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Before the rover rolls
Guess and check
A trial can miss the target—and still help you move towards it.
A Mars adventure begins with a rehearsal on Earth. The children learn from what their attempts reveal.
Li Wei moves the marker after the rover’s first run. Amara is preparing the science centre’s Mars-rehearsal game. CKSprout is their coach. Their loads, controls and travel rules are invented for the game; they can test those rules exactly.
Li Wei “That attempt didn’t do what we wanted. Should we start again?”
CKSprout “What did you find out that could help your next attempt?”
Amara “We can keep the useful information. Then change something for a reason.”
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.
Guess, check, improve
A model load must contain exactly eight cargo pods. Each pod has a mass of either 2 kg or 5 kg. The pod masses must total exactly 31 kg. Counts are whole numbers; either kind may be absent.
How many pods of each kind should be used? Explain why no other pair of counts works.
????????
Available pod masses
2 kg or 5 kg each
Required load
8 whole pods
31 kg exactly
Each slot is one pod. The question asks for counts of both kinds, not an ordering of the pods.A question to help me begin
Choose a legal load and check it. What stays fixed if you replace one pod with a different kind?
Try with the CKSTEM Thinking Compass
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How many pods of each kind should be used? Explain why no other pair of counts works.
Guess, check, improve
A model load must contain exactly eight cargo pods. Each pod has a mass of either 2 kg or 5 kg. The pod masses must total exactly 31 kg. Counts are whole numbers; either kind may be absent.
????????
Available pod masses
2 kg or 5 kg each
Required load
8 whole pods
31 kg exactly
Each slot is one pod. The question asks for counts of both kinds, not an ordering of the pods.
Your answer
Enter both pod counts. Also check the mass and explain uniqueness.
These quantities fit. Check both pod counts. Also check the mass and explain uniqueness.
Break it into steps
Step 1 of 3
How much mass does the three-heavy, five-light trial carry?
\(3\times5+5\times2=25\) kg, six short. A light-to-heavy swap adds three without adding a pod.
Step 2 of 3
What is your answer for 5 kg pods?
5 kg pods: 5. Compare this quantity with the complete worked reasoning; the number alone does not check your argument.
Step 3 of 3
What is your answer for 2 kg pods?
2 kg pods: 3. Compare this quantity with the complete worked reasoning; the number alone does not check your argument.
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 cargo load
Answer:Five 5 kg pods and three 2 kg pods. These are the only possible counts.
Li Wei tries three heavier pods and five lighter ones. That keeps eight pods, but its mass is only \(3\times5+5\times2=25\) kg. It is 6 kg short.
3 heavy · 5 light
5 kg5 kg5 kg2 kg2 kg2 kg2 kg2 kg
25 kg
6 kg short
5 heavy · 3 light
5 kg5 kg5 kg5 kg5 kg2 kg2 kg2 kg
31 kg
Exact mass
Two light-to-heavy replacements add six kilograms while preserving the eight-pod count.
Li Wei “My load is six kilograms too light. I could just try a very different load.”
CKSprout “What does the size of that difference tell you?”
Amara “Each swap adds three kilograms without adding a pod. Two swaps should add the six we need.”
Check: \(5+3=8\) pods and \(5\times5+3\times2=31\) kg. Each extra heavier pod adds exactly 3 kg while the count stays eight. The total strictly increases, so a different heavier-pod count cannot give the same mass.
You could also begin with eight light pods and calculate the extra mass needed. An equation works too. The useful part of this trial is that its mismatch tells us how to improve it.
Guess and check
Try a permitted choice, work out what it produces, and compare it with the whole request. Keep what the attempt teaches you. Then choose a better trial—or change your approach.
Choose. Does this trial obey the rules?
Check. What fits, and what misses?
Learn. What does the result justify trying next?
A trial is useful when checking it is easier than seeing the answer straight away. If a short deduction already settles the problem, use it. Finding one working choice does not automatically mean it is the only one.
Between two loads
A separate load must again have exactly eight pods, each 2 kg or 5 kg, with whole counts and either kind allowed to be absent. This time, the required total is exactly 30 kg.
Is this load possible? Explain why your answer covers every allowed whole count, rather than just a few attempts.
