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A problem-solving strategy
Work backwards
A missing beginning. A known finish. A trail of clues.
Mina, CKRobo and CKRobi at the stall earlier in the day. The clues below tell us what happened by closing time.
The school garden fair is over. Mina’s seed jar is empty, and there are $18 in the money tray. She wants to record how many seeds she brought—but the number on the jar’s label has rubbed off.
Mina “We can’t count the seeds now. They’ve all left the stall!”
CKRobi “What do we still know for certain?”
They still have their notes. Perhaps the end of the day can help them recover its beginning.
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
The seed-jar mystery
Four clues describe the day—in the order they happened.
Pack. Mina used all the seeds to make full packets, with 4 seeds in each packet.
Donate. She gave 3 packets to the school garden.
Sell. She sold every remaining packet for $2 each.
Buy. She spent $6 of the sales money on signs.
The stall began with no money and had no other money added or taken away. After the signs were paid for, $18 remained.
Before calculating, picture what is missing. We need the original number of seeds, but the number we know at the finish is an amount of money.
Why begin at the finish?
Mina could guess a seed count, follow all four clues and see whether $18 is left. But each guess would make her repeat the whole story.
The finish gives her a firmer starting point: there really are $18 in the tray.
Mina “At $2 a packet, is that nine packets?”
CKRobo “What does that $18 describe?”
Mina “What was left! We’d already paid for the signs.”
Work backwards means starting with a known result or a goal and asking what must—or could—come just before it. Each answer gives us a new fact to work from.
02
Follow the clues back to the seeds
Ask the same small question at each step: “What was true just before this?”
Which quantity could we recover first—and why?
← Reasoning backwardsChecking forwards →
Check your answer with the CKSTEM Thinking Compass
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Your answer
Check your reasoning against every clue in the question.
Break it into steps
Step 1 of 4
Before paying 6 dollars for the signs, how much sales money was there?
Put back the cost of the signs: \(18+6=24\) dollars. This recovers sales money, not a packet count.
Step 2 of 4
Each sold packet brought in 2 dollars. How many packets were sold?
\(24\div2=12\) packets were sold. Check: \(12\times2=24\) dollars.
Step 3 of 4
How many full packets had the team made before any were donated?
Restore the donated packets: \(12+3=15\) full packets made.
Step 4 of 4
How many seeds were in Mina’s jar at the start?
\(15\times4=60\) seeds. This answers the original question. Replay the complete story below to check every clue.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning
Answer: There were 60 seeds in the jar at the start.
1
Before buying the signs: recover the sales money
The $18 is what remains after spending $6. Restore that expense to the count.
\[18+6=24\text{ dollars}\]
Mina collected $24 from sales. This is money, not a packet count yet.
2
Before selling: recover the number of packets sold
Each packet brought in $2. How many groups of $2 make $24?
\[24\div2=12\text{ packets}\]
She sold 12 packets.
CKRobo “Which packets actually brought in sales money?”
Mina “Those twelve earned the sales money. The three donated packets never brought in money.”
3
Before donating: recover all the packets she made
The sold packets do not include the three given to the garden. Restore those packets once.
\[12+3=15\text{ packets}\]
Mina originally made 15 packets.
4
Before packing: recover the seeds
Each of the 15 packets held four seeds. Bring those equal groups back together.
\[15\times4=60\text{ seeds}\]
The jar originally contained 60 seeds.
That distinction changes the first move. Mina must recover the money before the last expense, not divide the money left afterwards.
Reasoning backwards · Recover the seed count
$18Money left
+ 6dollars←↓
$24Sales money
÷ 2dollars per packet←↓
12Packets sold
+ 3packets←↓
15Packets made
× 4seeds per packet←↓
60Seeds in jar
Read from the known finish towards the beginning. Each operation recovers an earlier fact, and each label says what that number counts.
CKRobi “What helped you choose each operation?”
Mina “I kept asking what the number counted: dollars, packets or seeds.”
Replay the whole day forwards.
Sixty seeds make \(60\div4=15\) packets. Donate three: \(15-3=12\) remain to sell. Selling them brings \(12\times2=24\) dollars. Spend six: \(24-6=18\) dollars remain.
