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A problem-solving strategy
Look for a pattern
Notice a rule. Discover why it works. Reach beyond what you can see.
A workshop journey to the Moon: Amara builds, Li Wei tests, and CKRobo listens to their ideas.
Tonight, the science club is travelling to the Moon—without leaving its workshop. Amara is designing a model base that visitors can explore. Li Wei is helping its miniature rover find its way. CKRobo has brought a box of tiny lights.
Amara “I want visitors to understand how we built it, not just watch it glow.”
Li Wei “Then our explanations have to work even when the model gets bigger.”
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.
01
Lights for a lunar doorway
Amara builds a stair-shaped triangular sign. Its top row has one light, and every new row below has one more light than the previous row. All rows start at the same straight edge. No lights are shared between rows.
1 row1 light
2 rows3 lights
3 rows6 lights
4 rows10 lights
The first four signs, with their row counts and total lights. Use the same construction for the finished sign.
The finished sign needs eight rows. How many lights will it need, and will the 36 lights in one box be enough?
Try with the CKSTEM Thinking Compass
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The finished sign needs eight rows. How many lights will it need, and will the 36 lights in one box be enough?
Amara builds a stair-shaped triangular sign. Its top row has one light, and every new row below has one more light than the previous row. All rows start at the same straight edge. No lights are shared between rows.
1 row1 light
2 rows3 lights
3 rows6 lights
4 rows10 lights
The first four signs, with their row counts and total lights. Use the same construction for the finished sign.
Li Wei “The totals go 1, 3, 6, 10. Could we keep adding three?”
CKRobo “How could you test that idea against the way the sign is built?”
Your answer
The Compass checks the number only. Also answer every other part of the question and explain why your pattern continues.
This number fits. Compare the rest of your answer and reasoning with the complete worked solution.
Break it into steps
Step 1 of 3
How many lights are in the first four rows?
\(1+2+3+4=10\) lights.
Step 2 of 3
How many lights do rows five through eight add?
\(5+6+7+8=26\). These rows are longer than the early rows.
Step 3 of 3
How many lights does the eight-row sign need?
\(10+26=36\). Also answer whether the box is enough and explain why the counting rule continues.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Li Wei “The totals go 1, 3, 6, 10. Could we keep adding three?”
CKRobo “How could you test that idea against the way the sign is built?”
Show the answer and reasoning: eight rows
Answer:Yes. Eight rows need exactly 36 lights, so one box is enough.
Amara “Adding three fits the jump from three to six. But the next row has four lights, and the row after that has five.”
Li Wei “The new rows get longer. The amount we add grows too.”
The eight-row sign has rows of one through eight lights, with no shared lights.
Each row from one to eight is counted once. The total uses the whole box, with no missing row and no lights left over. We can continue because we know how the sign is built—not just because the first totals look familiar.
What makes a pattern useful?
Look for a pattern means comparing cases, noticing a relationship, and finding a reason that relationship continues. The pattern can help you predict, count or avoid repeating the same work.
Use it when a situation repeats, when a design grows in a regular way, or when a large question is built from smaller cases. Ask “What changes, what stays the same, and why?”
Noticing is the beginning. Explaining is what makes the shortcut dependable.
Can we count a much taller sign?
The next sign needs twelve rows. Its rows start at the same straight edge: one light in the first row, then one more in each new row below. No lights are shared. Count all twelve rows.
How many lights will it need? Can you find the total without adding every row separately?
Try with the CKSTEM Thinking Compass
Interactive checking needs JavaScript. The complete question and worked answer remain available.
How many lights will it need? Can you find the total without adding every row separately?
Can we count a much taller sign?
The next sign needs twelve rows. Its rows start at the same straight edge: one light in the first row, then one more in each new row below. No lights are shared. Count all twelve rows.
Your answer
The Compass checks the number only. Also answer every other part of the question and explain why your pattern continues.
This number fits. Compare the rest of your answer and reasoning with the complete worked solution.
Break it into steps
Step 1 of 3
When two twelve-row triangles join, how many lights are in each combined row?
\(1+12=2+11=\cdots=13\) lights per joined row.
Step 2 of 3
How many lights are in both twelve-row triangles together?
\(12\times13=156\) lights belong to two equal triangles.
Step 3 of 3
How many lights are in one twelve-row sign?
\(156\div2=78\). Check that each original row has exactly one partner.
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 could you make with two copies of the same triangle?
Show the answer and reasoning: twelve rows
Answer: A twelve-row sign needs 78 lights.
