
CKSprout “Does the proof tell you which session is shared, or only that some session must be?”
Rowan “Only that some session must be shared. Which one depends on the bookings.”
Some is different from every. All seven teams might choose one session. The principle does not promise that every session is used, or identify the crowded one.
The afternoon workshops
How many teams would force four into one session?
The afternoon has five sessions. Each team books exactly one and can choose any session.
What is the smallest total number of teams that makes four or more teams in some session certain, whatever they choose?
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What is the smallest total number of teams that makes four or more teams in some session certain, whatever they choose?
The afternoon workshops
How many teams would force four into one session?
The afternoon has five sessions. Each team books exactly one and can choose any session.
Rowan “Let’s build the biggest booking list that still avoids four.”
CKSprout “What is the most one session can hold while still avoiding four?”
Your answer
The Compass checks these counts. You still need to choose useful categories, explain the guarantee and check any claim that the count is smallest.
This count fits. Compare your guarantee, avoiding example and explanation with the complete worked answer.
Break it into steps
Step 1 of 3
How many teams can one session hold while still avoiding four?
Three teams avoid four in one session.
Step 2 of 3
What is the largest total across five sessions while all avoid four?
\(5\times3=15\), attained by three teams in each session.
Step 3 of 3
What smallest team total guarantees four in some session?
\(15+1=16\). The proof covers every booking; the fifteen-team arrangement establishes smallest.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Rowan “Let’s build the biggest booking list that still avoids four.”
CKSprout “What is the most one session can hold while still avoiding four?”
Answer and reasoning: the smallest team total
Answer: The smallest total is 16 teams.
Rowan “Three teams can fit in each session without reaching four.”
CKSprout “How can you check that your total is the smallest?”
The largest arrangement with no session reaching four
Session 1
123
3 teams
Session 2
456
3 teams
Session 3
789
3 teams
Session 4
101112
3 teams
Session 5
131415
3 teams
Five groups of three avoid the target. Adding any sixteenth team makes some session reach at least four.
Three teams in each of five sessions gives fifteen teams, with no session reaching four.
\[5\times3=15\]
One more team must make some session reach four.
\[15+1=\boxed{16}\]
Both halves matter. Fifteen can avoid the target; sixteen cannot. Together, these show that sixteen is the smallest number that guarantees it.
With sixteen teams, the booking counts could be 4, 3, 3, 3, 3. So four in some session is guaranteed, while five in some session is not.
Optional extension: the same idea with any number of categories
Suppose there are \(b\) categories and you want \(r\) objects in some category. Here \(b\) and \(r\) are positive whole numbers, and every object may use any category.
Keeping every category at \(r-1\) objects avoids the target. The next object forces it:
\[b(r-1)+1\]
This is the smallest threshold when that arrangement is allowed. If the categories have extra restrictions, check whether the proposed avoiding arrangement is actually possible.
There is also a useful occupancy statement. If \(n\) objects are assigned to \(b\) categories, some category holds at least \(\lceil n/b\rceil\). The ceiling symbol means round up to the next whole number, keeping a whole-number result unchanged.
\[\left\lceil\frac{16}{5}\right\rceil=4\]
The arrangement 4, 3, 3, 3, 3 shows that four is all we can guarantee for sixteen teams. With fifteen teams, the guarantee is three: 3, 3, 3, 3, 3 meets it exactly.
What if the safe limits are different?
Three demonstration rooms can comfortably host 2, 3 and 4 teams respectively. Every team goes to exactly one room; any room may be chosen.
What is the smallest total number of teams that makes some room go over its own limit, whatever rooms the teams use?
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What is the smallest total number of teams that makes some room go over its own limit, whatever rooms the teams use?
What if the safe limits are different?
Three demonstration rooms can comfortably host 2, 3 and 4 teams respectively. Every team goes to exactly one room; any room may be chosen.
Your answer
The Compass checks these counts. You still need to choose useful categories, explain the guarantee and check any claim that the count is smallest.
This count fits. Compare your guarantee, avoiding example and explanation with the complete worked answer.
