Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 1 · Chapter companion part
Describe repeated and growing patterns, apply an explicit sequence rule and explain prime sieving and digit reversals.
Class 8 English Version: part A of the chapter 1 companion. Eight teaching sections, 30 written tasks and a notebook challenge. Selected companion coverage, not complete textbook coverage; use all three parts and review independent proofs and drawings with a teacher.
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A pattern may repeat a tile orientation, add a fixed amount or follow another stated relationship. Describe what repeats or changes, not merely that the picture looks nice. A linear number pattern in this chapter has a constant difference between consecutive terms. A finite list can fit more than one continuation rule. Our automatically checked questions state the intended rule; a different well-explained continuation belongs in open review.
In the supplied triangle each outside edge is 1 and each interior entry is the sum of the two entries immediately above it. For example the row 1, 3, 3, 1 is followed by 1, 4, 6, 4, 1. Trace the two parents of the 6: 3 + 3. This is commonly called Pascal’s triangle. A row number and the value of an entry are different things; state whether you count the top row as row zero or row one.
For the rule start at 4 and add 6, the terms are 4, 10, 16, 22, 28. Each difference is 6. For start at 3 and double, the terms are 3, 6, 12, 24, 48: the multiplier is fixed but the differences grow. Doubling is not a linear constant-difference pattern. A useful written answer names the starting value and operation, then applies it consistently.
The positive square numbers 1, 4, 9, 16, 25 are n² for n = 1, 2, 3, 4, 5. Their consecutive differences are 3, 5, 7, 9. For a Fibonacci-type rule, give two starting values and make every next term the sum of the previous two. Starting 2, 3 gives 2, 3, 5, 8, 13, 21. These are different rules; do not call every increasing list linear.
A prime is a whole number greater than 1 whose only positive divisors are 1 and itself. The number 1 is neither prime nor composite; a composite whole number greater than 1 has additional positive divisors. The source’s prime sentence omits itself, so read it with this correction. Two is the only even prime. Odd does not automatically mean prime: 9 = 3 × 3 is composite.
For a sieve from 1 to 100, exclude 1, keep 2 and cross out its proper multiples, then keep 3, 5 and 7 while crossing their remaining multiples. Starting at p² avoids repeating smaller multiples already crossed by smaller primes. Every composite up to 100 has a prime factor no larger than 10, so testing prime bases 2, 3, 5 and 7 is sufficient. There are 25 primes in this interval; that is a count for this exact range, not a prediction about any other interval.
Let a be the tens digit (1–9) and b the units digit (0–9). The number is 10a + b and its reversal has value 10b + a, even when b = 0. Adding gives 11(a + b), a multiple of 11. Their signed difference is 9(a − b); taking larger minus smaller gives 9 times the non-negative digit difference. Equal digits give zero, which is divisible by 9; no nonzero quotient is promised.
For digits a, b, c, write the original value as 100a + 10b + c and the reversal as 100c + 10b + a. Their difference is 99(a − c); the middle-digit terms cancel. Example: 743 − 347 = 396 = 99 × 4. A reversal beginning with zero still has a numeric value, such as 120 → 021 = 21. This is a place-value proof, not a calculator coincidence.
For 46 and 64, adding gives 110 = 11 × 10. The place-value explanation is 11 × (4 + 6). Testing one example illustrates the rule; the algebra explains why it holds for all allowed digits.
Printed page 2 omits itself in its prime definition and says cross out all multiples. Keep the chosen prime and cross its proper multiples; a prime has divisors 1 and itself.
Adding a complete Pascal row gives powers of 2 when rows start with row zero: 1,2,4,8,16. This is an additional observation; interior entries still follow the two-parent sum rule.
For the printed 4, 4, 5, 6, 8, 11, subtract 3 from every term to obtain 1, 1, 2, 3, 5, 8. Continuing that Fibonacci rule gives 13, then adding 3 back gives 16. Equivalently each new original term is the previous two added, minus 3: 8 + 11 − 3 = 16. Simply adding 8 + 11 gives 19 and fails the earlier terms. A finite prefix can fit other rules, so state which rule you use and review alternatives with a teacher.
State the starting values and the rule before deciding what comes next.
| Rule | Early terms | How to check |
|---|---|---|
| Start 4; add 6 | 4, 10, 16, 22, 28 | Every difference is 6 |
| Start 3; double | 3, 6, 12, 24, 48 | Every term is twice its predecessor |
| Positive squares n² | 1, 4, 9, 16, 25 | Differences are 3, 5, 7, 9 |
| Start 2,3; sum previous two | 2, 3, 5, 8, 13 | Each new term adds two previous terms |
A finite prefix can admit other rules. These rows explicitly choose a rule; different justified alternatives belong in open review.
Use the list as a text alternative to the coloured grid. Explain one retained or excluded value with factors.
| Interval | Retained primes |
|---|---|
| 1–20 | 2,3,5,7,11,13,17,19 |
| 21–40 | 23,29,31,37 |
| 41–60 | 41,43,47,53,59 |
| 61–80 | 61,67,71,73,79 |
| 81–100 | 83,89,97 |
There are 25 values in this exact range. One is excluded, and two is retained. This is a completed sieve count, not an estimate for another interval.
