Why the same digit sum also settles divisibility by 3

What to assume they know

• The digit-sum test for 9 and the run-of-nines decomposition of place values (The digit-sum test for 9, and the algebra underneath it)
• Place value, and writing a number as a sum of its place values
• Factors and multiples, and the rule that a number divisible by k is divisible by every factor of k (The four divisibility facts you can prove, and how to use them)
• Prime factorisation
• Remainders, and the fact that a remainder is less than the divisor

What they should be able to do

• Identify the exact step of the nines argument where the number 9 was used, and check whether 3 can replace it there
• Find the remainder left by each power of ten on division by 3, and state the pattern
• Derive the digit test for 3 from the same place-value split that gave the test for 9
• Explain why a multiple of 9 must be a multiple of 3, and give a number showing that the reverse fails
• Read both verdicts — divisibility by 3 and by 9 — from a single digit total
• Use the digit total to give the remainder on division by 3
• Combine the digit test with a second test to settle divisibility by a composite such as 18 or 36

Where it usually goes wrong

• "Two separate rules to memorise." It is one argument with a different divisor substituted at one step. A student who has to remember both has not understood either.
• "If the digit total divides by three then the number divides by nine." The test reports what it reports. A total of 12 is a multiple of three and not of nine, and the number behaves the same way. The chapter's three printed counter-instances exist for this.
• "Divisible by three twice over means divisible by nine." Two factors of three do give nine, but "the digit total is a multiple of three" is one piece of information, not two. Sum the digits again if you want the finer verdict.
• "The test for six is the test for three." Six needs two as well. The chapter makes this its next question on Part I p.129, and the reason belongs to Divisibility by 6 and other numbers, checked through their factors.
• "The rule for three works only one way, because that is how the book says it." The printed sentence is one-directional and the fact is not. Say so plainly; a student who thinks the reverse might fail will not trust the test when it says no.
• "Powers of ten leave different remainders under three, so this cannot be as neat as the nines case." They all leave one, for the same reason they all leave one under nine. Work two or three of them rather than asserting it.

Questions to check understanding

• Decide whether a given number is a multiple of three without dividing, and give the remainder
• Decide, from one digit total, both whether a number is a multiple of three and whether it is a multiple of nine
• Explain why the nines argument still works when three is put in its place
• Give a number that is a multiple of three and not of nine, and one that is both
• Find a missing digit making a number a multiple of three, and say how many values work
• Settle divisibility by a composite such as 18 or 36 by combining tests, and justify the choice of tests
• State the test for three as a two-way claim and justify both directions

Examples worth working on the board

Inputs. This is the shortest passage in the chapter — half a page — so most of what the explanation needs is the argument, not fresh data.

• The chapter's opening claim (Part I, §5.2, p.126, subheading "A Shortcut for Divisibility by 3"). Printed: a number that is a multiple of nine is also a multiple of three. This is the chapter's fourth divisibility fact in miniature — a divisor carries all its own factors.
• The three counter-instances, printed: 15, 33 and 87. Each is a multiple of three and not of nine. Three of them, not one, because the point is not that there is an exception but that there is a whole further population.6, 6 and 15 — each a multiple of three that is not a multiple of nine. Note that the third does not "stop short" of nine at all; what the three have in common is failing the nine test, not being smaller than nine.
• The chapter's instruction, printed (same page): explore the remainders left when the powers of ten are divided by three, and explain from that why the method works. The chapter gives no working. Everything in sections 4 to 6 is the explanation carrying out an instruction the book issues and leaves open, and it should be presented that way.
• The place-value identities to reuse (Part I, §5.2, p.125): 10 as nine plus one, 100 as ninety-nine plus one, 1000 as nine hundred and ninety-nine plus one, and so on. Each run of nines is three times a run of threes — 9 as 3 × 3, 99 as 3 × 33, 999 as 3 × 333 — and that is the single new line. Use these three products as inputs; they are working added here, not the chapter's.
• The statement as printed (same page): a number divides by three if its digit total divides by three. The wording is one-directional, unlike the "if and only if" the chapter uses for nine on Part I p.125. Both directions hold. Section 11 is about that gap and should not gloss over it.
• Where the test is put to work later in the chapter.
  • Part I p.129, the fill-in table: ten numbers, each to be judged against nine divisors, with 3 among them. The three-column is fastest done with this test, and its 6- and 9-columns lean on it too.
  • Part I p.129: divisibility by 6 checked through 2 and 3, on the numbers 38, 225, 186 and 64. Part I p.130: divisibility by 24 checked through 3 and 8.
  • Part I p.132 no. 5: a six-digit number written 48, then a letter a, then 23, then a letter b, said to be a multiple of 18. Since 18 is 2 times 9, this needs the nines test and an even-number test together.
  • Part I p.133 no. 8: five multiples of 36 wanted between 45,000 and 47,000, with the approach to be shared with the class. 36 is 4 times 9.
  • Part I p.133 no. 12 (i): whether a multiple of 6 times a multiple of 3 must be a multiple of 9.
  • Part I p.133 no. 13: when the total of any three chosen numbers divides by 3.
  • Part I p.131 no. 4 (ii): the relationship between a number's digital root and its remainder on division by 3 or by 9 — the bridge to Digital roots, and what survives repeated digit-summing.
• The chapter prints no answers to any of these.

Figures to have open

• A nesting diagram: multiples of three as an outer region with multiples of nine inside it, and the three printed counter-instances placed in the gap between. Standard schematic. The chapter draws nothing here, and the nesting is the whole argument of sections 7 and 8. It also rehearses the Venn item on Part I p.134.
• A run-of-nines block dividing cleanly into three run-of-threes blocks, matched to the tile convention used for the nines test on Part I pp.124–125 so the two videos share one visual language.
• A powers-of-ten column with the leftover after division by three marked beside each row. Standard schematic.
• A two-verdict readout: one digit total, two answer slots. Standard schematic.
• No photograph is needed.

Where this sits in the book

• NCERT Ganita Prakash Class 8, Part I, printed Chapter 5, "Number Play", §5.2 "Checking Divisibility Quickly", printed subheading "A Shortcut for Divisibility by 3", Part I p.126. The passage occupies roughly a third of one page; it is the shortest exposition in the chapter.
• Backward pointers inside the same chapter: the place-value identities and the worked nines argument, Part I pp.124–125; the rule that a divisor carries its own factors, Part I p.120.
• Forward pointers inside the same chapter: the fill-in divisibility table, Part I p.129; divisibility through factors, Part I pp.129–130; digital roots and their link to remainders under three and nine, Part I pp.130–131.
• Chapter-end "Figure it Out", Part I pp.132–133, items 5, 8, 12 (i) and 13.
• The chapter's SUMMARY, Part I p.134, names 3 alongside 9 and 11 as the divisors whose shortcuts were explained.