Always, sometimes, or never: one counterexample settles it

What to assume they know

• Factors, multiples, divisibility and remainders
• The four divisibility facts the chapter proves (The four divisibility facts you can prove, and how to use them), since this topic is about how they were earned
• Parity, and reading an expression's parity from its parts (Deciding parity without computing the answer)
• Letter-numbers, and forming an equation from a claim so that it can be tested
• LCM and prime factorisation

What they should be able to do

• Classify a divisibility statement as always, sometimes or never true, and name the evidence the verdict rests on
• Produce a single counterexample that refutes a stated universal claim, and explain why one is enough
• Produce an instance and a counterinstance to establish a verdict of "sometimes true"
• Build an impossibility argument for a "never true" verdict, using a parity or divisibility clash
• Explain why substituting values can refute a general claim but cannot confirm one
• Set up a claim about two unrelated numbers using two different letters, and say why one letter would narrow the claim
• Examine a conjecture attributed to another student, decide it, and write the justification

Where it usually goes wrong

• "Sometimes true means I'm not sure." It is a definite verdict with a definite proof obligation: exhibit one case that works and one that does not. Students routinely use it as a hedge, and the chapter's two worked "sometimes" statements each come with exactly that pair.
• "I tried twenty numbers and it worked every time, so it is always true." Twenty is not every number. The chapter says outright on Part I p.115 that the supply of cases is unlimited. Show a claim that survives many small numbers and dies on a larger one.
• "A counterexample only shows the rule usually works." It shows the rule as stated is false. What may survive is a corrected rule with a condition added — which is the productive response, and worth modelling on statement 7.
• "Never true just means nobody has found an example." Not searching is not an argument. A "never" verdict needs a clash: the chapter's is between an even quantity and an odd one.
• "Any even and any odd number can be written 2n and 2n + 1." That forces them to be neighbours. Two independent numbers need two independent letters. The chapter's own character asks this on Part I p.121.
• "If the statement is false, its reverse must be true." Item (ii) of Part I p.122 no. 3 turns a true divisibility fact around and gets something that is not always true. Reversing a true statement is a new claim needing a new verdict.
• "A conjecture with a name on it is probably right." Four of the exercises put claims in students' mouths. Some hold and some do not, and the naming is deliberate: it makes examining a claim feel like ordinary conversation rather than an attack.

Questions to check understanding

• Give a verdict of always, sometimes or never for a stated divisibility claim, with a justification in algebra
• Refute a stated claim with a single counterexample, and say why one suffices
• Supply an instance and a counterinstance to support a verdict of "sometimes"
• Show that a stated claim can never hold, by deriving a contradiction from it
• Examine a conjecture attributed to a named student and write a reasoned reply
• Repair a false claim by adding the condition that makes it true
• Explain why a claim about two arbitrary numbers cannot be set up with a single letter

Examples worth working on the board

Inputs. The chapter prints verdicts for statements 1 to 5 and 8, and leaves 6 and 7 to the class.