Break it into steps
Step 1 of 2
How many teams fit while all three room limits are respected?
\(2+3+4=9\) teams can fit, with each room at its own limit.
Step 2 of 2
What smallest total makes some room exceed its own limit?
A tenth team exceeds the combined nine-team capacity, whichever room it uses. Nine is a real safe arrangement.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Answer and reasoning: when must a room exceed its limit?
Answer: The smallest total is 10 teams.
Room 1
Limit: 2 teams
12
Room 2
Limit: 3 teams
345
Room 3
Limit: 4 teams
6789
These nine teams exactly fill the three stated limits. A tenth team makes some room exceed its own limit, whichever room is chosen.
\[2+3+4=9\]
Nine teams can fit: two in the first room, three in the second and four in the third. A tenth team cannot fit within all three limits.
\[9+1=\boxed{10}\]
CKSprout “Why does your argument cover every allowed room assignment?”
Amara “I added what all three rooms can hold while their limits are respected. Their limits do not need to be equal.”
The organisers now know when they must change the plan or add a room.
A test for the light designers
When must two designs match?
Visitors have made seven light strips. Every strip has three lights in a row. Each light is either amber or teal. The strips have identical ends: you may turn a whole strip end for end to compare it with another.
You may not rearrange individual lights or exchange the colours. Two strips match if their colours line up as they are, or after turning one end for end.
Must two of the seven strips match? What is the smallest number of three-light strips that guarantees a match?
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Interactive checking needs JavaScript. The complete question and worked answer remain available.
Must two of the seven strips match? What is the smallest number of three-light strips that guarantees a match?
A test for the light designers
When must two designs match?
Visitors have made seven light strips. Every strip has three lights in a row. Each light is either amber or teal. The strips have identical ends: you may turn a whole strip end for end to compare it with another.
You may not rearrange individual lights or exchange the colours. Two strips match if their colours line up as they are, or after turning one end for end.
A amberT teal
???
Seven strips · three lights each · either end may face left
Each position can be amber or teal. The question allows turning a whole strip; individual lights stay in order.
CKSprout “What are you allowed to do when comparing two strips? Could two orders on your list really be the same design?”
Rowan “We may turn a strip end for end. Different colour orders could still match.”
Before deciding, check what “match” allows.
Use symmetry
Turn the whole strip end for end. It keeps the same lights and colours, but reverses their order. We can use this allowed turn to compare designs.
Your answer
The Compass checks these counts. You still need to choose useful categories, explain the guarantee and check any claim that the count is smallest.
This count fits. Compare your guarantee, avoiding example and explanation with the complete worked answer.
Break it into steps
Step 1 of 5
How many raw three-light colour orders are possible before matching?
\(2\times2\times2=8\) orders. Different raw orders may still match after a turn.
Step 2 of 5
How many orders already match themselves after an end-for-end turn?
There are two choices for the ends and two for the middle: four fixed orders.
Step 3 of 5
How many categories do the other four raw orders form?
The four orders form two pairs. Reversing twice returns the original, so no order belongs to two pairs.
Step 4 of 5
How many matching categories are there altogether?
\(4+2=6\) categories. Pick one from each to build a six-strip avoiding example.
Step 5 of 5
What smallest number of three-light strips guarantees a match?
Seven strips exceed six categories and must match somewhere. Also answer the original Yes/No question and explain why six can avoid a match.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
A amberT teal
???
Seven strips · three lights each · either end may face left
Each position can be amber or teal. The question allows turning a whole strip; individual lights stay in order.
CKSprout “What are you allowed to do when comparing two strips? Could two orders on your list really be the same design?”
Rowan “We may turn a strip end for end. Different colour orders could still match.”
Before deciding, check what “match” allows.
First make the complete list
Choose amber or teal for the first light, then the second, then the third. This gives \(2\times2\times2=8\) colour orders.
A amberT teal
AAA
AAT
ATA
ATT
TAA
TAT
TTA
TTT
Fix the first colour, then the second, then the third. Every one of the eight colour orders appears once.
