Before the stars appear

Bound the size

You may know enough to decide, even when you do not know the exact answer.

Amara unrolls a ground covering while Arjun hangs an unlit sample lantern. Sky-blue CKRobo sits nearby, listening as their coach.
Earlier preparations at the camp. The diagrams below show the quantities and plans they must still decide.

Amara and Arjun are helping at a dark-sky camp. They want to welcome everyone, then switch off the decorations and enjoy the stars. CKRobo joins them as their coach.

Amara “The cases are sealed. Must we weigh every one before choosing a cart?”

CKRobo “What do you already know that could help you decide?”

Arjun “The labels give limits, even though they don’t give exact masses.”

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?

The direction reflects your self-check. Use the lesson’s answer and reasoning to check further when you are ready.

Choose a cart before opening the cases

All 7 equipment cases must travel on one cart. Each case may have any mass from 3 kg to 4 kg, including both limits. Any combination of these masses is allowed. There is no other load.

One cart carries at most 20 kg; the other carries at most 28 kg. A load exactly at a cart’s limit is allowed.

Which cart can certainly carry all seven cases? Can the other cart possibly carry them?

Case 1
3–4 kgCase 2
3–4 kgCase 3
3–4 kgCase 4
3–4 kgCase 5
3–4 kgCase 6
3–4 kgCase 7
3–4 kg
20 kg cart

Maximum load

28 kg cart

Maximum load

Each case has an inclusive promised range; none has been weighed exactly.
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Show the answer and reasoning: the carts

Answer: The 28 kg cart certainly works. The 20 kg cart cannot work: even the smallest allowed load is 21 kg.

Arjun “Seven lightest cases would already weigh 21 kg. We can stop considering the small cart.”

CKRobo “What still needs checking about your other choice?”

Amara “The heaviest load. Seven cases at 4 kg give 28 kg, so the other cart covers every allowed mix.”

20 kg cart (kg)212820 limitFilled ends include both limits.28 kg cart (kg)212828 limitFilled ends include both limits.
The same possible load range is compared with each capacity. Filled ends include the limits; the dashed line is the cart capacity.
\[7\times3=21\ \text{kg},\qquad7\times4=28\ \text{kg}.\]

Check: every allowed total lies from 21 kg to 28 kg. The small limit is below that entire range; the large limit includes its upper end. Exact weighing is unnecessary for this decision.

Bound the size

A lower bound is a limit the answer cannot fall below. An upper bound is a limit it cannot exceed. Choose a limit with a reason covering every allowed case, then compare it with the actual goal.

Use this when being too big or too small could settle a decision, or when a limit could end a search. A close guess is not automatically a safe bound. A range crossing the decision point may leave the answer open.

Start by trying an allowed extreme. Ask whether the same argument covers everything between the extremes. Tighten the limit only as much as the problem needs.

Will their money be enough?

Binocular hire costs $24 to $25; camera hire costs $17 to $18. Add exactly $4 for delivery, then subtract a discount of $5 to $6. Both ends of every range are included, any combination within the ranges is possible, and every tax and fee is already included. The children have $40.

Later, the shop confirms that binocular hire is at most $24.50, camera hire is at most $17.50, and the discount is at least $6. The original limits still apply.

Find the smallest and largest possible bill before and after the confirmation. At each stage, can they be certain that $40 pays the whole bill?

Add: binoculars

$24–$25

Add: camera

$17–$18

Add: delivery

Exactly $4

Subtract: discount

$5–$6

Original inclusive quotes. Read the later confirmation in the question as well; it tightens these ranges.
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Show the answer and reasoning: before and after

Answer: Initially the bill ranges from $39 to $42, so payment is not certain. After confirmation it ranges from $39 to $40, so $40 certainly pays.

Bound above and below

To make the bill large, increase what is charged and decrease what is taken off. A bigger discount makes the final bill smaller.

Arjun “The highest hires and a $6 discount give $41. Is that the largest bill?”

CKRobo “How could you check that against the other allowed quotes?”

