Series route: This guide belongs to the complete PSLE Learner’s Guide volume directory. The original PSLE Learning Guide supplies the wider English, Mathematics and Science route, while PSLE Mathematics provide verified subject-specific continuation.
Find your route through this guide
This volume solves one precise learner problem: the learner treats 1625 as an awkward four-digit multiplier, copies a shortcut for 1375 or 1125 without deriving it, or loses the 1000 factor after dividing by eight. The aim is not to memorise a slogan. It is to understand the mechanism well enough to recognise the idea when the wording, numbers, diagram or context changes. A Primary 6 learner should be able to explain the decision, perform it under time pressure, and detect a tempting answer that only looks familiar.
- Understand the mechanism before memorising a form.
- Study the worked examples and compare near-misses.
- Repair common misconceptions with targeted tests.
- Complete graded independent practice before reading the answers.
- Use delayed transfer and examination control to make the skill durable.
- Use the parent and tutor notes only after the learner has attempted the work.
The problem beneath the mistake
The multiplier 1625 is thirteen groups of 125. Since eight groups of 125 make 1000, 1625 is thirteen eighths of 1000.
For 24×1625, divide 24 by 8 to obtain 3, multiply 3 by 13 to obtain 39, then restore the factor 1000. The product is 39,000.
The same number is 1000+625. The distributive route gives 24,000+15,000 because 24×625=15,000.
It is also 2000−375. For some multiplicands, doubling a thousand product and subtracting three eighths of a thousand product is efficient.
A shortcut is not a chant. Every step must be an equality that could be reversed. Deriving 1625=13×125 protects the method from confusion with 1375=11×125.
Multiples of eight favour the thirteen-eighths route because division by eight stays in whole numbers.
A multiplicand such as 7 or 9 may favour 1000+625, 2000−375, or ten groups minus one group.
The 625 component has its own structure: 625=5×125=5/8×1000=10,000/16. Choose the version that simplifies the given number.
For a positive multiplicand, 1625 lies between 1500 and 1750. The product must lie in the corresponding scaled interval.
Last digits provide another check. Since 1625 ends in 25, a whole-number product ends in 25, 50, 75 or 00 according to the multiplicand.
Units still matter. Multiplication can be numerically commutative while the story meaning—items per box, cents per item or square centimetres—must be stated correctly.
The safe routine is derive two forms of 1625, inspect divisibility by 8, choose an exact route, label the thousand factor and verify with a different structure.
Core decision: Because 1625=1000+625=2000−375=13×125=(13/8)×1000, a product n×1625 can be reorganised through the representation that best fits n. If n is divisible by 8, n÷8×13×1000 is often shortest. Otherwise 1000+625, 2000−375 or 13×125 may keep the arithmetic exact. For positive n, the result must be greater than n×1500 and less than n×1750, so a magnitude corridor can detect missing factors and zeros.
A useful rule has three parts: a trigger, an operation and a check. The trigger tells you when the rule applies. The operation tells you what to do. The check tells you whether the result still matches the meaning of the question. Learners who remember only the operation often apply it in the wrong place; learners who remember only an example often freeze when the surface details change.
Mechanism: why the rule works
1. Build from 125
Eight lots of 125 make 1000, and thirteen lots make 1625.
Diagnostic question: Can the multiplicand be grouped into eights? Reliable response: Use n÷8×13×1000 when that order is exact and easy.
2. Distribute across 1000 plus 625
The distributive property gives n×1625=n×1000+n×625.
Diagnostic question: Can 625 be obtained from five eighths of 1000 or from 10,000÷16? Reliable response: Keep the two partial products labelled before adding.
3. Use a nearby round number
Because 1625=2000−375, subtraction from a simple double-thousand product can be efficient.
Diagnostic question: Is n×375 easier than n×625? Reliable response: Compute both terms exactly and subtract in the correct direction.
4. Protect the factor chain
Dividing n by 8 changes the scale and must remain linked to ×13×1000.
Diagnostic question: Where is the factor 1000 at every line? Reliable response: Write the equality chain before doing mental arithmetic.
5. Select rather than force
The shortest representation depends on n; no route is universally fastest.