Turn the whole strip end for end. It keeps the same lights and colours, but reverses their order. We can use this allowed turn to compare designs.
Answer and reasoning: must two light strips match?
Answer:Yes. Seven three-light strips force a matching pair; seven is the smallest guaranteed number.
AAT
↔turn end for end
TAA
The light order reverses under an allowed turn. These two orders belong to the same matching category.
CKSprout “What happens when we turn one of these orders end for end?”
Amara “Amber–amber–teal and teal–amber–amber belong in one category. Turning either strip makes it match the other.”
A amberT teal
Category 1
AAA
Already matches itself
Category 2
AAT
TAA
The two orders match after a turn
Category 3
ATA
Already matches itself
Category 4
ATT
TTA
The two orders match after a turn
Category 5
TAT
Already matches itself
Category 6
TTT
Already matches itself
Take one strip from each row: six strips can all differ under the matching rule. A seventh strip must match a category already used.
There are six matching categories. Four orders match themselves after a turn. The other four orders form two pairs.
\[4+\frac{4}{2}=6\]
Pigeonhole principle: seven strips must share six categories. Therefore two of the seven strips must match.
\[6+1=\boxed{7}\]
Taking one design from each category gives six strips with no matching pair. That checks that seven is the smallest guaranteed threshold.
Why not divide eight by two? A design that already looks the same after a turn does not acquire a different partner. Those four categories each contain only one colour order.
Make an organized list
Find all eight colour orders without missing one.
Use symmetry
Group orders that match under the allowed turn. Eight orders become six categories.
Pigeonhole principle
Seven strips exceed six categories, forcing a match.
The list finds every order. The turn shows which orders match. The count tells us when a match cannot be avoided.
A gap in the film programme
Can two films always fill the slot?
Amara needs to fill a 25-minute screening slot with exactly two films.
She has 13 films, all with different lengths.
Each film lasts a whole number of minutes, from 1 to 24 minutes, including both ends.
The two films will play back to back, with no extra time between them.
Can she be certain that a pair will fill all 25 minutes, whatever the thirteen film lengths turn out to be? What is the smallest number of different films from this length range that guarantees such a pair?
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Can she be certain that a pair will fill all 25 minutes, whatever the thirteen film lengths turn out to be? What is the smallest number of different films from this length range that guarantees such a pair?
A gap in the film programme
Can two films always fill the slot?
Amara needs to fill a 25-minute screening slot with exactly two films.
She has 13 films, all with different lengths.
Each film lasts a whole number of minutes, from 1 to 24 minutes, including both ends.
The two films will play back to back, with no extra time between them.
We do not know the film list yet. Decide whether every allowed list must contain a pair that fills the slot.
Your answer
Enter the smallest guaranteed film count. Also answer whether the original thirteen films always contain a pair, and explain why the distinct-length condition matters.
This count fits. Compare your guarantee, avoiding example and explanation with the complete worked answer.
Break it into steps
Step 1 of 3
How many complementary length categories cover lengths one through twenty-four?
The disjoint pairs (1,24), (2,23), …, (12,13) give twelve categories.
Step 2 of 3
How many different films can still avoid every twenty-five-minute pair?
Twelve can avoid: lengths 1 through 12 have no pair summing to 25.
Step 3 of 3
What smallest number of different films guarantees such a pair?
Thirteen different lengths force two partners in one category. Answer Yes to the original promise and explain why distinct lengths matter.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
We do not know the film list yet. Decide whether every allowed list must contain a pair that fills the slot.
Answer and reasoning: is a 25-minute pair guaranteed?
Answer:Yes. Every allowed set of thirteen films contains two whose lengths add to exactly 25 minutes. The smallest guarantee is 13 different films.
Make an organized list
Rowan “Trying one film list would not prove the promise for every list. I need groups where a shared group gives us the right pair.”
CKSprout “What relationship would make a group useful?”
Amara “A one-minute film needs a 24-minute partner. Then I can pair two with 23 and keep going.”
