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The mosaic suitcase
Find an invariant
When everything seems to change, look for something the rules keep fixed.
A mountain journey, a suitcase of pieces, and a gift that needs a better plan.
Mina and Amara are making a magnetic mosaic for a friend while their train crosses a mountain valley. Two little windows will hold pictures. The other squares must be covered with tiles.
Mina “I have enough tiles. Why do my last pieces keep getting stuck?”
CKRobo “What would tell us whether another arrangement could work?”
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.
Enough area, but will the pieces fit?
The board has 8 rows of 8 squares. The top-left square, (row 1, column 1), and bottom-right square, (row 8, column 8), are picture windows.
There are 31 straight domino tiles. Each covers exactly two whole squares sharing an edge. Tiles may lie horizontally or vertically. No cutting, overlap, covering a window or leaving any other square empty.
Can the 31 tiles cover every usable square exactly? Give a cover, or explain why no arrangement can work.
CKSTEM Thinking CompassPlan and check your reasoning.
Can the 31 tiles cover every usable square exactly? Give a cover, or explain why no arrangement can work.
Enough area, but will the pieces fit?
The board has 8 rows of 8 squares. The top-left square, (row 1, column 1), and bottom-right square, (row 8, column 8), are picture windows.
There are 31 straight domino tiles. Each covers exactly two whole squares sharing an edge. Tiles may lie horizontally or vertically. No cutting, overlap, covering a window or leaving any other square empty.
Rows count down from the top; columns count from the left. Crosses mark the two windows. All other cells must be covered.
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.
Rows count down from the top; columns count from the left. Crosses mark the two windows. All other cells must be covered.A question to help me begin
What does every permitted tile placement have in common? Could a mark or count help you compare all placements?
Show the answer and reasoning: the two windows
Answer:No arrangement can work. Every domino covers one square of each checkerboard colour, but the usable board has 30 dark and 32 light squares.
Area
The area check passes: \(8\times8-2=62\) usable squares, and \(31\times2=62\). That counts how much the pieces cover; it does not prove they fit.
Mina “What if I colour alternate squares? Every horizontal tile would cover one of each.”
CKRobo “Does your observation cover every permitted orientation?”
Amara “Vertical tiles do too. Both windows remove dark squares, so the two colours no longer have equal counts.”
Horizontal: one of each.
Vertical: one of each.
Both legal orientations cover opposite colours. Dark cells have a filled dot; light cells have an open ring.
30 dark · filled dot32 light · open ring
The full board had 32 of each. Both picture windows remove dark squares.
Imagine placing or picking up tiles. Each move changes the number of uncovered dark squares and light squares by the same amount. The excess of two light squares never changes. A fully covered board would have no uncovered squares and no excess.
For example, placing one domino leaves 31 light and 29 dark squares uncovered. The difference is still two:
The colours are our reasoning aid, not extra rules. They cover every legal position and both orientations. Moving the pieces again cannot fix this board; changing the design can.
Find an invariant
An invariant is a quantity or property that every allowed move keeps unchanged. The number of uncovered squares changed, but the difference between the two colour counts did not.
Choose what to track. Test it against every kind of legal move. Then compare it with the requested finish. A mismatch rules out every route at once. A match leaves more to check.
Both 30 and 32 are even; their exact difference still matters. An invariant can tell us more than even and odd alone.
Try a different outline
Amara tries the eight-square stencil below. Only the outlined squares belong to it. Use four straight dominoes, each covering exactly two outlined squares that share an edge, horizontally or vertically. No cutting, overlap or uncovered outlined squares.
Can the stencil be covered? Give a complete cover or prove that none exists.
CKSTEM Thinking CompassPlan and check your reasoning.
Can the stencil be covered? Give a complete cover or prove that none exists.
Try a different outline
Amara tries the eight-square stencil below. Only the outlined squares belong to it. Use four straight dominoes, each covering exactly two outlined squares that share an edge, horizontally or vertically. No cutting, overlap or uncovered outlined squares.
Only (1,2), (2,1), (2,2), (2,3), (2,4), (3,3), (4,3) and (4,4) belong to the stencil. Coordinates give (row, column); all other positions are absent.
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.