Checking forwards · Replay the day
60Seeds in jar
÷ 4seeds per packet→↓
15Packets made
− 3packets→↓
12Packets sold
× 2dollars per packet→↓
$24Sales money
− 6dollars→↓
$18Money left
Follow packing, donation, sales and the expense in their original order. Arriving at $18 checks the recovered beginning against every clue.
Every clue fits. The answer is more than a number: it is a beginning that explains the whole story.
Inverse operations
Inverse operations undo a change. If you spend six dollars, putting six back recovers the amount before spending. If each packet sells for two dollars, dividing the sales money by two recovers the packet count.
Pay $6− 6is undone by+ 6Restore the expense
Sell at $2 per packet× 2is undone by÷ 2Recover packets sold
Give away 3 packets− 3is undone by+ 3Restore the donated packets
Pack 4 seeds per packet÷ 4is undone by× 4Recover all the seeds
Reverse the order as well as the operations. Packing happened first in the story, so it is the last change we undo. The final expense is the first change we undo.
Follow the meaning too. Restoring three packets is different from restoring three seeds. An operation without its meaning can send us down the wrong path.
See the whole story as an equation
Let \(x\) stand for the original number of seeds.
\(x\div4\) is the number of full packets made.
\(x\div4-3\) is the number of packets sold.
Multiplying that count by 2 gives the sales money; subtracting 6 gives the money left.
\[2\left(\frac{x}{4}-3\right)-6=18\]
Doing the same operation to equal quantities keeps them equal. Follow Mina’s reasoning:
\[2\left(\frac{x}{4}-3\right)=24\]
\[\frac{x}{4}-3=12\]
\[\frac{x}{4}=15\]
\[x=60\]
The equation records a story we already understand. It is not a rule pulled out of nowhere.
03
A new stall, the same useful question
What if a clue tells us about a fraction of a collection?
At the neighbouring bake stall, the team sold one quarter of its first batch of muffins. Then 8 freshly baked muffins were added. There are now 26 muffins. No others were added or removed.
How many muffins were in the first batch?
A question to help me begin
Which change happened last? After undoing it, which part of the first batch would remain?
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Your answer
Check your reasoning against every clue in the question.
Break it into steps
Step 1 of 3
Before the 8 fresh muffins were added, how many muffins remained from the first batch?
Put aside the fresh muffins: \(26-8=18\) muffins remained from the first batch.
Step 2 of 3
How many muffins were in one quarter of the first batch?
Selling one original quarter leaves three of the four equal quarters. Split the remaining muffins equally: \(18\div3=6\) muffins in each original quarter.
Step 3 of 3
How many muffins were in the complete first batch?
\(6\times4=24\) muffins. Check: sell 6, then add 8: \(24-6+8=26\).
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning
Answer: The first batch contained 24 muffins.
CKRobo “Which change happened last?”
Mina “The eight new muffins arrived. I’ll remove those from the count first.”
\[26-8=18\]
Those 18 muffins are what remained after one quarter was sold. If one of four equal parts was sold, three equal parts remained.
The four equal parts represent the first batch, before any muffins were sold.
Remaining quarter
?muffins
Remaining quarter
?muffins
Remaining quarter
?muffins
Sold quarter
?muffins
The three remaining parts contain 18 muffins.
What must be in one equal part? Find that before rebuilding all four parts.
Find one equal part before rebuilding the whole
Three equal parts contain 18 muffins. So one part contains:
\[18\div3=6\]
The first batch had four of those parts:
\[4\times6=24\]
Now the first batch can be reconstructed.
Remaining quarter
6 muffins
Remaining quarter
6 muffins
Remaining quarter
6 muffins
Sold part recovered
6 muffins
Four equal quarters of 6 muffins make the original batch of 24.
Check forwards.
One quarter of 24 is 6. Sell those six: 18 remain. Add eight: 26. The first batch contained 24 muffins.
\[24-\frac14\times24+8=26\]
Why not add one quarter of 18?