Amara first tries the eight-row sign they already know.
Amara “What if I made a second copy and turned it around?”
Li Wei “The shortest row fits beside the longest. Every joined row has nine lights!”
● First copy● Turned copy
8 rows × 9 lights = 72 lights in two equal copies
In every joined row the two counts add to nine: 1+8, 2+7, 3+6, and so on. Each colour contains one whole 36-light triangle.
Eight joined rows of nine contain \(8\times 9=72\) lights. The two triangles are equal, so one has \(72\div 2=36\).
For twelve rows, the paired row counts are \(1+12\), \(2+11\), and so on. Every joined row has thirteen lights. Twelve such rows hold two equal triangles:
\[12\times 13\div 2=78\text{ lights}\]
Every original row has one partner; none is missed or counted twice within either triangle. As a separate check, \(1+\cdots+6=21\) and \(7+\cdots+12=57\), giving \(21+57=78\).
A rule for any number of rows
Let \(n\) be the number of rows. Each of the \(n\) joined rows has \(n+1\) lights. Two equal triangles contain \(n(n+1)\), so:
The construction explains every row count; a few matching totals alone would not prove the rule.
A bigger sign: does doubling work?
The triangular sign has one light in its first row and one more in every new row below, all starting at the same straight edge. No lights are shared.
Eight rows need 36 lights. How many lights will sixteen rows need? Is doubling 36 a valid shortcut?
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Eight rows need 36 lights. How many lights will sixteen rows need? Is doubling 36 a valid shortcut?
A bigger sign: does doubling work?
The triangular sign has one light in its first row and one more in every new row below, all starting at the same straight edge. No lights are shared.
Your answer
The Compass checks the number only. Also answer every other part of the question and explain why your pattern continues.
This number fits. Compare the rest of your answer and reasoning with the complete worked solution.
Break it into steps
Step 1 of 2
How many lights are added by rows nine through sixteen?
There are four pairs of 25: \(9+16,10+15,11+14,12+13\). They add 100 lights.
Step 2 of 2
How many lights does the sixteen-row sign need?
\(36+100=136\). The extra rows are longer; explain why doubling the height does not double the total.
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: sixteen rows
Answer:No. Sixteen rows need 136 lights.
Two sixteen-row copies make sixteen rows of seventeen: \(16\times 17\div 2=136\). Doubling the height does not double the count: the extra rows are longer than the early ones.
Check from the original eight-row sign. The new rows 9 through 16 contribute 100 lights:
Beside the doorway, a beacon follows a fixed program: White → Amber → Blue → Green → White, over and over. At move 0, before anything moves, it is White. It changes once on every rover move.
How many moves pass before the beacon first returns to White? Do not count its starting state. How is this pattern different from the growing sign?
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How many moves pass before the beacon first returns to White? Do not count its starting state. How is this pattern different from the growing sign?
Beside the doorway, a beacon follows a fixed program: White → Amber → Blue → Green → White, over and over. At move 0, before anything moves, it is White. It changes once on every rover move.
Your answer
The Compass checks the number only. Also answer every other part of the question and explain why your pattern continues.
This number fits. Compare the rest of your answer and reasoning with the complete worked solution.
Break it into steps
Step 1 of 1
After how many moves does White first return?
Amber, Blue, Green, then White: four moves. A fixed whole block repeats; the triangular sign grows.
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: the beacon
Answer: The first return to White is after 4 moves. Its four-colour program repeats; the sign gains longer rows and grows.
CKRobo “What information would help you predict a much later colour?”
Li Wei “Our place in the four-colour repeat. We don’t keep adding new colours.”
Move 0White
Move 1Amber
Move 2Blue
Move 3Green
Move 4White
Move 0 is the starting state. After four changes, the program returns to White and repeats the same order.
Starting from White, four changes pass Amber, Blue and Green, then return to White.
The fixed program now gives the same next colours again.
Three numbers do not tell the whole story
CKRobo finds an unfinished number card: 1, 2, 4, …. Only these three entries are known; no rule is supplied.
Must the next number be eight?
CKSTEM Thinking CompassPlan and check your reasoning.
Must the next number be eight?
Three numbers do not tell the whole story
CKRobo finds an unfinished number card: 1, 2, 4, …. Only these three entries are known; no rule is supplied.
Notice a pattern, then look for the reason it continues.
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 unfinished card
Answer:No. Eight is possible, but it is not forced without a rule.