Only (1,2), (2,1), (2,2), (2,3), (2,4), (3,3), (4,3) and (4,4) belong to the stencil. Coordinates give (row, column); all other positions are absent.Show the answer and reasoning: the stencil
Answer:No. The colour counts match, but two different tip squares would both need the same neighbouring square.
Amara “Four dark and four light squares. Our colour test doesn’t reject it.”
CKRobo “What does that test leave undecided?”
Mina “Whether the tiles fit the actual boundary. These two tips each have only one possible partner—and it’s the same square.”
There are four cells of each colour. The two red-outlined tips both require the same partner, (2,2).
Square (1,2) can pair only with (2,2). Square (2,1) can also pair only with (2,2). Covering both tips would use (2,2) twice, which is forbidden. That settles the whole stencil.
Equal colour counts are necessary for a domino cover, but they are not sufficient for every shape. The invariant has done useful work; it has not answered every fitting question.
Change the pieces, check the argument again
A smaller board has 4 rows of 4 squares, with windows at (1,1) and (4,4). The new kit contains two L-shaped pieces, each made from three squares of a \(2\times2\) block, and four straight dominoes.
Use every piece. Rotations are allowed. All pieces must cover whole usable squares exactly, with no cutting, overlap, covered window or gap.
Does the old domino colour argument still rule out a cover? Can this new kit cover the board?
CKSTEM Thinking CompassPlan and check your reasoning.
Does the old domino colour argument still rule out a cover? Can this new kit cover the board?
Change the pieces, check the argument again
A smaller board has 4 rows of 4 squares, with windows at (1,1) and (4,4). The new kit contains two L-shaped pieces, each made from three squares of a \(2\times2\) block, and four straight dominoes.
Use every piece. Rotations are allowed. All pieces must cover whole usable squares exactly, with no cutting, overlap, covered window or gap.
2 L-shaped pieces
4 dominoes
The displayed shapes define the pieces. Use two copies of the L and four copies of the domino.
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.
2 L-shaped pieces
4 dominoes
The displayed shapes define the pieces. Use two copies of the L and four copies of the domino.A question to help me begin
If you colour alternate squares, what can one L-shaped piece cover? Does the old claim about every piece still hold?
Show the answer and reasoning: the changed kit
Answer: The old invariant does not survive the new rule. Yes, this kit covers the board: put one L around each window and four dominoes in the remaining two blocks.
Mina “An L can cover two light squares and one dark square. The difference can change now.”
CKRobo “How will you check the whole proposed cover?”
Amara “I’ll account for every piece and every usable square, including those beside both windows.”
Matching letters identify one piece. A and B are the Ls; C–F are the four dominoes. Windows remain clear.
The upper-left L covers (1,2), (2,1), (2,2). The lower-right L covers (3,3), (3,4), (4,3). The four vertical dominoes fill the upper-right and lower-left \(2\times2\) blocks.
\[
2\times3+4\times2=14=4\times4-2.
\]
Every usable square appears once; both windows stay clear. Each L here removes one more light square than dark, together removing the board’s original excess of two. A property is invariant under specified moves, not under every rule we might invent.
During the journey: what stays in the tins?
The train enters a tunnel. The children pause the mosaic and share their snacks.
Three tins begin with numbers of fruit pieces in the ratio 4 : 7 : 3. That means 4, 7 and 3 equal-size counting units; the number of pieces in one unit is unknown.
First, the children move the same whole number of pieces from the middle tin to each outer tin, making one gift to each. The first and middle tins then have equal counts. At this checkpoint, the third tin has 28 pieces.
Later, the children move pieces between the tins several more times. These amounts are not recorded. Throughout all the sharing, nothing is eaten, lost or added.
How many fruit pieces are there altogether after all the sharing?
First tin
4 equal-size units
Middle tin
7 equal-size units
Third tin
3 equal-size units
The same gift goes from the middle to each outer tin. Recorded checkpoint: first = middle; third = 28 pieces. Later transfers between the tins are unrecorded.
Each small block represents the same unknown number of pieces. No pieces enter or leave the three tins.Try with the CKSTEM Thinking Compass
Interactive checking needs JavaScript. The complete question and worked answer remain available.
How many fruit pieces are there altogether after all the sharing?
During the journey: what stays in the tins?