CKSTEM Thinking CompassPlan and check your reasoning.
Is this load possible? Explain why your answer covers every allowed whole count, rather than just a few attempts.
Between two loads
A separate load must again have exactly eight pods, each 2 kg or 5 kg, with whole counts and either kind allowed to be absent. This time, the required total is exactly 30 kg.
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 gap
Answer:No. No allowed eight-pod load has mass exactly 30 kg.
4 heavy · 4 light
5 kg5 kg5 kg5 kg2 kg2 kg2 kg2 kg
28 kg
2 kg short
5 heavy · 3 light
5 kg5 kg5 kg5 kg5 kg2 kg2 kg2 kg
31 kg
1 kg too heavy
The target lies between these adjacent whole-count loads. A pod cannot be split to fill the mass gap.
Four heavy pods give 28 kg. Five give 31 kg. There is no whole heavier-pod count between four and five. Every count below four gives still less mass; every count above five gives still more. Together, those facts cover all counts from zero to eight.
Check: the target is strictly between the two adjacent attainable totals. A fractional number of pods would break the rules. Two close trials prove the gap here because the input must be whole and each replacement increases the total.
Which way should the next trial go?
A rectangular surface mat has 20 unit lengths around all four edges. Its front width and depth are positive whole numbers of units. Its area must be 24 unit squares.
The front edge is labelled. Swapping width and depth counts as a different oriented design. A trial with width 8 and depth 2 has area 16.
Find every allowed width-and-depth pair. Must we increase the width because the trial area is too small? Explain.
Supplied trial · area 16
The question gives this trial. Its front width and depth are different roles; check both the perimeter and the requested area.Try with the CKSTEM Thinking Compass
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Find every allowed width-and-depth pair. Must we increase the width because the trial area is too small? Explain.
Which way should the next trial go?
A rectangular surface mat has 20 unit lengths around all four edges. Its front width and depth are positive whole numbers of units. Its area must be 24 unit squares.
The front edge is labelled. Swapping width and depth counts as a different oriented design. A trial with width 8 and depth 2 has area 16.
Supplied trial · area 16
The question gives this trial. Its front width and depth are different roles; check both the perimeter and the requested area.
Your answer
Build the complete list of allowed ordered width/depth pairs. The front edge distinguishes the two roles. Add or remove rows as needed. Explain the direction of the next useful trial separately.
The complete ordered pair set fits. Explain why the initial small area does not require increasing width.
Break it into steps
Help step 1
What must width plus depth equal?
Half of twenty is ten. List positive whole pairs with that sum and check area.
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 surface mat
Answer:Width 4, depth 6, or width 6, depth 4. No, a too-small area does not always mean “increase the width”.
Two widths and two depths make 20, so one width plus one depth makes 10. Once a width is chosen, its depth is fixed. The only possible whole widths are 1 through 9.
All positive whole width choices
Width
Depth
Area
1
9
9
2
8
16
3
7
21
4
6
24
5
5
25
6
4
24
7
3
21
8
2
16
9
1
9
Width plus depth is ten. Every allowed width appears exactly once; the area does not keep increasing with width.
Width 4 · depth 6
Width 6 · depth 4
Both oriented designs use twenty edge units and twenty-four unit squares.
Amara “Eight by two is too small. I was going to make the width larger.”
CKSprout “What else changes when you change the width?”
Li Wei “The depth gets smaller. Nine by one has even less area. We need to check the complete shape.”
Check: both answers have perimeter \(2(4+6)=20\) and area \(4\times6=24\). Every legal width appears in the comparison. The areas rise towards the square and then fall; there is no single larger-width direction that increases the area throughout this domain.
What have the tests really established?
A separate model controller has settings 1, 2, 3, 4 and 5. Each setting produces a fixed whole output from 0 to 30. The rule is unknown; no formula is given.
The only recorded tests are 1 gives 4; 2 gives 8; 3 gives 12.
Must setting 4 give 16? Support your answer with two complete possible output tables that agree with every recorded test but differ at setting 4.
CKSTEM Thinking CompassPlan and check your reasoning.
Must setting 4 give 16? Support your answer with two complete possible output tables that agree with every recorded test but differ at setting 4.