Double each number
1 → 2 → 4 → 8
×2 each time
Add successively more
1 → 2 → 4 → 7
+1, then +2, then +3
These are two possible rules among many. Both agree with the given prefix; their next terms differ.
Both rules fit all three given numbers. They give different next numbers, so neither continuation is forced.
CKRobo “What makes our sign and beacon more dependable than this unfinished card?”
Amara “We know how every new row is built and how the whole beacon program runs. Here we have only three numbers.”
Notice a pattern, then look for the reason it continues.
A familiar cycle before launch
The club plans a celebration 100 days from today. Today is Monday, day 0; tomorrow is day 1. The weekdays repeat in their usual seven-day order.
Day 0Monday
Day 1Tuesday
Day 2Wednesday
Day 3Thursday
Day 4Friday
Day 5Saturday
Day 6Sunday
Day 7Monday
Day 0 is today. After seven elapsed days, the weekday returns to Monday.
What day of the week will the celebration be?
CKSTEM Thinking CompassPlan and check your reasoning.
What day of the week will the celebration be?
A familiar cycle before launch
The club plans a celebration 100 days from today. Today is Monday, day 0; tomorrow is day 1. The weekdays repeat in their usual seven-day order.
Day 0Monday
Day 1Tuesday
Day 2Wednesday
Day 3Thursday
Day 4Friday
Day 5Saturday
Day 6Sunday
Day 7Monday
Day 0 is today. After seven elapsed days, the weekday returns to Monday.
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: 100 days later
Answer:Wednesday.
Every complete week brings us back to Monday. Fourteen weeks account for 98 days; two more days reach Tuesday, then Wednesday.
\[100=7\times 14+2\]
Check: day 98 is Monday, day 99 Tuesday and day 100 Wednesday. Day 0 is today, so it is not counted as a day that has passed.
03
Two doorways, one box of lights
The workshop has 123 lights for two model doorways: a main doorway and a side doorway. Each doorway has two identical stair-shaped triangles and one height strip. A triangle starts with one light, with one more light in each row below; all rows start at the same edge. The strip has one light per row. No lights are shared.
Within each doorway, both triangles and its strip have the same positive whole number of rows. The two doorways may have different heights. The main doorway must have more rows than the side doorway. Use all 123 lights.
Main doorway
Two equal triangles + one strip
Unknown height · more rows
Side doorway
Two equal triangles + one strip
Unknown positive whole height
123 lights shared between both doorways
Within each doorway, all three parts have the same height. The drawing labels the requirements; it does not show the answer heights.
Find every possible pair of row counts: main doorway first, side doorway second. Do these clues identify one design?
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Find every possible pair of row counts: main doorway first, side doorway second. Do these clues identify one design?
The workshop has 123 lights for two model doorways: a main doorway and a side doorway. Each doorway has two identical stair-shaped triangles and one height strip. A triangle starts with one light, with one more light in each row below; all rows start at the same edge. The strip has one light per row. No lights are shared.
Within each doorway, both triangles and its strip have the same positive whole number of rows. The two doorways may have different heights. The main doorway must have more rows than the side doorway. Use all 123 lights.
Main doorway
Two equal triangles + one strip
Unknown height · more rows
Side doorway
Two equal triangles + one strip
Unknown positive whole height
123 lights shared between both doorways
Within each doorway, all three parts have the same height. The drawing labels the requirements; it does not show the answer heights.
Your answer
Add a row for every possible design, keeping the main and side doorway heights together. Explain why the list is complete and whether the clues identify one design.
Your complete set of designs fits. Explain why no design is missing and whether the clues choose one of them.
Break it into steps
Help step 1
How many lights does one four-row doorway use?
\(10+10+4=24\), also \(4\times6=24\).
Help step 2
How many lights would a seven-row side doorway use?
\(7\times9=63\) lights.
Help step 3
What would the smallest taller main doorway use with side height seven?
\(8\times10=80\). Together 143 exceeds 123, and taller choices cost more.
Help step 4
What is the last side height that still needs testing?
Test side heights 1 through 6. Each fixes the remaining main-doorway light budget.
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
How could knowing the cost of one height help you search for two heights?
Show every design and why the search is complete
Answer: There are two designs: (10 rows, 1 row) and (9 rows, 4 rows), with the main doorway listed first. No: the clues do not choose one of them.
CKRobo “Which part of this new problem can you already count?”
Amara “A pair of equal triangles makes a rectangle. Its strip adds one light to every row.”