The train enters a tunnel. The children pause the mosaic and share their snacks.
Three tins begin with numbers of fruit pieces in the ratio 4 : 7 : 3. That means 4, 7 and 3 equal-size counting units; the number of pieces in one unit is unknown.
First, the children move the same whole number of pieces from the middle tin to each outer tin, making one gift to each. The first and middle tins then have equal counts. At this checkpoint, the third tin has 28 pieces.
Later, the children move pieces between the tins several more times. These amounts are not recorded. Throughout all the sharing, nothing is eaten, lost or added.
First tin
4 equal-size units
Middle tin
7 equal-size units
Third tin
3 equal-size units
The same gift goes from the middle to each outer tin. Recorded checkpoint: first = middle; third = 28 pieces. Later transfers between the tins are unrecorded.
Each small block represents the same unknown number of pieces. No pieces enter or leave the three tins.
Your answer
Enter the total after all the sharing, including the unrecorded transfers. Explain why the whole can be known even when the final tins cannot.
This value fits. Compare the chosen invariant, every original rule and the unscored parts with the full answer and reasoning.
Break it into steps
Step 1 of 5
Work in the original equal-size units. How many units must go in each equal gift when the first two tins become equal?
The original gap is \(7-4=3\) units. One gift closes it by three gift amounts: the first gains one and the middle loses two. Each gift is one original unit.
Step 2 of 5
How many original units are in the third tin after those gifts?
The third tin finishes with \(3+1=4\) original units.
Step 3 of 5
At the recorded checkpoint the third tin has 28 pieces. How many pieces are in one original unit?
\(28\div4=7\) pieces per unit.
Step 4 of 5
How many original units are there across all three tins together?
\(4+7+3=14\) units. All those pieces are still in the three tins after sharing.
Step 5 of 5
How many pieces are in the three tins altogether after all the sharing?
\(14\times7=98\) pieces. Later transfers are unrecorded, so the final individual counts are unknown. The conserved whole still answers the question.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
A question to help me begin
The first tin receives one gift. How many gifts does the middle tin give? How does this change the gap between them?
Show the answer and reasoning: the final snack total
Answer: There are 98 pieces altogether after all the sharing. Moving pieces between tins preserves the total.
Amara “During the two equal gifts, the first tin gains one gift; the middle loses two. That closes their gap by three gifts.”
CKRobo “How does that help identify the amount moved?”
Amara “Their original gap is three units, so each gift is one unit. At that checkpoint the third tin has grown from three units to four.”
Mina “Four units make 28, so one unit is seven. But the later transfers aren’t recorded. I can’t reconstruct each final tin.”
CKRobo “What can those later moves change—and what can they leave unchanged?”
Mina “They change where the pieces are, but none leave the three tins. Our original fourteen units are still there altogether.”
First gains 1 unit ←Middle gives 2 units→ Third gains 1 unit
First at checkpoint
5 units
Middle at checkpoint
5 units
Third at checkpoint
4 units
At the recorded checkpoint, two equal gifts have closed the first–middle gap. The 5 + 5 + 4 units still total 14. Later transfers change the individual counts, with the whole unchanged.
\[
\begin{aligned}28\div4&=7,\\4+7+3&=14,\\14\times7&=98.\end{aligned}
\]
Check the recorded checkpoint: \((28,49,21)\to(35,35,28)\). The middle tin gives seven to each outer tin, has enough to do so, and becomes equal to the first. Both totals are 98.
The later transfers may change all three individual counts, so those final counts are not determined. Every piece still belongs to one of the tins. The total stays 98 throughout, which answers the question even though the final arrangement is unknown. The gap between individual tins is not the invariant.
A design across two halves
Outside the window, mountains and their reflection meet at the lake. Inside the train, the children try a new board for their gift.
A 5-row, 6-column board has windows at (3,2) and (3,5). Cover every other square with straight dominoes. Each domino covers two whole squares sharing an edge, horizontally or vertically. No cutting, overlap, window coverage or gap.
The marked middle line lies between columns 3 and 4. A domino crosses it only when it covers one square on each side. The completed cover may have any arrangement.
What is the fewest dominoes that can cross the middle line in a complete cover? Show a complete cover attaining that number.