CKSprout “How will you know when every allowed length has a place?”
All possible partner categories · minutes
1+24= 25
2+23= 25
3+22= 25
4+21= 25
5+20= 25
6+19= 25
7+18= 25
8+17= 25
9+16= 25
10+15= 25
11+14= 25
12+13= 25
These are all the possible partner pairs. Each film belongs to the pair containing its length.
These twelve pairs use every length from 1 to 24 exactly once. Two different films in the same pair fill all 25 minutes.
Pigeonhole principle: assign each film to the pair containing its length. Thirteen films go into twelve categories, so some category contains at least two films.
The lengths are different. A category contains only two possible lengths, so the two films must use the two partners. Their lengths add to 25 minutes.
\[13>12\]
Is thirteen necessary? Twelve films of lengths 1, 2, …, 12 minutes have no pair adding to 25. Their two longest lengths add to only 23. So twelve does not guarantee success.
Why use both tools? The list finds every useful pair. Pigeonhole reasoning shows that thirteen different films must fill one of those pairs.
The different lengths matter. Thirteen one-minute films would never make 25 minutes in a pair. That is why the question says all the lengths are different.
The CIC display challenge
Build the biggest display without a matching pair
The team is planning a display of five-light strips. Each of the five lights is either amber or teal. Any colour order may be made, and different visitors may make the same order.
The strip ends are identical. You may turn a whole strip end for end; two strips match if their colours line up as they are or after that turn. Individual lights cannot be rearranged, and amber cannot be exchanged with teal.
A amberT teal
?????
Five positions · amber or teal at each · either end may face left
These are unknown colours, not a proposed answer. Only a whole end-for-end turn is allowed when comparing strips.
What is the smallest number of strips that guarantees a match? Also build the largest display with no matching pair: how many strips can it have, and what is every design in that display? Explain why a bigger nonmatching display is impossible.
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What is the smallest number of strips that guarantees a match? Also build the largest display with no matching pair: how many strips can it have, and what is every design in that display? Explain why a bigger nonmatching display is impossible.
The CIC display challenge
Build the biggest display without a matching pair
The team is planning a display of five-light strips. Each of the five lights is either amber or teal. Any colour order may be made, and different visitors may make the same order.
The strip ends are identical. You may turn a whole strip end for end; two strips match if their colours line up as they are or after that turn. Individual lights cannot be rearranged, and amber cannot be exchanged with teal.
A amberT teal
?????
Five positions · amber or teal at each · either end may face left
These are unknown colours, not a proposed answer. Only a whole end-for-end turn is allowed when comparing strips.
Rowan “A count would tell us how big. We also need the actual designs to make.”
CKSprout “What would let you check that the whole display follows the matching rule?”
Your answer
Enter the forcing threshold and the size of a largest nonmatching display. Also supply every design in that display and prove maximality; those parts are not graded by this numeric check.
The two counts fit. Check your actual twenty designs, their matching categories and the proof that no larger nonmatching display is possible.
Break it into steps
Step 1 of 6
How many raw five-light colour orders are possible?
\(2^5=32\) raw orders.
Step 2 of 6
How many orders match themselves after reversal?
Three independent colour choices give \(2^3=8\) fixed orders.
Step 3 of 6
How many categories are formed by the remaining raw orders?
\((32-8)\div2=12\) two-order categories.
Step 4 of 6
How many matching categories exist altogether?
\(8+12=20\). A count alone does not give the requested actual display: use the complete design list to choose representatives.
Step 5 of 6
What smallest strip total guarantees a match?
Twenty-one strips exceed twenty categories. The matching pair need not be known in advance.
Step 6 of 6
How many strips belong in a largest nonmatching display?
Twenty is attained by the first strip in each of the twenty rows. Check every chosen design and explain why no two can match.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Rowan “A count would tell us how big. We also need the actual designs to make.”
CKSprout “What would let you check that the whole display follows the matching rule?”
Show both numbers, the full display and why it is largest
Answer:21 strips guarantee a match. The largest nonmatching display has 20 strips. Build it by taking the first strip in every row of the complete design list below.