The sold quarter was a quarter of the first batch, not a quarter of the 18 that remained. In the model, 18 fills three equal parts. The missing part must match one of those parts.
Working backwards is more than swapping signs. Sometimes the key is seeing which whole a fraction belongs to.
Does every finish reveal one beginning?
As the fair closes, two friends put coins into a jar. Each adds at least one coin, and together they add 10 coins. Nothing else changes the number in the jar.
Does the total tell us exactly how many coins each friend added?
CKSTEM Thinking CompassPlan and check your reasoning.
Does the total tell us exactly how many coins each friend added?
Does every finish reveal one beginning?
As the fair closes, two friends put coins into a jar. Each adds at least one coin, and together they add 10 coins. Nothing else changes the number in the jar.
Ask: “What could have led here?” Then: “Does the information force one answer?”
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.
A question to help me begin
Try one pair of contributions. Could a different pair reach the same finish?
Show the answer and reasoning
Answer:No. Ten coins do not tell us one unique pair of contributions.
First friend: 3
+
Second friend: 7
=10
First friend: 4
+
Second friend: 6
=10
Different contributions produce the same total. Ten coins do not identify one unique beginning.
Three and seven works; four and six also works. Both friends add coins in each example. These two ways already show that the total does not force one beginning.
Mina “I can work back to more than one pair.”
CKRobi “What would you need to choose between them?”
Mina “Another clue about the contributions.”
Working backwards cannot create a missing clue. In the seed mystery, without a packet price, even the sales total would not tell us the packet count.
Ask: “What could have led here?” Then: “Does the information force one answer?”
A ribbon puzzle to think through
Mina cuts a ribbon into 3 equal pieces for the display. Then she trims 2 centimetres from each piece. Every finished piece is 6 centimetres long. The cuts between the three original pieces use no ribbon.
How long was the original ribbon?
Picture the finished pieces. What would you undo first?
A question to help me begin
What was the last change to each piece? Before that trimming, how long was each one?
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Your answer
Check your reasoning against every clue in the question.
Break it into steps
Step 1 of 2
How long was each piece before it was trimmed?
Restore the trim to each piece: \(6+2=8\) centimetres.
Step 2 of 2
How long was the original ribbon before the equal pieces were cut?
\(3\times8=24\) centimetres. Each cut uses no ribbon; trimming each piece leaves 6 centimetres.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning
Answer: The original ribbon was 24 centimetres long.
Restore two centimetres to each finished piece:
\[6+2=8\text{ cm}\]
Then bring the three original pieces together:
\[3\times8=24\text{ cm}\]
Check forwards: three 8-centimetre pieces can each lose two centimetres and finish at six.
Compare the actions. Mina donated three packets once from the whole collection of packets. The ribbon was trimmed separately in three places. Ask exactly what each action affected.
04
The sticker-sharing trail
After the fair, Mina shares her stickers. First, she gives Li Wei half her stickers plus one more. Then she gives Rowan half the stickers now left plus one more. She finishes with 3 stickers. All stickers and gifts are whole; there are no other changes.
How many stickers did Mina have before either gift?
A question to help me begin
Imagine the last gift just happened. What does returning the one extra sticker leave you with?
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Your answer
Check your reasoning against every clue in the question.
Break it into steps
Step 1 of 2
How many stickers did Mina have just before giving stickers to Rowan?
Return the extra one, then double: \((3+1)\times2=8\) stickers.
Step 2 of 2
How many stickers did Mina have before either gift?
\((8+1)\times2=18\) stickers. Replay: give 10, then 5, leaving 3.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning
Answer: Mina started with 18 stickers.
CKRobi “Which part of the gift can you put back first?”
Mina “The extra sticker. Then what I have is half the earlier pile. I can double that.”
Before the gift to Rowan, put back one sticker, then double:
\[(3+1)\times2=8\text{ stickers}\]
Use the same reasoning for the earlier gift to Li Wei:
\[(8+1)\times2=18\text{ stickers}\]
Reasoning backwards · Return the last gift first
3Stickers left
+ 1, then × 2Return one sticker, then double the pile←↓
8Stickers before giving to Rowan
+ 1, then × 2Return one sticker, then double the pile←↓
18Stickers before giving to Li Wei
Each number counts stickers. Returning the extra sticker leaves half the earlier pile; doubling rebuilds that pile.