The dashed line lies between columns 3 and 4. A crossing domino uses one cell on each side.Try with the CKSTEM Thinking Compass
Interactive checking needs JavaScript. The complete question and worked answer remain available.
What is the fewest dominoes that can cross the middle line in a complete cover? Show a complete cover attaining that number.
A design across two halves
Outside the window, mountains and their reflection meet at the lake. Inside the train, the children try a new board for their gift.
A 5-row, 6-column board has windows at (3,2) and (3,5). Cover every other square with straight dominoes. Each domino covers two whole squares sharing an edge, horizontally or vertically. No cutting, overlap, window coverage or gap.
The marked middle line lies between columns 3 and 4. A domino crosses it only when it covers one square on each side. The completed cover may have any arrangement.
The dashed line lies between columns 3 and 4. A crossing domino uses one cell on each side.
Your answer
Enter the crossing minimum. Your complete cover and reason that fewer cannot work remain unscored; compare both with the full answer.
This value fits. Compare the chosen invariant, every original rule and the unscored parts with the full answer and reasoning.
Break it into steps
Step 1 of 6
Colour alternate squares, starting with a dark top-left square. How many usable dark squares are in columns 1–3 before placing tiles?
The left half contains eight usable dark squares.
Step 2 of 6
With that same colouring, how many usable light squares are in columns 1–3?
Seven light squares were in the full left rectangle; its window leaves six usable light squares.
Step 3 of 6
Track left squares until a wholly-left domino covers them, keeping crossing endpoints counted. What dark-minus-light difference stays fixed?
The difference is \(8-6=2\). Wholly-left placements remove one of each; the other placements leave both tracked counts unchanged. At completion the tracked cells are exactly the left crossing endpoints.
Step 4 of 6
With the same colouring, how many dark squares in column 3 could serve as left ends of crossing dominoes? Choose positions and check a whole cover yourself; those choices are unscored.
Rows 1, 3 and 5 offer three dark left endpoints. To attain the bound with two crossings, choose two dark endpoints and construct the remaining cover. Counting candidates alone does not show that a chosen pair works.
Step 5 of 6
How many dominoes does any complete cover of this board use?
\(28\div2=14\) dominoes are needed for any complete cover. Use this total to check an attaining cover. It does not choose a layout or require symmetry; check the actual positions too.
Step 6 of 6
What is the least number of crossings that a complete legal cover can attain?
The ledger requires at least two crossings, and the shown cover has two. Both parts are needed for the minimum proof; the Compass checks only the number.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
A question to help me begin
What can a piece wholly on one side cover? How is a piece crossing the line different? Check any proposed count against both kinds of placement.
Show the answer and reasoning: the minimum crossings
Answer: The minimum is two crossing dominoes. The complete 14-domino cover below attains it; reflection is a useful choice for building it.
Find an invariant
Left half only: eight dark and six light. Within-half dominoes cannot remove that excess of two dark cells.
Amara “The left half has eight dark squares and six light ones. A tile entirely on the left always takes one of each.”
Mina “But a crossing tile takes just one left square. The uncovered colour difference could change.”
CKRobo “What will you count so that your claim covers those placements too?”
Amara “I’ll keep a left square in our count until a tile lies entirely on the left over it. So crossing tiles leave their left squares in the count.”
Track the left squares not covered by a wholly-left domino: squares still uncovered and left squares covered by crossing dominoes. A wholly-left tile removes one dark and one light from this count. A right-hand or crossing tile changes neither count. Picking up a tile reverses the same changes. The dark-minus-light difference stays 2 under every allowed move.
Track left cells until a wholly-left domino covers them.
Legal placement
Tracked dark
Tracked light
Difference changes by
Wholly left
−1
−1
0
Wholly right
0
0
0
Crossing the line
0
0
0
Keep a crossing tile’s left cell in this ledger. All three placement types preserve the initial difference of 2; removing a tile reverses the same changes.
Mina “When the cover is finished, our tracked squares are exactly the left ends of the crossing tiles.”
Amara “They must include two more dark squares than light ones. So there must be at least two of them—and one crossing tile supplies only one left end.”
CKRobo “Can you build a whole cover that meets that lower bound?”