Amara “A $5 discount is allowed. Less comes off, so the bill can reach $42.”

\[\begin{aligned}\text{Smallest initially: }24+17+4-6&=39,\\\text{Largest initially: }25+18+4-5&=42.\end{aligned}\]
Initially allowed: fits

$24 + $17 + $4 − $6 = $39

Initially allowed: fails

$25 + $18 + $4 − $5 = $42

Both initial outcomes are possible; this is not a prediction of the actual bill.

These are both possible before confirmation: one fits the money and one fails. The information does not determine the actual initial bill.

Afterwards the discount is exactly $6: the old upper limit of $6 still applies to the new lower limit of $6.

\[\begin{aligned}\text{Smallest later: }24+17+4-6&=39,\\\text{Largest later: }24.50+17.50+4-6&=40.\end{aligned}\]
Before confirmation (dollars)394240 limitFilled ends include both limits.After confirmation (dollars)394040 limitFilled ends include both limits.
A common scale keeps the same forty-dollar decision line. The later range includes equality but never exceeds it.

Check: every later-compatible bill is at most $40, including equality. The earlier $42 example is no longer compatible with the confirmed prices. A guessed middle price would not establish either guarantee.

A memory card for the camera

Arjun’s star-map picture has exactly 397 rows, each containing 603 pixels. The camera’s storage card holds at most 230,000 pixels. Every pixel must be stored; resizing or compression is not allowed in this model.

Can the card hold the picture? How many digits does its total pixel count have in base-ten notation, without leading zeros? Explain how your method settles both decisions.

Picture array

397 rows

603 pixels in each row

Storage limit

At most 230,000 pixels

These are the actual dimensions and capacity. No comparison dimensions have been selected yet.
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Show the answer and reasoning: the storage decision

Answer: No, the picture will not fit. Its total pixel count has six digits.

Amara “There are more than 390 rows, with more than 600 pixels in each row. Even that smaller picture is too big for the card.”

CKRobo “What does that settle? What else would settle the digit count?”

Arjun “An upper limit too. There are fewer than 400 rows and fewer than 610 pixels in each row.”

Smaller array

390 rows × 600 pixels

234,000 pixels

Actual picture

397 rows × 603 pixels

Exact total not needed

Larger array

400 rows × 610 pixels

244,000 pixels

Both positive dimensions change in the same direction, making reliable product bounds.

Let \(N\) be the complete pixel count. Both dimensions are positive, so decreasing both decreases the product; increasing both increases it.

\[390\times600=234{,}000\lt N\lt244{,}000=400\times610.\]

The lower limit already exceeds 230,000, which settles storage without the exact multiplication.

Count the digits

Six-digit positive whole numbers run from 100,000 through 999,999. Our entire range lies there, so the digit count is fixed.

Check: the picture has an exact count that could be calculated; we simply do not need it. That differs from the genuinely unknown hire prices. Mixing one upward and one downward change, such as \(400\times600\), would not by itself prove a direction for the error.

Make room on the ground

Use all 400 identical square tiles, each exactly 10 cm by 10 cm, to make one rectangle. Use positive whole numbers of tiles along its sides, with no cuts, gaps or overlaps.

The clear viewing area is a 5 m square. The whole rectangle must fit inside that area, with its sides parallel to the sides of the area. Ignore the thickness of the edging. Exactly 10 m of edging is available for its four outside sides. Using exactly that or leaving some edging unused is allowed.

Width runs across the fixed front entrance; depth runs towards the rear. A different width/depth pair counts as a different layout, even if it is a turn of another. Count dimensions, not different positions on the ground.

Find every allowed width × depth pair in centimetres. How many layouts are there?

Tile stock

400 tiles, each 10 cm × 10 cm

Edging

10 m for the full outside boundary

Viewing site

5 m × 5 m square

Rectangle sides parallel to the site

Layout identity

Front width × rear depth

Different dimension pairs count separately.

Use every whole tile. Dimensions are requested in centimetres; sliding a shape does not create a new dimension pair.
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Show the answer and reasoning: all the dimensions

Answer: Five layouts: width × depth in centimetres is 100 × 400, 160 × 250, 200 × 200, 250 × 160, or 400 × 100.