Why add before doubling? The extra one was given away too. Returning it makes the remaining pile exactly half the earlier pile. Doubling first would double a pile that is still missing that sticker.
Replay both gifts.
From 18, give half (9) plus one: \(18-(9+1)=8\). From 8, give half (4) plus one: \(8-(4+1)=3\). Both gifts use whole stickers, and the finish is exactly three.
05
The missing packing record
Mina is checking a later fundraiser. The sale record survived, but the original packing record is missing. This time the small packets sell for $3 each and the large packets for $5 each.
Her notes give the order of events:
She packed the small and large packets, then donated one quarter of each type she had packed to the school garden.
From what remained, she saved 2 large packets for a display. She saved no small packets.
She sold every other packet: 22 packets altogether.
She spent $8 of the sales money on signs and had $78 left.
Packets stayed whole, including the donated quarters. No other packets were added or removed. The stall started with no money, and no other money was added or taken away.
How many small packets and how many large packets had Mina originally packed, before the donation?
A question to help me begin
Which fact gives you a certain place to start? Keep track of what moment each number describes.
Check your answer with the CKSTEM Thinking Compass
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Your answer
Check your reasoning against every clue in the question.
Break it into steps
Step 1 of 6
How much money did all the packet sales bring in, before the sign expense?
\(78+8=86\) dollars from sales.
Step 2 of 6
Of the 22 packets sold, how many were large? Use both prices and the sales total.
If \(p\) large packets were sold, \(22-p\) were small.
How many large packets remained just after the donation, before any were saved for the display?
Restore the display packets: \(10+2=12\) large packets.
Step 5 of 6
How many small packets had Mina originally packed, before donating a quarter?
\(12\div3\times4=16\) small packets originally packed.
Step 6 of 6
How many large packets had Mina originally packed, including donated and saved packets?
\(12\div3\times4=16\) large packets originally packed. Check both original counts in the full forward replay below.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning
Answer: Mina originally packed 16 small packets and 16 large packets.
Mina “I can put back the $8 first. That gives $86 from sales.”
CKRobi “What does each of your known totals describe?”
Mina “Twenty-two packets were sold, and those sales brought in $86. Neither total describes everything I packed.”
\[78+8=86\text{ dollars}\]
Mina “Two prices made that money, so one division won’t give both counts. If I name the number of large packets sold, I can write the number of small ones using the total of twenty-two.”
CKRobi “Show how that choice keeps both clues together.”
Use a variable
Use a variable means naming an unknown quantity with a symbol, then expressing what the clues say about it.
Mina calls the number of large packets sold \(p\). Then \(22-p\) small packets were sold. Here \(p\) is a whole number from 0 to 22. Both counts refer to the sales stage.
Two unknown sold counts must explain both the packet total and the sales money.
Packets sold
Count
Money
Small
\(22-p\)
\(3(22-p)\)
Large
\(p\)
\(5p\)
Total
22
$86
Two unknown sold counts must explain both the packet total and the sales money.
Packets sold: Small
Count
\(22-p\)
Money
\(3(22-p)\)
Packets sold: Large
Count
\(p\)
Money
\(5p\)
Packets sold: Total
Count
22
Money
$86
The variable counts large packets sold. Subtracting it from the sold total gives the small count; the money expressions use their respective prices.
The two contributions must make the recovered sales money:
So Mina sold 10 large packets and \(22-10=12\) small packets.
CKRobo “Have you answered the original question?”
Mina “Not yet. Those are the sold packets. I still need to put back the display packets and the donation.”
Recover what was packed
Restore the 2 saved large packets first. That gives \(10+2=12\) large packets after donation. There were also 12 small packets at that stage, because none of those was saved for the display.
CKRobi “How does that stock relate to the original packing?”
Mina “The twelve of each type are the three quarters that remained. One quarter has four packets, so four quarters have sixteen.”