Zero or one crossing tile cannot supply that difference. This is a lower bound for every legal cover, whether reflected or not. A passing count still does not put the other tiles in place.
You can also read the finished cover as a still picture: wholly-left tiles cover equal numbers of the two colours, so the excess of two dark squares must appear among the crossing ends.
Amara “I can cross in the first and last rows, using two dark squares on the left. Then I can cover the rest of that half with six tiles.”
Mina “The board and windows match across the middle. What if we reflect those six tiles to finish the other half?”
CKRobo “Check that every reflected piece stays legal—and which pieces you should count only once.”
Use symmetry
Reflecting the six legal left-hand pieces supplies the right half because the board, windows and edge-neighbour rule are preserved. The two crossing pieces reflect onto themselves, so keep each once. This reduces the construction work; the problem did not require a reflected design.
A–F cover the left; G–L are their reflected partners; M and N cross the axis. Letters are checking aids, not markings required to reflect.
The crossings cover (1,3)–(1,4) and (5,3)–(5,4). Each of the six left-hand pieces has a separate reflected partner:
The model checks every usable square and leaves both windows empty. Its two crossings meet the lower bound, proving the minimum. A different complete two-crossing cover would be equally valid; the reflected construction is a way to find and check one efficiently.
Choose the windows for the postcard
The gift is ready. Mina wants to add a reusable puzzle postcard, so their friend can choose the picture windows.
Use a 3-row, 4-column board. Choose two distinct squares as windows. At least one must be one of the four corners: (1,1), (1,4), (3,1) or (3,4). Choosing two corners is allowed.
Five straight dominoes must cover the other ten squares exactly. Each covers two whole squares sharing an edge, horizontally or vertically. No cutting, overlap, covered window or gap.
We count window pairs. Swapping their order gives the same pair. Different tilings of the same pair do not count again.
How many window pairs work? Explain why every pair you count really has a cover.
Choose two windows; at least one must be one of the four marked corners.
No windows have been chosen yet. Domino positions do not count as additional window pairs.Try with the CKSTEM Thinking Compass
Interactive checking needs JavaScript. The complete question and worked answer remain available.
How many window pairs work? Explain why every pair you count really has a cover.
Choose the windows for the postcard
The gift is ready. Mina wants to add a reusable puzzle postcard, so their friend can choose the picture windows.
Use a 3-row, 4-column board. Choose two distinct squares as windows. At least one must be one of the four corners: (1,1), (1,4), (3,1) or (3,4). Choosing two corners is allowed.
Five straight dominoes must cover the other ten squares exactly. Each covers two whole squares sharing an edge, horizontally or vertically. No cutting, overlap, covered window or gap.
We count window pairs. Swapping their order gives the same pair. Different tilings of the same pair do not count again.
Choose two windows; at least one must be one of the four marked corners.
No windows have been chosen yet. Domino positions do not count as additional window pairs.
Your answer
Enter the number of valid window pairs. The full construction and completeness proof remain unscored; check them in the complete answer.
This value fits. Compare the chosen invariant, every original rule and the unscored parts with the full answer and reasoning.
Break it into steps
Step 1 of 7
Colour the whole 3-by-4 board alternately from a dark top-left. How many dark squares are there before choosing windows?
Each row has two dark squares, giving six in the whole board.
Step 2 of 7
With that same colouring, how many light squares are there?
There are six light squares too.
Step 3 of 7
How many opposite-colour candidate window pairs does a dark-first list contain?
\(6\times6=36\) candidates, without reversed duplicates. The complete-loop argument in the answer is still needed to prove every candidate tileable.
Step 4 of 7
After excluding all four corners from the choices, how many dark squares remain?
\(6-2=4\) noncorner dark choices remain.
Step 5 of 7
After excluding all four corners, how many light squares remain?
\(6-2=4\) noncorner light choices remain.
Step 6 of 7
How many opposite-colour window pairs use neither corner?
\(4\times4=16\) unwanted pairs. They are a disjoint part of the same 36-pair whole.
Step 7 of 7
How many valid window pairs use at least one of the four corners?
\(36-16=20\) window pairs. Check the whole loop construction and completeness argument in the answer; those are not graded by this number.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
A question to help me begin
Which pairs can you rule out? What would establish that the remaining pairs fit? Then decide how to count the corner condition without repeats.