One tile edge is 10 cm. The 10 m edging therefore equals 100 tile edges. We temporarily label the shorter side first to search; this does not say the actual front width must be shorter.

Bound the size

Arjun “If the shorter side had 21 tiles, the other side would have at least 21 too. That needs more tiles than we have.”

CKRobo “What does that let you rule out before trying dimensions?”

Amara “Every shorter side above 20. We can finish the search instead of wondering how far to go.”

\[21\times21=441\gt400.\]
Whole-tile pairs after testing shorter sides 1–20
Side counts in tilesEdging in tile edgesDecision
1 × 400802Too much edging
2 × 200404Too much edging
4 × 100208Too much edging
5 × 80170Too much edging
8 × 50116Too much edging
10 × 40100Fits
16 × 2582Fits
20 × 2080Fits
The other tested shorter-side counts (3, 6, 7, 9, 11, 12, 13, 14, 15, 17, 18, 19) do not divide 400 into a whole-number other side. The available edging equals 100 tile edges.

Testing the shorter side from 1 through 20 leaves three shapes. The smaller divisor choices all need more edging. The surviving shorter-side counts are 10, 16 and 20; these came from the complete search, not an assumed lower cutoff.

Use symmetry

A quarter-turn swaps width and depth. It keeps the tile count, whole-tile condition and edging unchanged. Every surviving dimension is at most 4 m, so all fit the 5 m square, including after the turn. The site excludes no additional candidate.

10 × 40 tilesFront width 100 cmDepth 400 cm

100 × 400 cm

Two ordered layouts: exchange width and depth

16 × 25 tilesFront width 160 cmDepth 250 cm

160 × 250 cm

Two ordered layouts: exchange width and depth

20 × 20 tilesFront width 200 cmDepth 200 cm

200 × 200 cm

One ordered layout: fixed under the turn

Every small square is one tile. All three drawings use the same scale; the square is a fixed case, not a second new orientation.

Amara “Turning 10 by 40 gives 40 by 10. But the 20 by 20 square hasn’t changed.”

CKRobo “How will that affect your final count?”

Arjun “Two orientations for each non-square, and just one for the square. Five altogether.”

Each non-square shape gives two ordered layouts. The square stays the same under the turn, so it gives one. Doubling every shape would count the square twice.

\[2+2+1=5\ \text{layouts}.\]

Check: every rectangle has a shorter side in our tested range, and turning recovers either front orientation. These dimensions use exactly 400 tiles and need 10.0 m, 8.2 m or 8.0 m of edging. Checking every ordered width directly is another valid route; the turn lets us reuse half the search.

Explore Use symmetry in other settings.

Every camper’s lantern

The campers have prepared 15 identical paper lanterns. They want every one on display, spread across as many windows as possible, before switching them off for observing.

There are 22 windows in one straight row, numbered from left to right. Windows 5, 12 and 18 hold notice boards and cannot be used. Each of the remaining 19 windows has a hanging frame. Lanterns may be displayed only in frames.

Frames at windows 1, 3, 6, 8, 13, 17, 19 and 21 hold exactly two lanterns when used. Each usable frame not in that list holds exactly one. A frame is either empty or completely filled; do not half-fill a two-lantern frame.

No two occupied windows may be consecutive. Two lanterns within one frame occupy that single window. All 15 lanterns must be displayed.

What is the largest possible number of occupied windows? Give one complete placement and explain why a larger number cannot work.

12
21
32
41
5—
62
71
82
91
101
111
The same row continues: 11 is next to 12 ↓
12—
132
141
151
161
172
18—
192
201
212
221
One straight row, continued below. Below each window number is its frame capacity, 1 or 2; a dash marks an unavailable notice window. All frames are still unchosen.
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Show the answer and reasoning: a complete welcome display

Answer: 10 occupied windows is the maximum. One placement uses 1, 4, 6, 9, 11, 13, 15, 17, 19 and 22. Fully filling those frames displays all 15 lanterns.

Divide and conquer

Amara “The notice boards leave four shorter stretches to think about.”