What have the tests really established?
A separate model controller has settings 1, 2, 3, 4 and 5. Each setting produces a fixed whole output from 0 to 30. The rule is unknown; no formula is given.
The only recorded tests are 1 gives 4; 2 gives 8; 3 gives 12.
Setting 14
Setting 28
Setting 312
Setting 4?
Setting 5?
Only the first three outputs are known. No formula for the untested settings has been supplied.
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.
Setting 14
Setting 28
Setting 312
Setting 4?
Setting 5?
Only the first three outputs are known. No formula for the untested settings has been supplied.Show the answer and reasoning: the untested setting
Answer:No. Setting 4 could give 16, but the three tests do not force it. These two complete possibilities fit all the supplied information.
Complete possibility 1
Setting 14
Setting 28
Setting 312
Setting 416
Setting 520
Complete possibility 2
Setting 14
Setting 28
Setting 312
Setting 415
Setting 520
Both complete tables preserve the three recorded outputs and stay in the allowed output range, but disagree at setting four.
Check: both tables give 4, 8 and 12 at the tested settings. All their outputs are whole numbers from 0 to 30, yet they disagree at setting 4. The tests suggest a possible rule; they do not establish an unstated rule. Checking setting 4 would provide new evidence.
Choose a signal board
A signal routine must contain exactly ten pulses lasting exactly 65 seconds altogether, with no gaps or overlapping pulses.
Choose one board and use only its pulse lengths: Board A: 4 or 10 seconds; Board B: 3 or 8 seconds. Counts must be whole; either length may be absent.
Which board works, and how many pulses of each length does it need? Explain why no other board-and-count choice works. You do not need to specify pulse order.
Board A
4 seconds or 10 seconds
Board B
3 seconds or 8 seconds
Choose one board. Every pulse must use one of its two stated durations; ten pulse durations must total sixty-five seconds.Try with the CKSTEM Thinking Compass
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Which board works, and how many pulses of each length does it need? Explain why no other board-and-count choice works. You do not need to specify pulse order.
Choose a signal board
A signal routine must contain exactly ten pulses lasting exactly 65 seconds altogether, with no gaps or overlapping pulses.
Choose one board and use only its pulse lengths: Board A: 4 or 10 seconds; Board B: 3 or 8 seconds. Counts must be whole; either length may be absent.
Board A
4 seconds or 10 seconds
Board B
3 seconds or 8 seconds
Choose one board. Every pulse must use one of its two stated durations; ten pulse durations must total sixty-five seconds.
Your answer
Enter the working signal’s counts. Also choose its board and prove why the other board cannot work; this number check does not grade the board decision.
These quantities fit. Check the working signal’s counts. Also choose its board and prove why the other board cannot work; this number check does not grade the board decision.
Break it into steps
Step 1 of 3
How long is a trial with five 3-second and five 8-second pulses?
\(5\times3+5\times8=55\). A short-to-long replacement adds five seconds, so two replacements add the missing ten.
Step 2 of 3
What is your answer for Short pulses?
Short pulses: 3. Compare this quantity with the complete worked reasoning; the number alone does not check your argument.
Step 3 of 3
What is your answer for Long pulses?
Long pulses: 7. Compare this quantity with the complete worked reasoning; the number alone does not check your argument.
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 signal
Answer:Board B, with three 3-second pulses and seven 8-second pulses. No other board-and-count choice works.
Use even and odd
Every pulse on Board A lasts an even number of seconds. Adding ten even lengths still gives an even total. It cannot give the odd target 65, however its pulses are chosen.
Guess and check can now focus on Board B. Five short and five long pulses last \(5\times3+5\times8=55\) seconds. Replacing a short pulse with a long one adds 5 seconds while keeping ten pulses. Two replacements supply the missing 10 seconds.
5 short · 5 long
3 s3 s3 s3 s3 s8 s8 s8 s8 s8 s
55 seconds
3 short · 7 long
3 s3 s3 s8 s8 s8 s8 s8 s8 s8 s
65 seconds
Each card represents one pulse, with its duration written. Two short-to-long replacements add ten seconds without changing the pulse count.