For each type, one quarter is \(12\div3=4\) packets. The original four quarters contain \(4\times4=16\) packets.
Reasoning backwards · Return to the original stock
12 small 10 largeSold
+ 2 largeRestore the saved display packets←↓
12 small 12 largeAfter donation
÷ 3, then × 4Rebuild the four equal quarters of each type←↓
16 small 16 largeOriginally packed
Restore the display packets before undoing donation. The 12 of each type after donation are three quarters of that type’s original stock.
Restoring the donation before the saved packets would mix two moments. The two saved large packets belonged to the three quarters left after donation.
Replay from the original stock.
Start with 16 small and 16 large. Donate 4 of each: 12 of each remain. Save 2 large: 12 small and 10 large are sold, making 22 packets.
The small packets bring \(3\times12=36\) dollars; the large packets bring \(5\times10=50\) dollars.
All packets and donated quarters are whole. The counts fit every event and the final money. The sold-count equation has only one solution, and reversing the earlier events fixes one original stock.
How the strategies shared the work
Use a variable linked two sold counts to the total money. Work backwards recovered the earlier stages, including the original stock the question asked for. Finding the sales mix was an intermediate step.
06
The missing display plan
For their own Canadian Innovation Contests (CIC) display, Mina and the team prepared a separate batch of packets. They sketched three possible display arrangements. One was used, but its card was never ticked.
The records say:
They packed small and large packets, then donated one quarter of each original type to the school garden.
From the remaining packets, they saved exactly the packets on one of the three display cards below.
They sold every other packet: 27 packets altogether, at $4 per small packet and $7 per large packet.
They spent $9 on display materials and had $126 left.
Before packing, they had only 18 large-packet envelopes. Every large packet they originally packed needed one, including packets later donated or saved for the display. They had no extra envelopes and did not reuse them. There were enough small envelopes.
All packets stayed whole, including the donated quarters. No other packets were added or removed. They began with no money, and no other money was added or taken away.
Exactly one of these display cards was used. These counts were saved after the donation.
Plan
Small saved
Large saved
A
3
3
B
2
2
C
0
6
Exactly one of these display cards was used. These counts were saved after the donation.
Plan A
Small saved
3
Large saved
3
Plan B
Small saved
2
Large saved
2
Plan C
Small saved
0
Large saved
6
Each card states packets kept for the display. Every packet not donated or saved was sold.
Which display plan fits, and how many packets of each type were originally packed? Could a different answer fit every clue?
A question to help me begin
Which facts are certain even though the display card is unknown? What still has to be explained after using those facts?
Check your answer with the CKSTEM Thinking Compass
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Your answer
Also name the display card and explain why neither other card can work.
Break it into steps
Step 1 of 5
How much money came from packet sales before the materials were paid for?
\(27-9=18\) small packets were sold. Every possible display plan must explain this same sold mix.
Step 4 of 5
After testing the display cards against every clue, how many small packets were originally packed?
Only plan A fits both conditions: B cannot form whole original quarters; C would require 20 large envelopes, more than 18. For A, \(18+3=21\) small packets remained after donation, so \(21\div3\times4=28\) were originally packed. The full case checks below explain each rejection.
Step 5 of 5
How many large packets were originally packed for that feasible display plan?
\(9+3=12\) large packets remained after donation; \(12\div3\times4=16\) were originally packed. These use 16 of the 18 large envelopes. Explain why no other card fits, using the full reasoning below.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning
Answer: They used Plan A and originally packed 28 small packets and 16 large packets. No different answer fits every clue.
Mina “The display card is missing, but the sales money isn’t. Putting back $9 gives $135.”
CKRobi “What could you settle before choosing a card?”
Mina “The sold mix. Its count and prices are known, whichever card was used.”
\[126+9=135\text{ dollars}\]
Mina again names the number of large packets sold \(p\), a whole number from 0 to 27. Then \(27-p\) small packets were sold:
They sold 9 large packets and \(27-9=18\) small packets. This conclusion is shared by all three display plans; we need not solve it three times.