CKRobo “Which rules can you check separately, and which still connect the choices?”

Arjun “A notice window separates neighbours in different stretches. But we still need fifteen lanterns altogether.”

Cut the adjacency problem at the notice boards into windows 1–4, 6–11, 13–17 and 19–22. We can check separation within each run; the exact total of fifteen lanterns still connects the choices.

Windows 1–4

4 available windows

Windows 6–11

6 available windows

Windows 13–17

5 available windows

Windows 19–22

4 available windows

Fixed empty notice windows separate adjacency checks. They do not remove the shared requirement to display exactly fifteen lanterns.

Bound the size

Pair consecutive windows within each run. Each pair permits at most one occupied window; an unpaired final window permits at most one. The four run limits are 2, 3, 3 and 2. They cover every usable window, so no allowed display can occupy more than ten windows, whatever the frame capacities.

Windows 1–4
1 and 2At most 1 occupied window
3 and 4At most 1 occupied window
Windows 6–11
6 and 7At most 1 occupied window
8 and 9At most 1 occupied window
10 and 11At most 1 occupied window
Windows 13–17
13 and 14At most 1 occupied window
15 and 16At most 1 occupied window
17At most 1 occupied window
Windows 19–22
19 and 20At most 1 occupied window
21 and 22At most 1 occupied window
Every usable window occurs in one consecutive pair or singleton. Each such group contributes at most one occupied window, independent of lantern capacities.
\[2+3+3+2=10\ \text{occupied windows at most}.\]

To reach ten occupied windows, every run must reach its own limit. If one run falls short, no other run can exceed its limit to make up the difference.

More generally, \(k\) occupied windows in a run need their own \(k\) places and at least \(k-1\) intervening empty places. Thus \(2k-1\le n\) in a run of \(n\) windows.

Bound, then build the best

Ten is a valid upper limit, but we must build a permitted ten-window display using exactly fifteen lanterns before calling it the maximum.

Ten windows, too few lanterns

Frames 2, 4, 7, 9, 11, 13, 15, 17, 20, 22

12 lanterns in ten windows

Ten windows, too many lanterns

Frames 1, 3, 6, 8, 10, 13, 15, 17, 19, 21

18 lanterns in ten windows

Both sketches meet the window-count ceiling, but neither displays exactly the required fifteen lanterns.

Arjun “My ten-window sketch uses eighteen lanterns. Another uses only twelve. Both miss our goal.”

CKRobo “What could the smaller parts tell you about a construction that uses fifteen?”

Amara “Keep each stretch at its window limit, then choose how many lanterns its full frames hold.”

One example for each lantern total, keeping each run at its occupied-window limit
RunOccupied windowsLantern total → example frames
1–422 → (2, 4)3 → (1, 4)4 → (1, 3)
6–1133 → (7, 9, 11)4 → (6, 9, 11)5 → (6, 8, 10)
13–1735 → (13, 15, 17)
19–2222 → (20, 22)3 → (19, 22)4 → (19, 21)
This table supplies usable component choices for a construction. It is not a catalogue of every global placement.

Three occupied windows in the five-window run must be 13, 15 and 17: their three positions and two separating empty positions use all five places. Its full frames therefore supply five lanterns. We need ten more across the other three runs. Choose attainable totals of three, four and three:

\[3+4+5+3=15\ \text{lanterns}.\]
12★★
21
32
41★
5—
62★★
71
82
91★
101
111★
The same row continues: 11 is next to 12 ↓
12—
132★★
141
151★
161
172★★
18—
192★★
201
212
221★
Windows occupied

10, matching the universal limit

Lanterns displayed

15, five double frames and five single frames

Stars show full chosen frames. All notice windows remain empty and neighbouring windows are never both occupied.

Check: windows 1, 6, 13, 17 and 19 have full two-lantern frames; the other five chosen frames hold one each. That is \(5\times2+5\times1=15\) lanterns in ten occupied windows. No notice board is used and every gap between chosen window numbers is at least two. The construction reaches the universal bound, proving the maximum. There are other valid maximum placements; one checked construction is enough for the requested possibility.