Use place value to explain why the divisibility rule holds for every allowed digit pair.
(10a + b) + (10b + a) = 11(a + b) (10a + b) − (10b + a) = 9(a − b)
The signed algebraic difference may be negative; larger minus smaller gives its non-negative magnitude. Equal digits give zero.
Write each digit at its actual place value, then simplify before choosing numbers.
(100a + 10b + c) − (100c + 10b + a) = 99(a − c)
The proof concerns numeric values. It still works for a leading-zero reversal and gives zero for a palindrome.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original repeat arrangement, not a photograph or copied textile design. Colours support reading the rule; they are not the rule itself.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original layout of a standard mathematical construction. A finite row does not establish a unique rule unless that rule is supplied.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original completed grid for the stated interval. Colour alone is not required: the adjacent text defines the highlighted cases and the panel lists the retained primes.
Show your working on paper. Check your answer, then choose Next.
34
28 + 6 = 34; the rule is explicitly stated.
6
Every consecutive increase is six.
48
24 × 2 = 48.
36
6² = 36.
21
The previous terms are 8 and 13; 8 + 13 = 21.
10
The entry is the sum of the two parents.
2
They are 1 and the prime itself.
neither
One is not greater than one and has only one positive divisor.
2
Every larger even number has divisor 2 as well as 1 and itself.
49
7² = 49; smaller multiples have smaller prime factors.
composite
The stated factorisation supplies divisors besides 1 and 91.
10
110 ÷ 11 = 10, also the digit sum 4 + 6.
5
45 ÷ 9 = 5, also the digit difference 8 − 3.
4
396 ÷ 99 = 4.
25
This counts primes in the stated interval only.
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Name the two orientations and the arrangement rule. State which orientation comes first and how neighbouring blocks alternate. A useful instruction describes relationships, not only colours.
Check that another learner can reproduce your chosen repeat.
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Different rules can agree on a finite prefix and then diverge. State a rule, test it on all given terms and explain the next term under that rule; ask for clarification if the question requires one unique answer without enough information.
Do not mark every different but justified continuation wrong.
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Rows zero to four are 1; 1,1; 1,2,1; 1,3,3,1; 1,4,6,4,1. Edges are 1 and each interior number sums its two parents.
Check row convention, both edges and one explained addition.
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The first adds 6 at every step and is linear in the chapter’s sense. The second multiplies by 2, so its successive differences are 3,6,12 and are not constant.
Name both operations and show the difference calculation.
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The next square is 6² = 36. The differences between consecutive positive squares are 3,5,7,9,11, so 25 + 11 = 36.
Use two consistent routes, not a bare answer.
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The next ten terms are 55,89,144,233,377,610,987,1597,2584,4181. For example 21 + 34 = 55 and 34 + 55 = 89.
Each new value must sum the previous two; check all ten terms.
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A prime greater than 1 has exactly two positive divisors: 1 and itself. Omitting itself would wrongly exclude every prime.
Define the term and use an example such as 7.
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Two is the selected prime. Cross proper multiples beginning at 4, while retaining the base itself. Similarly keep later prime bases while excluding their composite multiples.
Distinguish the base from its proper multiples.
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A composite number has a factor pair with at least one factor no larger than its square root. Up to 100 that smaller factor is at most 10, and any prime factor there is 2,3,5 or 7.
Explain the factor-pair argument; do not simply cite the finished grid.
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For example 15 = 3 × 5. It is odd but has divisors 3 and 5 as well as 1 and 15.
A different correct example is acceptable with its factors.
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(10a + b) + (10b + a) = 11a + 11b = 11(a + b). The sum is therefore divisible by 11.
Define both digits and connect factor 11 to divisibility.
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The signed difference is (10a + b) − (10b + a) = 9(a − b). Larger minus smaller makes the result non-negative, with magnitude 9 times the digit difference.
Handle a smaller or equal digit without a negative-length interpretation.
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(100a + 10b + c) − (100c + 10b + a) = 99(a − c). The two 10b terms cancel, leaving a multiple of 99.
Give the place values and the cancellation, not just a tested number.
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The reversal of 120 is written 021 but has value 21; the difference is 99. A palindrome equals its reversal, so the difference is 0, also divisible by 99.
Leading-zero value and zero difference must both be handled.
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A fixed-rule example can start at 5 and add 4. An open example can show a prefix and ask learners to propose and test a continuation rule. The open response needs human review.
Separate a uniquely specified calculation from creative rule-making.
Build a small sieve and a sequence-rule card set. Include a written digit-reversal proof.
Keep your work on paper or in a local file. This page does not upload it or automatically grade the project.
This original Kidemy practice follows the topic order of 2026 Class 8 Mathematics · English Version, chapter 1, printed pages 1–11. It does not replace the complete chapter or its practical teaching.
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