Show the answer and complete reasoning: the postcard
Answer:20 window pairs work. Each removes one square of each checkerboard colour and uses at least one of the four corners. The loop below supplies a cover for every counted pair.
Find an invariant
The full board has six dark and six light squares. Every domino uses one of each, so the difference in the uncovered colour counts stays fixed. A finished cover has difference zero: the windows must remove one dark and one light square. That rejects same-colour pairs, but we still need to prove the remaining pairs fit this particular board.
Mina “If I list a first window and then a second, the same pair could turn up in the opposite order.”
CKRobo “What could give each candidate pair one place in your list?”
Amara “Put its dark window first. Fix each dark square in row order, then list the light partners in row order. Every pair will have exactly one home.”
Make an organized list
Fix each of the six dark choices in turn and include all six light partners. The list has 36 candidate pairs, with no repeated unordered pair. Set the corner condition aside for a moment; a candidate still needs a legal cover.
Dark window (1,1)
(1,2)(1,4)(2,1)(2,3)(3,2)(3,4)
6 light partners
Dark window (1,3)
(1,2)(1,4)(2,1)(2,3)(3,2)(3,4)
6 light partners
Dark window (2,2)
(1,2)(1,4)(2,1)(2,3)(3,2)(3,4)
6 light partners
Dark window (2,4)
(1,2)(1,4)(2,1)(2,3)(3,2)(3,4)
6 light partners
Dark window (3,1)
(1,2)(1,4)(2,1)(2,3)(3,2)(3,4)
6 light partners
Dark window (3,3)
(1,2)(1,4)(2,1)(2,3)(3,2)(3,4)
6 light partners
Fix each of the six dark choices in turn, then include all six light partners. This lists 36 pairs without repeats.
Amara “I found a loop through every square, always stepping across an edge. The colours alternate all the way around.”
CKRobo “How could that help with every proposed dark–light pair?”
Mina “Remove one odd position and one even position. Each remaining run has an even number of squares, so we can pair neighbours along both runs.”
Follow 1 → 2 → … → 12 → 1. Each step crosses a shared edge.
Odd loop positions are dark; even positions are light. Removing one of each splits the loop into two even-length runs.
Along this 12-position loop, an odd position and an even position are an odd number of steps apart in either direction. Removing the two endpoints leaves an even number of squares in each run. Pair consecutive squares in the runs. Every pair shares an edge, no square repeats, and all ten remaining cells are covered. An empty run needs no tiles.
Example: remove loop positions 1 and 4. Pair the two cells in one run and the eight in the other: five legal dominoes.
The example removes the two top corners. The same construction works for all 36 dark–light pairs, so these form our verified whole. This uses the complete loop on this rectangle; equal colour counts alone did not settle the earlier stencil.
Mina “I could make four lists, one for each corner. But a pair of two corners would be in two lists.”
CKRobo “Is there a simpler group you could leave out of a complete count?”
Amara “Pairs using no corner. There are four noncorner dark squares and four noncorner light squares.”
Count the complement
Within the same 36 verified pairs, the exact opposite of “at least one corner” is neither window is a corner. Those unwanted pairs have four dark choices and four light choices, giving \(4\times4=16\). This avoids merging four overlapping corner-first lists.
Whole: 36 workable pairs
One dark window and one light window.
16 unwantedNeither window is a corner
20 wantedAt least one corner
These two groups are disjoint and exhaust the same 36-pair universe. Of the six pairs using two corners, two same-colour pairs were rejected. The four opposite-colour pairs belong to the wanted group once.
\[
36-16=\boxed{20}.
\]
Another valid count
First count pairs whose dark window is a corner: \(2\times6=12\). For the rest, the dark window is a noncorner and the light window must be a corner: \(4\times2=8\). These disjoint groups give \(12+8=20\). This direct count is valid too. The complement makes the unwanted group particularly easy to describe.
Check the boundary: choosing two of the four corners gives six pairs. Two pairs use the same colour and were already rejected: (1,1) with (3,1), and (1,4) with (3,4). The other \(2\times2=4\) opposite-colour corner pairs are among our 20. Asking for exactly one corner would exclude those four workable pairs and give 16, which is a different question. We have counted pairs, not their order or their possible tilings.