Diagnostic question: Which operation leaves the fewest awkward fractions or large partial products? Reliable response: Compare two setup lines before calculating.
6. Verify independently
A reliable check changes representation or reverses the operation.
Diagnostic question: Can the answer be rebuilt with 1000+625 after using thirteen eighths? Reliable response: Use a second route, interval bound, last digits or division by 1625.
Worked examples: make the decision visible
Do not race through these examples. Before reading each explanation, name the evidence that matters and the information that does not. Then compare the correct route with the tempting route. That comparison is where the transferable learning sits.
Worked example 1: Eight groups
Task. Find 8×1625.
Reasoning. Eight groups of thirteen eighths of 1000 make thirteen thousands. The controlling idea is the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 8×1625=13,000.
Independent check. Direct addition 1625×4=6500, doubled, agrees. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 2: Twenty-four groups
Task. Find 24×1625.
Reasoning. 24÷8=3; 3×13×1000=39,000. The controlling idea is the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 24×1625=39,000.
Independent check. Alternative: 24,000+24×625=24,000+15,000. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 3: Thirty-two groups
Task. Find 32×1625.
Reasoning. 32÷8=4; 4×13×1000=52,000. The controlling idea is the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 32×1625=52,000.
Independent check. The result lies between 48,000 and 56,000. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 4: Twelve groups
Task. Find 12×1625.
Reasoning. Use 12×1000+12×625. Since 12×625=7500, add 12,000. The controlling idea is the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 12×1625=19,500.
Independent check. Sixteen groups are 26,000; three quarters gives 19,500. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 5: Seven groups
Task. Find 7×1625.
Reasoning. Use 7×(1000+625)=7000+4375. The controlling idea is the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 7×1625=11,375.
Independent check. The last two digits 75 match 7×25. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 6: Nine groups
Task. Find 9×1625.
Reasoning. Ten groups minus one group gives 16,250−1625. The controlling idea is the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 9×1625=14,625.
Independent check. Use 9000+5625 as a second route. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 7: Forty groups
Task. Find 40×1625.
Reasoning. 40÷8=5; 5×13×1000=65,000. The controlling idea is the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 40×1625=65,000.
Independent check. Four groups are 6500, then multiply by ten. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Worked example 8: Decimal multiplicand
Task. Find 2.4×1625.
Reasoning. Use 24×1625=39,000, then divide by 10. The controlling idea is the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition. This prevents a learner from choosing an answer merely because one word, number or observation resembles a remembered example.
Answer. 2.4×1625=3,900.
Independent check. Check 2×1625+0.4×1625=3250+650. A good check approaches the conclusion from a different direction. Repeating the same steps more slowly may repeat the same hidden assumption.
Deepening lab: compare cases, conditions and limits
A method becomes independent when you can say where it applies, where it stops and which change would alter the conclusion. The cases below deliberately vary one structural feature at a time. For each case, pause before the analysis and predict whether the core rule remains available. Then name the exact condition that controls the decision. This is harder than repeating a worked solution, but it is also closer to what an unfamiliar examination item demands.
Deepening case 1: Sixteen
Case. Two groups of eight.
Analysis. 16×1625=26,000. Notice that the answer depends on the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition; changing a decorative detail would not be enough to change the reasoning.
Challenge. Verify with 16,000+10,000. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 2: Forty-eight
Case. Six groups of eight.
Analysis. 48×1625=78,000. Notice that the answer depends on the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use a bound. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 3: Fifty-six
Case. Seven groups of eight.
Analysis. 56×1625=91,000. Notice that the answer depends on the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition; changing a decorative detail would not be enough to change the reasoning.
Challenge. Check 56×2000−56×375. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 4: Sixty-four
Case. Eight groups of eight.
Analysis. 64×1625=104,000. Notice that the answer depends on the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition; changing a decorative detail would not be enough to change the reasoning.
Challenge. Reverse by division. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 5: Seventy-two
Case. Nine groups of eight.
Analysis. 72×1625=117,000. Notice that the answer depends on the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition; changing a decorative detail would not be enough to change the reasoning.
Challenge. Compare with 72×1600. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 6: Eighty
Case. Ten groups of eight.