Check: \(3+7=10\) pulses and \(3\times3+7\times8=65\) seconds. Each extra long pulse strictly increases the total by 5 seconds, so no different count gives 65. The parity argument excluded every Board A choice; the complete check establishes the Board B solution.
Revisit Use even and odd when a quick necessary-condition check can rule out a whole set of attempts. Passing that check alone does not prove a plan works.
A rehearsal ready on time
The final simulation has more than one choice. A fast journey can use too much energy; a low-energy journey can return too late.
The rover starts at 10:00. The rehearsal must finish on or before 10:35. After returning, the rover must dock for 9 minutes, then upload for 6 minutes. These stages cannot overlap.
Choose exactly one route: A is 72 m each way; B is 90 m each way. The outward and return distances are equal within each route.
Choose a whole setting from 1 to 10. If the setting is \(v\), the outward speed is \(v\) metres per minute and the return speed is \(v+3\) metres per minute. Speeds are constant on each leg; there are no other journeys, stops or delays.
The game charges \(v^2\) energy tokens for the whole journey. The energy limit is 50 tokens.
Find every route-and-setting pair that satisfies both limits. Which feasible pair uses the least energy? Explain why no discarded pair can work. A plan may finish early.
Start and finish
Start 10:00
Finish by 10:35
After returning
Dock: 9 minutes
Then upload: 6 minutes
Route A
72 m outward
72 m return
Route B
90 m outward
90 m return
A whole setting
1 through 10
Outward speed: setting
Return speed: setting + 3
Speeds in m/min
Game energy rule
Setting squared, in tokens
At most 50 tokens
Both legs, the ordered finishing stages and the energy rule apply to every plan. The deadline allows an early finish.Try with the CKSTEM Thinking Compass
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Find every route-and-setting pair that satisfies both limits. Which feasible pair uses the least energy? Explain why no discarded pair can work. A plan may finish early.
A rehearsal ready on time
The final simulation has more than one choice. A fast journey can use too much energy; a low-energy journey can return too late.
The rover starts at 10:00. The rehearsal must finish on or before 10:35. After returning, the rover must dock for 9 minutes, then upload for 6 minutes. These stages cannot overlap.
Choose exactly one route: A is 72 m each way; B is 90 m each way. The outward and return distances are equal within each route.
Choose a whole setting from 1 to 10. If the setting is \(v\), the outward speed is \(v\) metres per minute and the return speed is \(v+3\) metres per minute. Speeds are constant on each leg; there are no other journeys, stops or delays.
The game charges \(v^2\) energy tokens for the whole journey. The energy limit is 50 tokens.
Start and finish
Start 10:00
Finish by 10:35
After returning
Dock: 9 minutes
Then upload: 6 minutes
Route A
72 m outward
72 m return
Route B
90 m outward
90 m return
A whole setting
1 through 10
Outward speed: setting
Return speed: setting + 3
Speeds in m/min
Game energy rule
Setting squared, in tokens
At most 50 tokens
Both legs, the ordered finishing stages and the energy rule apply to every plan. The deadline allows an early finish.
Your answer
List every feasible route/setting pair, then enter the least-energy pair and its energy. Route choices come from the original two routes. Explain feasibility and optimality separately.
The complete feasible set and least-energy pair fit. Check both route cases, the full timing chain and the energy bound.
Break it into steps
Help step 1
How many minutes are available for both travel legs?
Thirty-five minutes minus six minus nine leaves twenty. Judge each route’s trials against this complete travel budget.
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: every feasible plan
Answer:Route A at setting 6 and Route A at setting 7 are the only feasible pairs. A at 6 uses the least energy: 36 tokens, compared with 49.
Amara “A quick trip might still leave the upload unfinished. I need a limit for the travel, then a way to compare both routes.”
CKSprout “What must every plan complete, and which choices could you keep separate?”
Li Wei “I can trace back through the upload and docking, then test settings on each route without mixing their distances.”
Work backwards
A trial’s return time is not its finish time: docking and uploading still have to happen. Recover the latest return first, so every later trial is judged against the same complete deadline.
Undo the upload and docking times from the latest finish. The rover must return by 10:20. Starting at 10:00 leaves at most 20 minutes for both travel legs together.