Mina “Each card would mean putting back different display packets. I’ll keep all three possibilities and work each one back.”
CKRobo “What will make that a complete answer?”
Mina “Every card must be checked against the original rules, not just the money.”
Casework
Casework splits the allowed possibilities into cases, so every possibility is considered. Here exactly one of A, B or C was used. Restore that card’s saved packets, then recover the original four quarters.
Mina “Plan A puts back three of each. That makes twenty-one small and twelve large after donation. Each count is three quarters: the missing quarters are seven small and four large. So we began with twenty-eight small and sixteen large.”
CKRobi “Does that beginning fit the envelope supply?”
Mina “Sixteen large packets need sixteen envelopes. We had eighteen, so this beginning fits.”
CKRobi “What do the other cards allow?”
For A, the equal-quarter calculations are \(21\div3=7\) and \(12\div3=4\). The original four quarters give \(4\times7=28\) small and \(4\times4=16\) large packets.
Mina “B would leave twenty small and eleven large after donation. Neither splits into three equal whole-packet quarters.”
CKRobo “Could a different original packing make B work?”
Mina “No. The sold mix and B’s saved packets fix those remainders. Every possible beginning for B must explain those same three quarters.”
CKRobi “Which count belongs to the envelope clue?”
Mina “All the large packets originally packed. Plan C needs twenty, even though only nine were sold. That is more than the eighteen envelopes.”
For C, restoring six large display packets gives \(9+6=15\) after donation. Rebuilding four quarters needs \((15\div3)\times4=20\) original large packets, exceeding the 18 envelopes.
Restore each display choice, then recover the original stock and test every condition.
Plan
After donation
Originally packed
Check
A
21 small; 12 large
28 small; 16 large
All counts are whole; 16 large envelopes fit the supply of 18.
B
20 small; 11 large
No whole counts
The remaining counts cannot form three equal whole-packet quarters.
C
18 small; 15 large
24 small; 20 large
Needs 20 large envelopes; only 18 were available.
Restore each display choice, then recover the original stock and test every condition.
Plan A
After donation
21 small; 12 large
Originally packed
28 small; 16 large
Check
All counts are whole; 16 large envelopes fit the supply of 18.
Plan B
After donation
20 small; 11 large
Originally packed
No whole counts
Check
The remaining counts cannot form three equal whole-packet quarters.
Plan C
After donation
18 small; 15 large
Originally packed
24 small; 20 large
Check
Needs 20 large envelopes; only 18 were available.
All three cards share the same sold mix: 18 small and 9 large. Their different saved packets lead to different possible beginnings.
Replay Plan A from the beginning.
Pack 28 small and 16 large, using 16 of the 18 large envelopes. Donate 7 small and 4 large: 21 small and 12 large remain. Save 3 of each for Plan A: 18 small and 9 large are sold.
The sold counts give \(18+9=27\) packets. Small packets bring \(4\times18=72\) dollars; large packets bring \(7\times9=63\) dollars.
Every packet count and quarter is whole. B and C each break an original condition, so A is the only possible plan and its original stock is fixed.
How the strategies shared the work
Work backwards reconstructed the sales money and the earlier stock, in the right order.
Use a variable resolved the linked sold counts once for all three plans.
Casework covered the missing display choice and established that only one complete beginning fits.
A finish can point to several earlier stories. Work back as far as the facts allow, choose another tool for the uncertainty, then test the whole reconstructed story.
Your turn
Try each problem before opening its answer. Start with the finish, but keep asking what each step means.
A hidden starting number
Choose a whole number. Multiply by 3, then subtract 4, then divide by 2. The division is exact, and the finish is 10.
What was the starting number?
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Your answer
Check your reasoning against every clue in the question.
Break it into steps
Step 1 of 3
What number was there just before the final division?
\(10\times2=20\).
Step 2 of 3
What number was there just before the subtraction?
\(20+4=24\).
Step 3 of 3
What was the original starting number?
\(24\div3=8\). Check: multiply, subtract, then divide in the original order.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning
Answer: The starting number was 8.