Analysis. 80×1625=130,000. Notice that the answer depends on the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition; changing a decorative detail would not be enough to change the reasoning.
Challenge. Check zero placement. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 7: Ninety-six
Case. Twelve groups of eight.
Analysis. 96×1625=156,000. Notice that the answer depends on the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use 100 groups minus four groups. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 8: One hundred
Case. Place value is simplest.
Analysis. 100×1625=162,500. Notice that the answer depends on the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition; changing a decorative detail would not be enough to change the reasoning.
Challenge. Do not force eighths. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 9: Five
Case. Half of ten groups.
Analysis. 5×1625=8,125. Notice that the answer depends on the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use last digits. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 10: Eleven
Case. Ten groups plus one.
Analysis. 11×1625=17,875. Notice that the answer depends on the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition; changing a decorative detail would not be enough to change the reasoning.
Challenge. Check 11×1000+11×625. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 11: Twenty-five
Case. A quarter of one hundred groups.
Analysis. 25×1625=40,625. Notice that the answer depends on the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition; changing a decorative detail would not be enough to change the reasoning.
Challenge. Multiply the answer by four. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 12: Three
Case. Use 1000+625.
Analysis. 3×1625=4,875. Notice that the answer depends on the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition; changing a decorative detail would not be enough to change the reasoning.
Challenge. Estimate between 4500 and 5250. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 13: One half
Case. Half the multiplier.
Analysis. 0.5×1625=812.5. Notice that the answer depends on the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition; changing a decorative detail would not be enough to change the reasoning.
Challenge. Double to recover 1625. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 14: One and one half
Case. One group plus half.
Analysis. 1.5×1625=2437.5. Notice that the answer depends on the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition; changing a decorative detail would not be enough to change the reasoning.
Challenge. Check 1625+812.5. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 15: Reverse problem
Case. A total of 52,000 contains equal groups of 1625.
Analysis. 52,000÷1625=32. Notice that the answer depends on the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition; changing a decorative detail would not be enough to change the reasoning.
Challenge. Multiply back. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 16: Algebraic form
Case. Keep the multiplicand symbolic.
Analysis. 1625n=1000n+625n=(13/8)1000n. Notice that the answer depends on the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition; changing a decorative detail would not be enough to change the reasoning.
Challenge. State when n÷8 is integral. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 17: Money conversion
Case. 1625 cents is $16.25.
Analysis. A context may make decimal money multiplication easier. Notice that the answer depends on the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition; changing a decorative detail would not be enough to change the reasoning.
Challenge. Protect cents and dollars. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 18: Area context
Case. A rectangle 1625 mm by n mm has area 1625n mm².
Analysis. The arithmetic route does not remove squared units. Notice that the answer depends on the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition; changing a decorative detail would not be enough to change the reasoning.
Challenge. Attach the unit. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 19: Negative extension
Case. A negative multiplicand changes sign, not magnitude.
Analysis. −8×1625=−13,000. Notice that the answer depends on the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition; changing a decorative detail would not be enough to change the reasoning.
Challenge. Use only if negative numbers are in scope. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Deepening case 20: Complete audit
Case. Equality, route choice, factors, zeros, last digits, bounds and units agree.
Analysis. The shortcut is transparent rather than magical. Notice that the answer depends on the exact equality used to replace 1625, the divisibility and place-value structure of the multiplicand, and an independent bound or second decomposition; changing a decorative detail would not be enough to change the reasoning.
Challenge. Apply to an unfamiliar n. State the conclusion in one precise sentence, then add one sentence that names the evidence or relationship that makes it valid.
Finish the lab with a blank-page recall. Write the rule, its conditions, one confirming example, one near-miss and one independent check from memory. Compare that page with the article only after finishing. If the rule is correct but the condition is missing, the learning is not yet safe: an overgeneralised rule can produce confident errors. If the condition is correct but no example can be generated, practise construction rather than rereading.
Common misconceptions and targeted repairs
Misconception 1: Copy the 1375 rule
What the learner may think. Use eleven eighths because the digits look similar.