Work backwards · latest allowed times
10:35Latest finish
Undo 6 min upload←↓
10:29Latest upload start
Undo 9 min docking←↓
10:20Latest return
Returning by ten-twenty leaves time for both finishing stages. From ten o’clock, both travel legs together may take at most twenty minutes.
Casework
The same setting has different travel times on the two routes. Split by route A or B, then ask which whole settings survive inside each route. Every allowed plan belongs to exactly one route case; a fast but wasteful working trial does not settle the least-energy question.
Fix one choice
Hold the route fixed while investigating settings. That gives each trial a clear meaning. Find the feasible settings and best energy choice for that route, then compare the route results. Routes A and B are the complete set of route choices.
Settings 8, 9 and 10 already cost too much: \(8^2=64>50\), and their costs only increase. That leaves settings 1 through 7.
Guess and check now tests the full travel time. Divide each leg’s distance by its own speed, then add the two times. Do not use one average of the two speeds.
Route A · setting 5
Outward: \(14\frac{2}{5}\) min
Return: 9 min
Total: \(23\frac{2}{5}\) min
Energy: 25 tokens
Too late
Route A · setting 6
Outward: 12 min
Return: 8 min
Total: 20 min
Energy: 36 tokens
Fits both limits
Route A · setting 7
Outward: \(10\frac{2}{7}\) min
Return: \(7\frac{1}{5}\) min
Total: \(17\frac{17}{35}\) min
Energy: 49 tokens
Fits both limits
Route B · setting 7
Outward: \(12\frac{6}{7}\) min
Return: 9 min
Total: \(21\frac{6}{7}\) min
Energy: 49 tokens
Too late
Within each route, increasing the setting makes both legs faster. Use the energy limit as well as the time comparison; A at seven works but costs more than A at six.
On A, setting 5 takes \(72\div5+72\div8=23.4\) minutes: too slow. Setting 6 takes \(72\div6+72\div9=12+8=20\) minutes and costs 36 tokens. Setting 7 is faster and costs 49, so it works too.
On B, even setting 7—the fastest one the energy limit permits—takes \(90\div7+90\div10=\frac{153}{7}>20\) minutes. Lower settings take longer, so none can work on B.
Li Wei “A at six works. Have we finished?”
CKSprout “Which parts of the request does that answer, and which still need checking?”
Amara “We need every feasible plan and the best one. A at seven works too, but costs more. We also have to account for B.”
Why are no settings missing? For a fixed route, a greater setting makes both speeds greater, so each fixed-distance leg takes less time. A’s settings 1–5 are at least as slow as its failed setting 5; 6 and 7 work; 8–10 exceed the energy limit. B’s settings 1–7 all fail time; 8–10 fail energy. Those groups cover every allowed route-and-setting pair.
Check forwards · Route A at setting 6
10:00Start
20 min travel→↓
10:20Return
9 min docking→↓
10:29Docking finished
6 min upload→↓
10:35Upload finished
The least-energy feasible plan reaches the deadline exactly and costs thirty-six tokens. The other feasible plan finishes earlier at a higher energy cost.
Check the best plan: A at 6 returns at 10:20, docks until 10:29 and finishes uploading at 10:35. It uses 36 tokens, within 50. A at 7 finishes earlier but uses 49 tokens. Therefore A at 6 is the unique least-energy feasible pair, even though there are two feasible pairs.
How the tools fit: Work backwards recovers the travel budget. Fix one choice makes each route a manageable investigation and requires checking both route results. Informative trials and justified faster/slower comparisons find and verify the settings. A full table or algebra could also solve the problem.
Revisit Work backwards when a final deadline can reveal the time an earlier stage is allowed.
Rehearsals matter
Engineers test robotic systems in simulated terrain on Earth, including rehearsals for Mars exploration. A model lets people investigate an idea before relying on it somewhere harder to reach.
Our game’s exact rules make its calculations checkable. A real rover also needs measurements and tests of its actual equipment and surroundings. An attractive mathematical result does not replace those tests.
Your turn
Try these before opening the answers. What can a trial establish—and what still needs checking?
Sticker bundles
A shop uses exactly six unopened bundles, each containing 4 or 9 stickers, to make 39 stickers exactly. Whole bundle counts only; either kind may be absent.