• The eight statements (Part I, §5.1, pp.118–121). Numbered on the page, and the numbering should be kept so a teacher can point at the book.
  • Eight dividing two numbers separately must divide their sum. Verdict printed: always true. The page then asks whether subtraction behaves the same, and leaves that half open.
  • If a number is divisible by 8, then 8 divides any two numbers that add up to it. Verdict printed: sometimes true.
  • If a number is divisible by 7, every multiple of that number is divisible by 7. Verdict printed: always true.
  • A number divisible by 12 is divisible by every factor of 12. Verdict printed: always true.
  • A number divisible by 7 is divisible by every multiple of 7. Verdict printed: sometimes true.
  • Divisible by both 9 and 4 forces divisibility by 36. Flagged "Math Talk"; no verdict printed.
  • Divisible by both 6 and 4 forces divisibility by 24. Flagged "Math Talk"; no verdict printed.
  • An odd number added to an even number gives a multiple of 6. Verdict printed: never true.
• The witnesses the chapter supplies. For statement 2: 72 written as 48 + 24, annotated 8 × 9 = 8 × 6 + 8 × 3, and 72 written as 50 + 22 — one split where the divisor reaches both parts and one where it reaches neither. For statement 5: 42, written as 7 × 6, is divisible by 14, written as 7 × 2, and is not divisible by 28, written as 7 × 4. Each "sometimes" verdict arrives as a pair, and the pairing is the lesson.
• The counterexample for statement 7, printed later in the chapter. Part I p.130 states that 12 is divisible by both 4 and 6 and not by 24. Use the chapter's own number. It is worth pointing out that the book supplies the refutation nine pages after posing the question, which is exactly how conjecturing works.
• The never-true argument (Part I, §5.1, p.121). Two routes are printed. The short one: multiples of 6 are all even, while an odd added to an even is odd. The algebraic one: set an even number 2n plus an odd number 2m + 1 equal to a multiple of 6, written 6j, then rearrange to 2(n + m) = 6j − 1 and observe that an even value has been made equal to an odd one.
• The one-letter trap (Part I, §5.1, p.121, artwork). A student is drawn at the side of that argument, in a green checked shirt with a notebook and pen, asking in a speech bubble whether the even and odd numbers could be written 2n and 2n + 1 instead. The question deserves a real answer: that pair is forced to be consecutive, so it would only settle the special case, and a claim about any even and any odd needs two independent letters. Read the speech bubble on the printed page — the artwork carries it.
• Exercise items, §5.1 "Figure it Out" (Part I p.122). No. 3 offers five statements for verdicts, asking for examples and non-examples where they help and for algebra to back the claim. Compressed, so that the wording stays mine: (i) two even numbers, added — must the total be a multiple of 3? (ii) a number that fails division by 18 — must it fail division by 9 too? (iii) two numbers, neither one a multiple of 6 — can their total be one? (iv) a multiple of 6 added to a multiple of 9 — always a multiple of 3? (v) a multiple of 6 added to a multiple of 3 — always a multiple of 9? Item (ii) is the one to plan around: it turns a true divisibility fact inside out, and the inverted claim is a new claim. No. 6 is Tathagat's claim: he has numbers that leave 2 on division by 6, and says any three of them add to a multiple of 6.
• Exercise items, chapter-end "Figure it Out" (Part I pp.132–133). No. 2 is Snehal's claim, about a number leaving 8 under division by 12 added to a number 4 short of a multiple of 12, said to give a multiple of 8. No. 4 is Sreelatha's claim, that reversing the digits of a multiple of 9 keeps it a multiple of 9, with a follow-up about other digit shuffles. No. 11 is Deepak's claim, that some multiples of 11 stay multiples of 11 when doubled and others do not. No. 12 offers four more statements for verdicts, again compressed here: (i) a multiple of 6 times a multiple of 3 — must the product be a multiple of 9? (ii) three consecutive even numbers, added — must the total divide by 6? (iii) a six-digit multiple of 6 whose first two digits are swapped and whose third and fourth are swapped — is the new number still one? (iv) the expression 8(7b − 3) − 4(11b + 1) — always a multiple of 12? No. 13 asks when the total of any three chosen numbers divides by 3.
• The chapter prints no answers to any exercise item. Every verdict above that is not explicitly marked as printed is for the student.

Figures to have open

• A verdict board: three labelled columns with the evidence each verdict requires written beneath. Standard schematic; it is the organising image of the whole video and the chapter draws nothing like it.
• Prime-factor bars for 9 with 4 and for 6 with 4, so the shared factor is visible rather than asserted. Standard schematic.
• The student character asking the 2n and 2n + 1 question (Part I, §5.1, p.121). The question matters more than the drawing — a plain speech bubble is enough, and the printed artwork should not be reproduced.
• A split-number diagram for 72, showing both of the chapter's splits against a row of eights. Standard schematic.
• No photograph is needed.

Where this sits in the book

• NCERT Ganita Prakash Class 8, Part I, printed Chapter 5, "Number Play", §5.1 "Is This a Multiple Of?", printed subheading "Always, Sometimes, or Never", Part I pp.118–121.
• §5.1 "Figure it Out", Part I p.122, items 3 and 6.
• Chapter-end "Figure it Out", Part I pp.132–133, items 2, 4, 11, 12 and 13.
• The chapter's SUMMARY, Part I p.134, closes by naming examples and counterexamples alongside algebra and visualisation as the tools the chapter used. That sentence is this topic's warrant.
• Part I p.126 carries an owl-illustrated box saying that learning mathematics is about understanding why something works rather than following procedures; it sits in §5.2 but belongs to this argument.