Li Wei “So a doorway with four rows uses four rows of six lights. But we still need two different heights.”
● First triangle● Turned triangle○ Height strip
One four-row doorway: 4 × 6 = 24 lights
Each joined row has five triangle lights and one strip light. For any height, the row width is two more than the height. This counts one doorway, not both.
For a height of \(n\) rows, the triangles make \(n\) rows of \(n+1\) lights. The strip adds one per row:
\[\text{lights for one doorway}=n(n+2)\]
The construction explains the rule for every height. Each extra row adds lights, so a taller doorway always needs more. The rule counts one doorway; it does not yet tell us how to share 123 lights between two.
CKRobo “How will you search without missing a height pair?”
Li Wei “Fix the side doorway at one row, then two, then three. Subtract its lights and see whether the rest makes a taller main doorway.”
Amara “We can stop before side height seven. Even the smallest taller main doorway, height eight, would take us over budget.”
Seven side rows use \(7\times9=63\) lights. Eight main rows use \(8\times10=80\). Together that is \(143>123\). Larger choices need still more, so side heights 1 through 6 cover every possibility.
Fix the side height; test the remaining lights for a taller main doorway.
Side rows
Side lights
Lights left
Can the main doorway use them?
1
3
120
120 = 10 × 12 → main height 10
2
8
115
99 < 115 < 120 → no whole height
3
15
108
99 < 108 < 120 → no whole height
4
24
99
99 = 9 × 11 → main height 9
5
35
88
80 < 88 < 99 → no whole height
6
48
75
63 < 75 < 80 → no whole height
One row for each allowed side height. Consecutive-height costs bracket every rejected remainder; 1 and 4 give the two valid designs.
In the table, 115 lies between the costs for heights 9 and 10; 108 does too. The remainders 88 and 75 fall between consecutive costs as well. A height must be a whole number, so those four rows cannot work. Heights 1 and 4 leave exactly enough for a taller whole-height doorway.
Check both designs in their original parts.
Main 10, side 1: \((55+55+10)+(1+1+1)=120+3=123\).
Main 9, side 4: \((45+45+9)+(10+10+4)=99+24=123\). Each design uses all the lights, has matching parts within each doorway, and has a taller main doorway.
Look for a pattern gives a dependable height-to-cost rule. Make an organized list then checks each possible side height once and keeps both valid designs. Finding one design would not answer “every”.
To choose the actual display, the club needs another clue or a design decision. The light budget alone leaves two choices.
04
The long rover rehearsal
The lights are ready. Visitors will run a long test of the rover and beacon. The team wants to predict how often the rover returns to its starting point with the starting colour showing.
The rover visits six stations in a fixed loop: A → B → C → D → E → F → A. Each move advances exactly one station.
The beacon advances one colour on every rover move: White → Amber → Blue → Green → White.
At move 0, before anything advances, the rover is at A and the beacon is White. Neither program changes or pauses during the simulation.
Rover starts at A
Beacon starts at White
White → Amber → Blue → Green → back to White
Both advance once per move. Start counting at move 0.
Both programs run together from move 0. Neither program pauses or changes its rule.
During moves 1 through 2,026, how many times will the rover be at A while the beacon is White? Also, where will the rover be and what colour will show after move 2,026? Do not count the starting moment, move 0, as a visit.
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During moves 1 through 2,026, how many times will the rover be at A while the beacon is White? Also, where will the rover be and what colour will show after move 2,026? Do not count the starting moment, move 0, as a visit.
The lights are ready. Visitors will run a long test of the rover and beacon. The team wants to predict how often the rover returns to its starting point with the starting colour showing.
The rover visits six stations in a fixed loop: A → B → C → D → E → F → A. Each move advances exactly one station.
The beacon advances one colour on every rover move: White → Amber → Blue → Green → White.
At move 0, before anything advances, the rover is at A and the beacon is White. Neither program changes or pauses during the simulation.
Rover starts at A
Beacon starts at White
White → Amber → Blue → Green → back to White
Both advance once per move. Start counting at move 0.
Both programs run together from move 0. Neither program pauses or changes its rule.
Li Wei “We could run all the moves, but then we would have nothing to compare the test with.”
Amara “Let’s make a prediction we can explain, then use it to check the simulation.”
Your answer
Enter the visit count. Also give the final station and colour and explain the complete reset; those parts are not checked by this numeric Compass.
This number fits. Compare the rest of your answer and reasoning with the complete worked solution.