CKSprout “Which apparent differences survive an allowed turn?”
Amara “We must group an order with its reverse. But some orders already look the same when turned.”
Rowan “I will fix each position in order so I cover all the colour orders, then keep just one design from every matching group.”
Five positions with two choices each give \(2^5=32\) colour orders. An ordered five-position choice record covers every raw order once.
A self-matching order has the same colours in positions 1 and 5, and in positions 2 and 4. The middle light is free. Choose the first two colours and the middle colour; the other two are forced.
1Pick
2Pick
3Pick
4Copy 2
5Copy 1
Each “Pick” independently allows amber or teal.
Positions 4 and 5 copy the colours in positions 2 and 1. Exactly three independent choices create every self-matching five-light order.
\[2\times2\times2=8\]
The other \(32-8=24\) orders form twelve different pairs. Reversal has one definite result, and reversing again returns to the start, so these pairs do not overlap.
\[8+\frac{32-8}{2}=20\]
These are twenty matching categories. A colour order that matches itself occupies one category, not half a category.
CKSprout “How will the people building the display know which strips to make?”
Rowan “I will keep one representative of every category. In our fixed list, I keep the first order and cross off its reverse if it is different.”
Amara “That gives a real display, and a check that none of its strips can match.”
Fix amber or teal at position one, then position two, and continue through position five. Keep a new category only the first time it appears. In the complete list below, make the first strip in each row; the other strip, when present, belongs to the same category.
A amberT teal
Category 1
AAAAA
Already matches itself
Category 2
AAAAT
TAAAA
The two orders match after a turn
Category 3
AAATA
ATAAA
The two orders match after a turn
Category 4
AAATT
TTAAA
The two orders match after a turn
Category 5
AATAA
Already matches itself
Category 6
AATAT
TATAA
The two orders match after a turn
Category 7
AATTA
ATTAA
The two orders match after a turn
Category 8
AATTT
TTTAA
The two orders match after a turn
Category 9
ATAAT
TAATA
The two orders match after a turn
Category 10
ATATA
Already matches itself
Category 11
ATATT
TTATA
The two orders match after a turn
Category 12
ATTAT
TATTA
The two orders match after a turn
Category 13
ATTTA
Already matches itself
Category 14
ATTTT
TTTTA
The two orders match after a turn
Category 15
TAAAT
Already matches itself
Category 16
TAATT
TTAAT
The two orders match after a turn
Category 17
TATAT
Already matches itself
Category 18
TATTT
TTTAT
The two orders match after a turn
Category 19
TTATT
Already matches itself
Category 20
TTTTT
Already matches itself
The complete organized list has twenty categories and all thirty-two colour orders. Take the outlined first strip in each row to build twenty strips without a match.
Every one of the thirty-two orders has one row. The twenty chosen representatives are in different rows, so no allowed turn can make two of them match. This is an actual largest nonmatching display, not just a suggested count.
Pigeonhole principle: a display without a match can use at most one strip from each of the twenty categories. Twenty-one strips therefore force two into one category, and those two match.
\[20+1=\boxed{21\text{ strips}}\]
Check both requests. The list builds twenty legal nonmatching designs. The category limit rules out every larger nonmatching display and proves that any twenty-one strips force a match. The actual matching pair depends on the visitors' strips; the proof does not identify it in advance.
Make an organized list
Cover every allowed order and give builders one actual design from each category.
Use symmetry
Group only allowed reversals; count self-matching orders once and the other orders in pairs.
Pigeonhole principle
Prove the maximum nonmatching display and the next-strip guarantee.
A symmetry count gives the number of categories. The organized list supplies the requested full design plan. The capacity argument proves why no larger nonmatching display can exist.
Your turn
Try a question before opening its answer. You can think it through in your head. For a smallest guarantee, check both why it must work and how one fewer could fail.
A bag of marbles
A bag has 5 red, 4 blue and 2 green marbles. You draw without looking and do not put any back.
