The digit-sum test for 9, and the algebra underneath it

What to assume they know

• Place value in the Indian system, and reading a number as a sum of place values (Powers of 10, and place value written out for whole numbers and decimals)
• Letter-numbers, and the chapter's convention of writing the digits of a general number as separate letters
• Multiples, factors, division and remainders
• Taking a common factor out of a sum, and rearranging the terms of a sum
• The rule that a common divisor of several numbers divides their total (The four divisibility facts you can prove, and how to use them)

What they should be able to do

• Write a general multi-digit number as a sum of its place values using letter-numbers, and identify each digit's contribution
• Explain, from that expansion, why a units digit of zero settles divisibility by ten
• Find the remainder left by a round number such as forty or three hundred on division by nine, and see that it equals the count of tens or hundreds
• Decompose each place value as a run of nines plus one, and separate a number into a multiple-of-nine part and a digit-sum part
• Use the digit sum to find the remainder on division by nine, repeating the step until a single digit is reached
• State the test as an equivalence and explain why the converse also holds
• Judge the truth of a claim, its converse, its inverse and its contrapositive
• Use the test backwards to find a missing digit in a number known to be a multiple of nine

Where it usually goes wrong

• "Look at the last digit, as with 2, 5 and 10." The chapter breaks this deliberately on Part I p.124 with two numbers that both end in nine. Ten's test works because every higher place value is already a multiple of ten; nine has no such luck, and the whole rest of the section is the repair.
• "The digit sum is the remainder." It leaves the same remainder as the number, which is not the same thing. For 427 the digits add to 13, which is bigger than the divisor and has to be reduced again. Students who stop at the first sum will report remainders larger than 9.
• "Adding digits is a trick with no reason behind it." The reason is one line of place-value algebra, and the chapter prints it in full for a four-digit number. Do not present the rule before the reason; the chapter deliberately does not.
• "It only works for small numbers." Part I p.125 asks this and answers it: the identity holds for every place value, however high, because ten to any power is one past a run of nines.
• "If the digit total is a multiple of nine, the number might still not be." The test runs both ways. This is the whole content of the four-statement item, and it is why the chapter states the rule with an "if and only if".
• "Rearranging the digits could change the answer." The digit total is unchanged by reordering, so the verdict cannot change. Sreelatha's exercise is built on exactly this and asks the student to see it.
• "A missing digit will have one value." In a multiple-of-nine puzzle the digit total can often reach a multiple of nine at two different digit values. Part I p.132 no. 1 makes the point explicitly by asking why there are two.

Questions to check understanding

• Decide whether a given number is a multiple of nine without dividing, and state its remainder
• Write a general multi-digit number in expanded form using letter-numbers and read a divisibility test off the expansion
• Explain why the test works, using the run-of-nines decomposition of place values
• Find a missing digit that makes a number a multiple of nine, and say how many answers there are
• Find the multiple of nine nearest a given number, or count the multiples of nine in a stated range
• Judge a claim and its converse separately, and say whether a two-way test makes both true
• Decide the effect of rearranging a number's digits on its divisibility by nine

Examples worth working on the board

Inputs. Values marked "printed" are the chapter's own working.