Why it fails. 1625 is thirteen eighths, not eleven eighths, of 1000. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Derive from 125 each time. Use this short test immediately: How many 125s make 1625?
Misconception 2: Drop the thousand
What the learner may think. Write 1625=13/8.
Why it fails. The correct equality is 1625=(13/8)×1000. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Keep scale with fraction. Use this short test immediately: What power of ten is missing?
Misconception 3: Distribute to one term
What the learner may think. Write n×(1000+625)=1000n+625.
Why it fails. n multiplies both addends. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Write 1000n+625n. Use this short test immediately: Can n be factored back?
Misconception 4: Divide twice
What the learner may think. Use 13/8 and also turn 625 into five eighths in the same chain.
Why it fails. The eighth factor is then counted incorrectly. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Choose one representation. Use this short test immediately: How many divisions by eight should the equality contain?
Misconception 5: Round after dividing
What the learner may think. Round n÷8 before multiplying.
Why it fails. Rounding destroys the exact product. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Keep a fraction or choose another route. Use this short test immediately: Can a different decomposition avoid rounding?
Misconception 6: Ignore magnitude
What the learner may think. Accept an answer below n×1500 for positive n.
Why it fails. 1625 is greater than 1500. The error feels efficient because it removes a decision, but it also removes the evidence boundary that makes the answer valid.
Targeted repair. Use an interval corridor. Use this short test immediately: Does the answer lie between the two easy bounds?
Graded independent practice
Attempt each item without looking ahead. Write one line that identifies the controlling evidence before writing the final answer. Items 1–3 check recognition, 4–7 require explanation, and the later items change the surface form so that memory alone is insufficient.
Practice 1: Anchor
Question. Find 8×1625.
Explained answer. 13,000.
Why this answer earns trust. 1625 is thirteen eighths of 1000. Transfer variation: Explain with 13×125.
Practice 2: Divisible by eight
Question. Find 56×1625.
Explained answer. 91,000.
Why this answer earns trust. 56÷8=7; 7×13×1000=91,000. Transfer variation: Verify with 56,000+35,000.
Practice 3: Decomposition
Question. Find 12×1625.
Explained answer. 19,500.
Why this answer earns trust. 12,000+12×625=12,000+7500. Transfer variation: Use half of 24 groups.
Practice 4: Near ten
Question. Find 9×1625.
Explained answer. 14,625.
Why this answer earns trust. Ten groups minus one group. Transfer variation: Check the last two digits.
Practice 5: Decimal
Question. Find 3.2×1625.
Explained answer. 5,200.
Why this answer earns trust. 32 groups make 52,000, then divide by ten. Transfer variation: Check 3×1625+0.2×1625.
Practice 6: Reverse
Question. What number times 1625 equals 104,000?
Explained answer. 64.
Why this answer earns trust. 104,000÷1000×8÷13=64. Transfer variation: Multiply back.
Practice 7: Error repair
Question. Repair 24×1625=24,000+625.
Explained answer. 24,000+24×625=39,000.
Why this answer earns trust. The 24 distributes to both terms. Transfer variation: Factor 24 back out.
Practice 8: Route choice
Question. Which route is shortest for 72×1625?
Explained answer. Thirteen eighths of 1000.
Why this answer earns trust. 72 divides exactly by 8. Transfer variation: Show a second route.
Practice 9: Bounds
Question. Give a quick bound for 37×1625.
Explained answer. Between 55,500 and 64,750.
Why this answer earns trust. Multiply 37 by 1500 and 1750. Transfer variation: Use it to reject 6,012.5.
Practice 10: Routine
Question. State the 1625 routine.
Explained answer. Derive 13×125, compare thirteen eighths, 1000+625 and 2000−375, choose an exact route, retain every factor and verify with bounds or a different form.
Why this answer earns trust. The method is derived, selected and checked. Transfer variation: Apply to 48.
Delayed transfer: prove that the learning lasts
Transfer test 1: Representation race
For twelve multiplicands, write two valid decompositions before calculating. Do this after a delay, without rereading the worked examples. The purpose is to test retrieval and decision-making, not immediate imitation.
Success criterion. The learner chooses a route from number structure rather than habit. If the criterion is not met, return to the smallest missing decision rather than repeating the entire chapter.