A trial of two large and four small bundles gives 34 stickers.
Use what the trial reveals to find the counts of both kinds. Is there another allowed pair?
Try with the CKSTEM Thinking Compass
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Use what the trial reveals to find the counts of both kinds. Is there another allowed pair?
Sticker bundles
A shop uses exactly six unopened bundles, each containing 4 or 9 stickers, to make 39 stickers exactly. Whole bundle counts only; either kind may be absent.
A trial of two large and four small bundles gives 34 stickers.
Your answer
Enter both bundle counts. Check the whole count and sticker total, and justify uniqueness.
These quantities fit. Check both bundle counts. Check the whole count and sticker total, and justify uniqueness.
Break it into steps
Step 1 of 3
How many stickers does a small-to-large bundle replacement add?
Nine minus four is five. The supplied trial is five short, so one replacement is enough.
Step 2 of 3
What is your answer for 9-sticker bundles?
9-sticker bundles: 3. Compare this quantity with the complete worked reasoning; the number alone does not check your argument.
Step 3 of 3
What is your answer for 4-sticker bundles?
4-sticker bundles: 3. Compare this quantity with the complete worked reasoning; the number alone does not check your argument.
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 bundles
Answer:Three 9-sticker bundles and three 4-sticker bundles. There is no other allowed pair.
The trial is 5 short. Replacing one small bundle with a large one adds exactly 5 stickers while keeping six bundles.
Check: \(3+3=6\) and \(3\times9+3\times4=39\). Each replacement increases the total by 5, so a different large-bundle count gives a different total.
A changed mat target
A new rectangular mat again has 20 unit lengths around its four edges and positive whole width and depth. Now the target area is 26 unit squares.
Can any allowed mat meet the new target? Explain why your answer covers all possible whole widths.
CKSTEM Thinking CompassPlan and check your reasoning.
Can any allowed mat meet the new target? Explain why your answer covers all possible whole widths.
A changed mat target
A new rectangular mat again has 20 unit lengths around its four edges and positive whole width and depth. Now the target area is 26 unit squares.
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 new area
Answer:No. The greatest allowed area is 25, for width 5 and depth 5.
Width plus depth is 10. The only possible widths 1–9 produce areas \(9,16,21,24,25,24,21,16,9\), respectively. None reaches 26.
Check: both dimensions stay positive whole numbers and every legal width is included. A design for the old target does not automatically solve the changed problem.
A code with a zero
A code has two digit slots. Each digit is from 0 to 4. The digits must add to 4 and their product must be 0. A zero in the first slot is allowed.
Find every code. Does finding 40 settle the whole request?
Try with the CKSTEM Thinking Compass
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Find every code. Does finding 40 settle the whole request?
A code with a zero
A code has two digit slots. Each digit is from 0 to 4. The digits must add to 4 and their product must be 0. A zero in the first slot is allowed.
Your answer
List every two-digit code. Keep a zero in its original slot; a leading zero is allowed. Add or remove rows as needed. Explain whether one found code settles the whole request.
The complete code set fits. Both slot orders matter; your explanation of completeness remains unscored.
Break it into steps
Help step 1
If one digit is zero, what must the other digit be?
The digit sum is four. The zero can occupy either allowed slot.
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: both codes
Answer:04 and 40. Finding 40 alone does not settle every code.
The complete sum-compatible pairs are \(0+4,1+3,2+2,3+1,4+0\). Only the first and last also have product zero.
Check: both codes use allowed digits, sum to 4 and have product 0. The visible rule permits the leading zero; ignoring that boundary would lose a solution.
What did the trial tell you?
Choose a legal trial, check every condition, and ask what the mismatch really tells you. An error can guide a useful change, reveal a gap, or show that another tool is needed. A first success answers “can this work?”; it may leave “what else works?” or “which is best?” open.
Next we will develop Fix one choice within Casework: hold one feature fixed, follow its consequences and account for every case. Casework has appeared before; this particular way of organising interacting choices is the new step.
Li Wei “That attempt didn’t do what we wanted. Should we start again?”
CKSprout “What did you find out that could help your next attempt?”
Amara “We can keep the useful information. Then change something for a reason.”