Break it into steps
Step 1 of 4
At what positive move does the entire A-and-White starting state first return?
The first full reset is move 12. Move 6 is A with Blue, so it is not a complete reset.
Step 2 of 4
How many complete twelve-move blocks fit into 2,026 moves?
\(168\times12=2016\); the next full block would end at 2028.
Step 3 of 4
How many moves remain after the complete blocks?
\(2026-2016=10\). Read row ten of the joint record for the final station and colour.
Step 4 of 4
How many positive-move visits are A with White during moves 1 through 2,026?
There are 168 such visits. Also give the final station and colour, and explain why the full reset repeats.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Li Wei “We could run all the moves, but then we would have nothing to compare the test with.”
Amara “Let’s make a prediction we can explain, then use it to check the simulation.”
Show both answers and reasoning: the rover rehearsal
Answer: There are 168 positive-move visits to A with White. After move 2,026, the rover is at E and the beacon is Blue.
Record the two programs together
CKRobo “What must a record keep if it is to answer both questions?”
Li Wei “The station and the colour in the same row. A colour list on its own loses the rover.”
Li Wei chooses to record every move from zero until the entire starting state returns.
Consecutive joint states, including the starting state and first complete return.
Move
Station
Beacon
0
A
White
1
B
Amber
2
C
Blue
3
D
Green
4
E
White
5
F
Amber
6
A
Blue
7
B
Green
8
C
White
9
D
Amber
10
E
Blue
11
F
Green
12
A
White
A and White return together at move 12. Earlier repeated stations or colours do not reset the complete joint state.
Li Wei “At move six the rover is back at A. Are we ready to repeat?”
Amara “The beacon is Blue then, not White. The whole situation has not reset.”
Explain why the joint pattern repeats
CKRobo “What would make the same block happen again?”
Amara “Move twelve has both starting states. Neither program changes, so their next moves have to match the first twelve.”
Li Wei “And our complete list has checked that no earlier move resets them both.”
Move 12 is the first complete reset. The unchanged programs must now produce the same twelve-state sequence again.
We need a complete reset, not just one repeated colour or one repeated station.
Modular reduction
When a full block of moves returns a system to the same state, remove complete blocks and keep the remainder—the moves left over. For this program, removing any complete block of twelve moves preserves the final station and colour.
\[2026=12\times 168+10\]
2,026 movesThe whole rehearsal
=
168 × 122,016 moves Complete resets → A / White
+
10 moreFollow row 10 E / Blue
Remove complete twelve-move blocks, each preserving both states. Only the ten leftover moves affect the final pair.
Li Wei “There are 168 full blocks: 2,016 moves. That leaves ten more moves, so I read row ten: E and Blue.”
Amara “Each full block ends with one A-and-White visit. The last ten moves do not finish another block. That gives 168 visits, with move zero left out.”
The simultaneous visits occur at 12,24,36,…,2,016: one at the end of every complete twelve-move block. There are 168 of these. The remaining ten moves do not complete another block, and move 0 was excluded, so the visit count is 168.
Check each program as well.
\(2026=6\times 337+4\): four moves beyond a rover reset reach E. Also \(2026=4\times 506+2\): two changes beyond White reach Blue. The two checks agree with the joint-state answer.
Answer and reasoning: what if the remainder is zero?
Answer: After 2,028 moves, the rover is at A and the beacon is White.
After 2,028 moves, \(2028=12\times 169+0\). There are no extra moves after the complete blocks. The rover is at A, and the beacon is White. Remainder zero means the starting state, which we labelled move 0.
Look for a pattern connected that reset to the unchanged programs, explaining why the same block repeats.
Modular reduction removed 168 complete blocks, leaving only ten moves to follow.
The separate-program check finds the final station and colour quickly. The joint record also identifies every simultaneous A/White visit, which is what the team needs to count.
Your turn
Try these before opening their answers. You can use a small drawing or a few cases to get started.
A row of square frames
A craft club makes equal square frames from sticks. One stick forms each side. The squares join in one straight row, each new square sharing exactly one complete side with its neighbour. Shared sides use one stick, not two; no other sides overlap.
1 square4 sticks
2 squares7 sticks
3 squares10 sticks
Each new square shares one full side with its neighbour. These are the first three constructions.
How many sticks are needed for eight joined squares? Explain what is added each time.
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How many sticks are needed for eight joined squares? Explain what is added each time.