How the ideas shared the work
Find an invariant rejected same-colour windows. A separate loop construction proved every remaining candidate tileable.
Make an organized list gave each unordered pair a unique place in the 36-pair whole.
Count the complement handled the corner condition by excluding the disjoint 16-pair unwanted group.
Mina puts the postcard beside the mosaic. Amara adds the mountain sketch. Their friend will get a gift, a puzzle with a checked promise, and room to discover an explanation of their own.
Choose what to track. Think through the whole request before opening an answer.
Two piles of books
Two piles start with 10 and 14 books. A move transfers exactly two books from either pile to the other whenever there are enough. No books are added or lost.
Can the piles become 12 and 12? Can they become 11 and 13? Give a legal route for each possible target and a reason for each impossible one.
CKSTEM Thinking CompassPlan and check your reasoning.
Can the piles become 12 and 12? Can they become 11 and 13? Give a legal route for each possible target and a reason for each impossible one.
Two piles of books
Two piles start with 10 and 14 books. A move transfers exactly two books from either pile to the other whenever there are enough. No books are added or lost.
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.
Show the answer and reasoning: the books
Answer:12 and 12 is possible; 11 and 13 is impossible. Move two books from the second pile to the first to reach 12 and 12.
The total stays 24, so both targets pass that test. But each pile starts even and changes by two, so each stays even. That rules out 11 and 13. The stated transfer constructs the possible target.
Four corner counters
A, B, C and D go around a square in that order. Their counts start at (2, 0, 0, 0). One move adds one to both ends of any edge: A–B, B–C, C–D or D–A. No subtraction or diagonal moves.
Can all four counts become 2? Can they become (3, 1, 1, 1), in A–B–C–D order? Justify both answers and give a route whenever possible.
CKSTEM Thinking CompassPlan and check your reasoning.
Can all four counts become 2? Can they become (3, 1, 1, 1), in A–B–C–D order? Justify both answers and give a route whenever possible.
Four corner counters
A, B, C and D go around a square in that order. Their counts start at (2, 0, 0, 0). One move adds one to both ends of any edge: A–B, B–C, C–D or D–A. No subtraction or diagonal moves.
Each move uses one drawn edge and adds one to both endpoint counts. No diagonal is an allowed edge.
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.
Each move uses one drawn edge and adds one to both endpoint counts. No diagonal is an allowed edge.Show the answer and reasoning: the counters
Answer:All four cannot become 2. The target (3, 1, 1, 1) is possible: choose edge A–B, then C–D.
Track \((A+C)-(B+D)\). Every edge adds one to the first bracket and one to the second, preserving their difference of 2. The all-2 target has difference 0, so it is impossible. After A–B, the counts are \((3,1,0,0)\). Then C–D gives \((3,1,1,1)\), checking the other target.
Two game scores
Two scores start at 18 and 8. Every round adds exactly one to each. Scores never decrease.
Can the first score become twice the second? Three times the second? Explain each answer and how many rounds any possible target needs.
CKSTEM Thinking CompassPlan and check your reasoning.
Can the first score become twice the second? Three times the second? Explain each answer and how many rounds any possible target needs.
Two game scores
Two scores start at 18 and 8. Every round adds exactly one to each. Scores never decrease.
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.
Show the answer and reasoning: the scores
Answer:Twice is possible after two rounds, at 20 and 10. Three times is impossible.
The difference always stays 10. For twice, the difference is one copy of the smaller score, so that score must be 10; two rounds produce 20 and 10. For three times, the difference is two copies of the smaller score, so it would have to be 5 and the first 15. Both are below their starts. Addition-only rounds cannot reach them.
Look back
The children stopped asking only where to put the next piece. They chose a property, checked it against the actual rules and asked what it could settle. A changed rule demanded a fresh check. A passing invariant still needed a legal plan.
For a new problem, ask: What changes? What might stay fixed? Have I checked every allowed move—and what remains to prove?
Mina “I have enough tiles. Why do my last pieces keep getting stuck?”
CKRobo “What would tell us whether another arrangement could work?”
Amara “Vertical tiles do too. Both windows remove dark squares, so the two colours no longer have equal counts.”