This is the whole starting bag. Each drawn marble stays out of the bag.
For each question, find the smallest number of draws that makes you certain.
How many draws make you certain of 3 marbles of some one colour?
How many make you certain of 3 red marbles?
How many make you certain of at least one of every colour?
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How many draws make you certain of 3 marbles of some one colour?
How many make you certain of 3 red marbles?
How many make you certain of at least one of every colour?
A bag of marbles
A bag has 5 red, 4 blue and 2 green marbles. You draw without looking and do not put any back.
This is the whole starting bag. Each drawn marble stays out of the bag.
For each question, find the smallest number of draws that makes you certain.
Your answer
The Compass checks these counts. You still need to choose useful categories, explain the guarantee and check any claim that the count is smallest.
These counts fit. Compare your guarantee, avoiding example and explanation with the complete worked answer.
Break it into steps
Step 1 of 6
What is the largest draw count with no colour appearing three times?
Two red, two blue and both green give six draws without three of a colour.
Step 2 of 6
How many draws could still avoid a third red?
Four blue, two green and two red give eight.
Step 3 of 6
What is the largest draw count that could still miss a colour completely?
All five red and four blue give nine draws while missing green.
Step 4 of 6
What smallest draw count guarantees three of some one colour?
Seven forces three red or three blue. Green has only two in the whole bag.
Step 5 of 6
What smallest count guarantees three red?
Nine draws force a third red.
Step 6 of 6
What smallest count guarantees at least one of every colour?
Ten cannot omit green, red or blue. Check each claim against the finite original bag.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answers and reasoning
Answer:7 draws for three of some colour; 9 for three red; 10 for at least one of every colour.
To avoid three of any colour, you can draw two red, two blue and both green: six marbles. Any seventh marble must be red or blue and makes three of that colour.
To avoid three red, you can draw all four blue, both green and two red: eight. The ninth must give you a third red.
You could miss green through nine draws by taking all five red and four blue. A tenth must include green. Ten draws cannot miss red or blue either: there are only six non-red and seven non-blue marbles.
Check: Each one-fewer example uses only marbles in the bag. The three questions ask for different promises, so they have different worst cases.
Number cards, two different promises
There is one card for each whole number from 1 to 10. You choose cards without replacing them. Each promise must hold whatever cards you get.
12345678910
One card for each number. No card can be chosen twice.
What is the smallest number of cards that guarantees two adding to 11?
A further challenge: what is the smallest number that guarantees two differing by 2?
Try with the CKSTEM Thinking Compass
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What is the smallest number of cards that guarantees two adding to 11?
A further challenge: what is the smallest number that guarantees two differing by 2?
Number cards, two different promises
There is one card for each whole number from 1 to 10. You choose cards without replacing them. Each promise must hold whatever cards you get.
12345678910
One card for each number. No card can be chosen twice.
Your answer
The Compass checks these counts. You still need to choose useful categories, explain the guarantee and check any claim that the count is smallest.
These counts fit. Compare your guarantee, avoiding example and explanation with the complete worked answer.
Break it into steps
Step 1 of 4
How many disjoint groups cover the cards and force sum eleven when a group is filled?
The five pairs are (1,10), (2,9), (3,8), (4,7), (5,6).
Step 2 of 4
How many groups are in the difference-two partition used in the worked answer?
The groups (1,3), (5,7), (9), (2,4), (6,8), (10) cover every card. A shared group cannot be a singleton.
Step 3 of 4
What smallest card count guarantees a pair summing to eleven?
Six forces a pair; 6,7,8,9,10 show five can fail.
Step 4 of 4
What smallest count guarantees a pair differing by two?
Seven forces a difference-two pair; 1,2,5,6,9,10 show six can fail.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answers and reasoning
Answer:6 cards force a sum of 11; 7 cards force a difference of 2.
For the sum, use the groups (1,10), (2,9), (3,8), (4,7), (5,6). Six cards must fill both places in one group. Five can avoid the target: 6,7,8,9,10 have no pair adding to 11.