A row of square frames
A craft club makes equal square frames from sticks. One stick forms each side. The squares join in one straight row, each new square sharing exactly one complete side with its neighbour. Shared sides use one stick, not two; no other sides overlap.
1 square4 sticks
2 squares7 sticks
3 squares10 sticks
Each new square shares one full side with its neighbour. These are the first three constructions.
Your answer
The Compass checks the number only. Also answer every other part of the question and explain why your pattern continues.
This number fits. Compare the rest of your answer and reasoning with the complete worked solution.
Break it into steps
Step 1 of 3
How many squares are added after the first square?
Eight squares contain the first square and seven further squares.
Step 2 of 3
How many sticks do those seven new squares add?
\(7\times3=21\) additional sticks.
Step 3 of 3
How many sticks make all eight joined square frames?
\(4+21=25\). Explain the three new sides per square and check by counting top, bottom and vertical sides.
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: square frames
Answer:25 sticks. Every new square adds three.
The first square needs 4 sticks. Each of the seven new squares shares its left side and needs only a top, a bottom and a right side: \(4+7\times 3=25\).
Check another way:8 top sticks,8 bottom sticks and 9 vertical sticks give \(8+8+9=25\). The shared sides are counted once.
A festival garland
A garland repeats the whole unit leaf, star, star, with no gaps or changes. Position 1 is a leaf; positions 2 and 3 are stars.
What symbol is in position 50? What whole unit repeats?
CKSTEM Thinking CompassPlan and check your reasoning.
What symbol is in position 50? What whole unit repeats?
A festival garland
A garland repeats the whole unit leaf, star, star, with no gaps or changes. Position 1 is a leaf; positions 2 and 3 are stars.
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 garland
Answer: A star. The repeated unit is leaf, star, star.
Sixteen whole units use 48 positions. Position 49 starts the next unit with a leaf, position 50 is its first star and 51 its second star. Check: \(50=3\times 16+2\), the second place in the unit.
Optional: a very long product
\(3^{2026}\) means multiplying 2,026 factors of 3. You do not need to find the whole number.
What is its last digit? Find a small pattern and explain why it keeps working.
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Interactive checking needs JavaScript. The complete question and worked answer remain available.
What is its last digit? Find a small pattern and explain why it keeps working.
Optional: a very long product
\(3^{2026}\) means multiplying 2,026 factors of 3. You do not need to find the whole number.
Your answer
The Compass checks the number only. Also answer every other part of the question and explain why your pattern continues.
This number fits. Compare the rest of your answer and reasoning with the complete worked solution.
Break it into steps
Step 1 of 3
How many last digits are in the complete repeating block for positive powers of three?
The block is 3,9,7,1. Multiplying by three sends the last digit through this same four-step loop.
Step 2 of 3
What is the remainder when 2,026 is divided by the block length?
\(2026=4\times506+2\): use the second position of the block.
Step 3 of 3
What is the last digit of the product?
The second last digit is 9. Also explain why the cycle continues; a few matching examples alone are not a proof.
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: the last digit
Answer:9.
The first four positive powers are 3,9,27,81. Their last digits are 3,9,7,1. Multiplying a number by 3 changes its last digit in the same way as multiplying that digit by 3:3→9→7→1→3. Returning to 3 therefore restarts the same four-step pattern.
\[2026=4\times 506+2\]
The exponent is in the second position of the four-power repeat, so the last digit is 9. Check the indexing: powers 1 and 5 both end in 3; powers 4 and 8 end in 1. A zero remainder would mean the fourth position, not an extra first position.
Look back
The visitors arrive to find a display they can question, predict and test. Amara’s building rule explained the growing lights. Li Wei’s careful record kept the rover and beacon together. Their shortcut saved work because they understood the structure first.
When a pattern catches your eye, ask: What is the rule? Why must it continue? What information must I keep? Those questions travel much further than a single answer.
Beyond the model: real light and real explorers
Our beacon makes its own light. Moonlight is reflected sunlight: a model’s coloured lights do not explain the real Moon. Explore NASA’s explanation of moonlight.
For the club’s wall of explorers, Amara chose Mae Jemison, an engineer and physician who became the first African American woman in space aboard Endeavour in 1992. Different skills can belong in the same crew. Meet Mae Jemison at NASA.
Explore another setting for Modular reduction: use a repeating structure to reduce a large task.
Amara “I want visitors to understand how we built it, not just watch it glow.”
Li Wei “Then our explanations have to work even when the model gets bigger.”
CKRobo “How could you test that idea against the way the sign is built?”