Undo the last division first: \(10\times2=20\). Restore the four that were subtracted: \(20+4=24\). Undo multiplication by three: \(24\div3=8\).
Check the original order.
\(8\times3=24\), then \(24-4=20\), then \(20\div2=10\). Every change is reversed in the opposite order.
Who should start the game?
Two players start with a running total of 0. Take turns adding 1, 2 or 3. You may not go past 20. The player who makes the total exactly 20 wins. Both players try to win.
If you can choose, should you play first or second? What plan guarantees a win, whatever the other player does?
CKSTEM Thinking CompassPlan and check your reasoning.
If you can choose, should you play first or second? What plan guarantees a win, whatever the other player does?
Who should start the game?
Two players start with a running total of 0. Take turns adding 1, 2 or 3. You may not go past 20. The player who makes the total exactly 20 wins. Both players try to win.
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
Answer: Choose second. Finish your turns at 4, 8, 12, 16, then 20.
Start at the finish. If you leave 16 after your turn, the other player can reach only 17, 18 or 19. You can then add 3, 2 or 1 to reach 20.
Use the same idea one step earlier: leave 12, and after the other player’s move you can reach 16. Working back again gives 8, then 4.
The first player can begin only at 1, 2 or 3. As the second player, add 3, 2 or 1 to make 4. After every later move, add enough to make the two moves total four. Each reply is allowed.
Check every reply.
If the other player adds 1, add 3; if they add 2, add 2; if they add 3, add 1. Your turns end at 4, 8, 12, 16 and 20. The other player never reaches 20 first, and neither player needs to go past it.
Ready for the demonstration
Mina’s science-club demonstration begins at 10:00 a.m. She needs 12 minutes to pack at home, then 18 minutes to walk to the venue, then 10 minutes to set up. The steps happen in that order, on the same morning, with no other stops or waiting.
What is the latest time she can start packing and still be ready at 10:00?
Check your answer with the CKSTEM Thinking Compass
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Your answer
Check your reasoning against every clue in the question.
Break it into steps
Step 1 of 3
By what time must Mina arrive to begin setup?
10:00 minus 10 minutes is 9:50 a.m.
Step 2 of 3
By what time must Mina leave home before the walk?
9:50 minus 18 minutes is 9:32 a.m.
Step 3 of 3
What is the latest time she can start packing?
9:32 minus 12 minutes is 9:20 a.m. Replay packing, walking and setup in their original order.
You have worked through these help steps. Return to Your answer when you are ready, and check every original clue.
Show the answer and reasoning
Answer: Start packing by 9:20 a.m.
Work back from being ready at 10:00. Set-up must start at 9:50. The walk must start eighteen minutes earlier, at 9:32. Packing must start twelve minutes before that, at 9:20.
Check the whole morning.
Pack from 9:20 to 9:32; walk until 9:50; set up until 10:00. Starting one minute later would finish at 10:01. Starting earlier leaves a cushion; 9:20 is the latest time that fits these steps.
Look back
Keep the starting question—and notice when to change strategy.
Mina first recovered a missing seed count by tracing back one fact at a time. Later, she put back an extra sticker before rebuilding each earlier pile. Two prices led to linked sold counts, but those were not the original stock. A variable helped with the sale; working backwards continued through the saved packets and donation. The missing display card needed every allowed possibility checked.
CKRobi “Where else could the finish give you a clue?”
Mina “A game’s final score. A recipe’s finished amount. Even when to start getting ready for a trip.”
You can use the strategy to recover a past beginning or to plan towards a future goal. Look for a known finish and reliable clues connecting it to what you need.
What must have happened just before this?
If you feel stuck, identify one certain finish and try one justified predecessor. If several fit, keep that uncertainty visible. Keep asking while the clues let you recover earlier facts. If the trail reaches linked unknowns or several allowed cases, choose a strategy that fits that obstacle. Then check by following the original story forwards.
If someone finds another route, compare the reasons. A careful question or a changed plan is progress; being the fastest is not the goal.
Mina “We can’t count the seeds now. They’ve all left the stall!”
CKRobi “What do we still know for certain?”
CKRobo “What does that $18 describe?”