• The general form of a number (Part I, §5.2, p.123). Printed: a five-digit number written with the letters e, d, c, b, a from the ten-thousands place down to the units, and expanded as e times ten thousand, plus d times a thousand, plus c times a hundred, plus b times ten, plus a. The general case is written with a leading ellipsis so the number of digits is not fixed. Printed instance: for 4075 the chapter gives d = 4, c = 0, b = 7, a = 5. Note that this instance is deliberately four digits, not five, so e is absent — a good moment to say what the ellipsis is doing.
• The easy test, worked (Part I, §5.2, p.123). Every place value above the units is already a multiple of ten, so only the units digit can stop a number from being a multiple of ten. The chapter then hands the same argument for 5, 2, 4 and 8 to the student.
• The all-nines-and-zeros case (Part I, §5.2, pp.123–124). Printed: 999, 909, 900, 90 and 990 offered as numbers to judge without calculating, with the chapter's own one-word verdict that every one of them qualifies. Printed expansion: 99009 set out as 9 times ten thousand, plus 9 times a thousand, plus 0 times a hundred, plus 0 times ten, plus 9.
• The pair that kills the units-digit idea (Part I, §5.2, p.124). Printed: 99 and 109, both ending in nine, only one of them a multiple of nine. The page says plainly that 9 is unlike 2, 5 and 10 in this respect.
• The leftovers pattern (Part I, §5.2, p.124). The chapter asks for the remainder when 10 is divided by 9, then for the multiples of ten in turn, and states the finding: the leftover equals the count of tens. It repeats the question for the multiples of a hundred and states that the leftover equals the count of hundreds.
• 427, worked on the page (Part I, §5.2, p.124). Printed: the number split as 400, 20 and 7 above a tile picture; four hundreds contribute a leftover of 4, two tens contribute 2, and the seven units contribute 7; those add to 13; one more group of nine can be taken out of 13, and the chapter states the final leftover as 4. The tile artwork below shows the leftover tiles gathered into a ring of nine with four outside it, arrowed and labelled. Read the tile labels off the printed page — they sit inside the artwork.
• The place-value identities (Part I, §5.2, p.125). Printed as a stack: 1 = 0 + 1; 10 = 9 + 1; 100 = 99 + 1; 1000 = 999 + 1; 10000 = 9999 + 1. The page states the consequence directly: each digit is the leftover from its own place value.
• 7309, worked in full (Part I, §5.2, p.125). Printed working, four lines: the expansion into place values; each place value replaced by its run-of-nines form; the brackets multiplied out; and the result grouped into two brackets, the first underlined and labelled as a multiple of nine, the second as the part that matters. The digit total is 19, then 1 + 9 gives 10, then 1 + 0 gives 1, and the chapter states the leftover as 1. Note a small printed slip: the marginal note beside the tile picture writes the digit total as 7 + 3 + 9 while the display line beneath writes 7 + 3 + 0 + 9. Both come to 19.
• The tile pictures on Part I p.125. Two panels. The left one shows the number's place values as stacks of tiles: seven purple thousand-tiles labelled 999 each carrying a small yellow 1, three teal hundred-tiles labelled 99 each with a 1, and nine loose yellow 1-tiles. The right one shows the same stacks with the 1s stripped off and gathered under a brace. A red tile labelled 9999 and a green tile labelled 9 appear at the top of the page as the other place values in the same family. Everything here is artwork lettering and must be read from the image.
• The statement of the test (Part I, §5.2, p.125). Printed as an "if and only if": the number is a multiple of nine exactly when its digit total is. The page adds that repeating the digit-summing to a single digit gives the leftover on division by nine.
• The four statements (Part I, §5.2, pp.125–126, items (i) to (iv), the question being which are correct and why). Compressed, so the wording stays mine: (i) a multiple of nine has a digit total that is a multiple of nine; (ii) the reverse of that; (iii) a number that is not a multiple of nine has a digit total that is not one; (iv) the reverse of that. Hand the four over as data. They are the original claim, its converse, its inverse and its contrapositive, and the interesting thing is that the equivalence makes all four behave alike — which is not what students expect from a set of four statements offered for judgement.
• Exercise inputs, §5.2 "Figure it Out" (Part I p.126). No. 1 gives five numbers to judge without dividing: 123, 405, 8888, 93547 and 358095. No. 2 asks for the smallest multiple of nine using no odd digits. No. 3 asks for the multiple of nine nearest to 6000. No. 4 asks how many multiples of nine lie between 4300 and 4400.
• Chapter-end exercise inputs (Part I p.132). No. 1 has a four-digit number written 31, then a letter z, then 5, said to be a multiple of nine; the item asks for z and, pointedly, why the item has two answers. No. 4 is Sreelatha's claim that reversing the digits of a multiple of nine leaves a multiple of nine, with a follow-up about other digit rearrangements. No. 5 has a six-digit number written 48, then a, then 23, then b, said to be a multiple of 18, and asks for every possible pair.
• The chapter prints no answers to any exercise item.

Figures to have open

• The tile figure: a place-value tile that visibly splits into a run-of-nines block and a single unit square, drawn for ten, a hundred, a thousand and ten thousand. This is the chapter's own device (Part I, §5.2, pp.124–125) and the argument is far harder without it. Redraw as a schematic; the printed art should not be reproduced.
• The two-heap separation diagram for a four-digit number: all the run-of-nines blocks swept into one pile, all the unit squares into another, with the second pile counted as the digit total. Standard schematic.
• A remainder ring: loose units gathered into groups of nine, with the ungrouped remainder highlighted, reusable for 427 and for 7309.
• A place-value strip with one digit replaced by a letter, for the missing-digit exercises. 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", Part I pp.123–126. The general form of a number and the tests for 10, 5, 2, 4 and 8 are on Part I p.123 under the section heading itself; the printed subheading "A Shortcut for Divisibility by 9" begins at the foot of Part I p.123 and runs to p.126.
• §5.2 "Figure it Out", Part I p.126, items 1 to 4.
• Chapter-end "Figure it Out", Part I p.132, items 1, 4 and 5.
• The chapter's SUMMARY, Part I p.134, names 3, 9 and 11 as the divisors whose shortcuts the chapter explains.
• Part I p.126 carries an owl-illustrated box on understanding why a method works rather than following a procedure; it sits immediately after this test and is effectively its moral.