Transfer test 2: Family discrimination
Mix multipliers 1125, 1375 and 1625 after a delay. Do this after a delay, without rereading the worked examples. The purpose is to test retrieval and decision-making, not immediate imitation.
Success criterion. The learner derives 9, 11 or 13 groups of 125 instead of attaching the wrong numerator. If the criterion is not met, return to the smallest missing decision rather than repeating the entire chapter.
Transfer test 3: Error reconstruction
Repair missing thousands, one-sided distribution and double division by eight. Do this after a delay, without rereading the worked examples. The purpose is to test retrieval and decision-making, not immediate imitation.
Success criterion. Each repair names the broken equality and supplies an independent check. If the criterion is not met, return to the smallest missing decision rather than repeating the entire chapter.
Examination control
1. Read the job before the story. Underline the command and the noun it governs. “Explain the change,” “state the word,” “find the fraction” and “identify the evidence” are different jobs even when the surrounding context is similar.
2. Separate given information from remembered information. Use only the facts supplied for the current task plus secure subject knowledge. Do not let a previous question, a familiar worksheet or an expected pattern silently become new evidence.
3. Write enough working to expose the decision. A marker or teacher should be able to see the relationship that justifies the answer. Extra sentences are not automatically stronger; the strongest working is concise, relevant and checkable.
4. Reserve a different-route check. For language, read the sentence for meaning and grammar. For mathematics, estimate or reverse the operation. For science, check the claim against the stated observation and a conservation or fair-test principle.
5. Stop when the requested job is complete. Unrequested speculation can contradict a correct answer. Add detail only when it explains the mechanism or satisfies another explicit part of the question.
Under time pressure, use a compact four-part margin routine: trigger → evidence → operation → check. Trigger names the feature that activates the method. Evidence identifies the exact words, values or observations allowed. Operation performs the language, mathematical or scientific step. Check asks whether the final statement answers the requested job, preserves units or referents, and avoids a stronger claim than the evidence supports.
A compact error log
After practice, do not record only “careless mistake.” Name the broken decision. Useful labels include trigger missed, evidence imported, relationship reversed, referent lost, unit dropped, operation correct but job wrong, and check repeated the same route. Add one prevention cue beside the label. A short accurate log is more useful than a long emotional account because it tells the learner what to do on the next item.
Review the log after two days and again after a week. Reattempt only one or two representative items first. If the method transfers, move to a mixed set. If it fails, repair the exact link between trigger and operation. This spacing matters because a solution can feel obvious while it is still visible and disappear when the page is closed.
Notes for parents and tutors
Ask the learner to build 1625 from 125 counters or a number line before showing the shortcut. Thirteen groups should be visible.
Alternate multiplicands divisible by eight with those better suited to 1000+625, so route selection remains a decision.
Require one magnitude estimate and one second-route check; speed without verification is not yet independent performance.
The adult’s most useful question is not “Why were you careless?” but “Which decision did the evidence require?” Ask the learner to point to the controlling word, value, relationship or observation. If the learner can explain that link, reduce support. If the learner cannot, model one example and then change the surface details for the next attempt.
Frequently asked questions
Why is 1625 thirteen eighths of 1000?
Because 1000÷8=125 and 13×125=1625.
When should I divide by eight first?
When the multiplicand is divisible by eight and this keeps the working exact and short.
What if it is not divisible by eight?
Use an exact fraction or choose 1000+625, 2000−375 or 13×125.
How do I avoid confusing 1625 with 1375?
Derive the number of 125s: 1625 has thirteen; 1375 has eleven.
How can I check the answer quickly?
For positive n, compare with n×1500 and n×1750, then inspect last digits or use another decomposition.
Is long multiplication wrong?
No. It is exact when performed correctly; decomposition reorganises the same product for efficiency and checking.
Next routes
Return to the complete volume directory to locate the series, use the original PSLE Learning Guide for the foundation route, and continue through PSLE Mathematics for verified subject guidance. The next useful task is a mixed set in which this idea is sometimes relevant and sometimes irrelevant; that is the point at which a learner must choose, not merely copy.