For the difference, those groups no longer help. Use (1,3), (5,7), (9), (2,4), (6,8), (10). Every card has one place. Seven cards force two into one group; a single-card group cannot hold two, so they fill a pair differing by 2.
Check the second limit: Six cards, 1,2,5,6,9,10, have no pair differing by 2. Both guarantees are smallest. A new promise needed new groups.
Birthday months
Every child has a birth month, one of the 12 months.
What is the smallest number of children that makes 3 sharing a birth month certain, whatever their months are?
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What is the smallest number of children that makes 3 sharing a birth month certain, whatever their months are?
Birthday months
Every child has a birth month, one of the 12 months.
Your answer
The Compass checks these counts. You still need to choose useful categories, explain the guarantee and check any claim that the count is smallest.
This count fits. Compare your guarantee, avoiding example and explanation with the complete worked answer.
Break it into steps
Step 1 of 3
How many children can share one month while avoiding three?
At most two per month avoid the target.
Step 2 of 3
How many children can twelve months hold at those avoiding limits?
\(12\times2=24\), with two children in each month.
Step 3 of 3
What smallest number guarantees three sharing a birth month?
Twenty-five must exceed a month’s two-child avoiding limit. Twenty-four can still avoid three.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning
Answer:25 children.
Two children in each month give \(12\times2=24\) with no three sharing. One more child must make some month reach three.
Check: The two-per-month arrangement is possible, so 24 does not guarantee three. We have not named the shared month; that depends on the children.
A contest card challenge
There is one card for each whole number from 15 to 43, including both ends. Cards are drawn without replacement.
What is the smallest number of draws that makes you certain of two cards whose product is divisible by 4?
Try with the CKSTEM Thinking Compass
Interactive checking needs JavaScript. The complete question and worked answer remain available.
What is the smallest number of draws that makes you certain of two cards whose product is divisible by 4?
A contest card challenge
There is one card for each whole number from 15 to 43, including both ends. Cards are drawn without replacement.
Your answer
The Compass checks these counts. You still need to choose useful categories, explain the guarantee and check any claim that the count is smallest.
This count fits. Compare your guarantee, avoiding example and explanation with the complete worked answer.
Break it into steps
Step 1 of 3
How many odd cards are there from fifteen through forty-three inclusive?
15,17,…,43 contain fifteen odd numbers.
Step 2 of 3
How many cards can the explicit product-avoiding example contain?
The sixteen cards avoid the target: odd×odd stays odd, and eighteen×odd has only one factor of two.
Step 3 of 3
What smallest draw count guarantees a product divisible by four?
Seventeen forces two even numbers; their product has at least two factors of two. The sixteen-card example proves smallest.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning
Answer:17 cards.
There are 15 odd numbers in the set. So 17 cards must include at least two even numbers. Their product contains two factors of 2, so it is divisible by 4.
Sixteen can still fail: take all the odd cards, 15,17,19,…,43, and the card 18. Two odd numbers have an odd product. The product of 18 and an odd number has only one factor of 2, so it is not divisible by 4.
Check: All 16 cards are different and lie between 15 and 43. This is a failing example with one fewer, while every set of 17 succeeds.
What forces the result?
How much can fit while the result I want is still avoided?
Name the guarantee. What exactly must happen?
Choose the objects and categories. Does every object have one category? Why does a shared category prove the target?
Find the avoiding limit. What is the largest total that could still fail?
Compare and check. Does the total exceed that limit? If you claim the smallest number, build a failing example with one fewer.
Amara “We didn’t know the bookings. But we could still be sure that some session would be shared.”
CKSprout “What made your categories useful in each problem? What would you check before trusting a new set of categories?”
When a new problem asks must, certain or how many to guarantee, keep this tool in mind. Your first useful action might be to draw categories and try to make the desired event fail for as long as possible.
Amara “The organisers need to plan for crowds, but visitors will choose their sessions.”
CKSprout “We don’t have the bookings yet. Can we still know that some teams will share a session?”
Rowan “What if they spread out as much as possible?”