Fifteen being between twelve and eighteen would not alone prove it is reachable. The chosen component options and actual windows provide that proof.

Continue to Bound, then build the best for more problems where a limit and an attainable result must meet.

A little less light, a little more sky

The welcome ends and the decorations go off. Necessary lights stay directed downwards. Light spilling upwards or sideways can scatter in the atmosphere and brighten the sky; shielding a light downwards reduces that spill.

The children’s limits settled some decisions. A checked construction finished the job that a limit alone could not.

Your turn

Try each question before opening its answer. Previous work stays available.

A carrier counts too

Nine boxes each have a mass from 200 g to 220 g, including both endpoints. Any combination is allowed. Their carrier adds exactly 100 g. The whole load must be at most 2 kg.

Can you determine whether this actual load fits from the information given?

CKSTEM Thinking CompassPlan and check your reasoning.

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?

The direction reflects your self-check. Use the lesson’s answer and reasoning to check further when you are ready.

Nine boxes

Each 200–220 g, inclusive

Carrier

Exactly 100 g

Whole-load limit

At most 2 kg

The carrier is part of the total. The actual box masses remain unknown.
Show the answer and reasoning: the carrier

Answer: No; the actual fit cannot be determined. A 1,900 g load fits, while a 2,080 g load does not, and both are allowed.

\[9\times200+100=1{,}900\ \text{g},\qquad9\times220+100=2{,}080\ \text{g}.\]

The limit is \(2\ \text{kg}=2{,}000\ \text{g}\). All boxes at the lower endpoint give the first outcome; all at the upper endpoint give the other. Forgetting the carrier would make the largest box-only total 1,980 g and change the decision.

A print with uncertain dimensions

A rectangular print has width from 10 cm to 12 cm and height from 14 cm to 16 cm. Endpoints are included and any pair of dimensions in these ranges is possible. The display requires area at least 195 cm².

What is the largest possible area of the print? Can any allowed print meet the requirement?

Width

10–12 cm

Height

14–16 cm

Required area

At least 195 cm²

Inclusive independent side ranges; the actual rectangle has not been selected.
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Show the answer and reasoning: the print

Answer: The largest allowed area is 192 cm², so no allowed print can meet 195 cm².

\[\text{area}\le12\times16=192\ \text{cm}^2\lt195\ \text{cm}^2.\]

Check: both positive dimensions are at most their stated upper limits. The permitted 12 cm by 16 cm print reaches 192 cm². Even it is too small; this question is settled although the actual dimensions remain unknown.

Now the positions form a circle

For a separate small display, there are 7 positions numbered around a circle. There are enough identical lanterns to use any number that fits. Each position holds at most one; adjacent positions cannot both be used. Position 7 is adjacent to position 1. None is reserved.

What is the largest lantern count, and one placement achieving it? Explain why a larger count cannot work.

1·2·3·4·5·6·7·
A separate display: at most one lantern per position, including across the closing 7-to-1 edge. No positions are chosen yet.
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Show the answer and reasoning: the circle

Answer: 3 lanterns is the maximum. Positions 1, 3 and 5 give one legal placement.

Let \(L\) be the number of lanterns. Each lantern needs the next position around the circle to be empty. Different lanterns have different next positions, so every lantern requires its own two distinct places: its occupied place and the following empty one.

\[2L\le7,\qquad L\text{ is a whole number},\qquad L\le3.\]
1★2·3★4·5★6·7·
Three separated positions form one witness. The independent circular bound rules out four.

Check: the closing 7-to-1 neighbour rule is respected. Three is attainable and four would need at least eight positions. Cutting this circle into a straight run would discard the connection across the cut.

Look back

Ask which kind of limit the goal needs. A lower limit above a capacity can rule something out; an upper limit within a capacity can guarantee it fits. A crossing range may require more information. When a best result is requested, a bound also needs an allowed construction reaching it.

Keep the condition behind the shortcut: a turn must preserve every rule, and smaller parts still have to satisfy their shared resources when combined.

CKSTEM Thinking Framework

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