Writing "leaves remainder r" as an algebraic expression

What to assume they know

• Division with a remainder, and the fact that a remainder is less than the divisor
• Multiples written as 5k, and letter-numbers standing for any whole number
• Collecting like terms, and taking a common factor out of a sum
• The four divisibility facts, especially that a common divisor divides a sum and a difference (The four divisibility facts you can prove, and how to use them)
• The two-class split of even numbers by their remainder on division by 4, from "Pairs to Make Fours" — this topic generalises it

What they should be able to do

• Generate several numbers satisfying a stated remainder condition, and recognise the regular step between them
• Write the family of numbers leaving a given remainder on division by a given number as an algebraic expression
• Test a candidate expression by substituting successive values and comparing the list produced with the family wanted
• Explain why two differently written expressions can name the same family, and state the condition on the starting value of the letter
• Explain why changing which number multiplies the letter changes the family entirely
• Represent a remainder condition as complete rows with a short row left over
• Deduce the remainder of a sum or difference from the remainders of the parts, and reduce a result that lands outside the permitted range
• Translate a word puzzle carrying several remainder conditions into a list of algebraic conditions

Where it usually goes wrong

• "5k + 3 and 3k + 5 are the same, because addition can be turned round." The number multiplying the letter is the divisor and sets the step; the other number is the offset. Swapping them produces a family stepping by 3, which is a different set entirely. Show both tables together.
• "Remainder 3 means the number is 3." Three is one member of the family, the one you get at the very start. The condition describes every number five steps along from it in both directions.
• "5k − 2 must be a different family, because it has a minus sign." Two short of a multiple of five and three past the previous multiple of five are the same position. The chapter prints both tables with identical entries for exactly this reason.
• "The letter always starts at 0." It starts wherever it must for the values to make sense in context. The chapter says explicitly on Part I p.122 that k is at least 1 in the second form. If anyone omits that line the two tables will not agree.
• "The remainder of a sum is the sum of the remainders." Only after reduction. Exercise 7 is built so the two remainders overshoot the divisor, and a student who stops before reducing gets a "remainder" larger than the number being divided by.
• "A remainder can come out negative." It cannot, but an intermediate calculation can, and the honest move is to add the divisor back. This idea returns with force in the test for 11 (The alternating-sum test for 11), where a negative total is exactly what the method produces.
• "Two remainder conditions can be combined by adding them." They constrain the same number simultaneously. The right move is to describe one family and then sieve it with the other condition.

Questions to check understanding

• Write an expression describing every number that leaves a given remainder under a given divisor
• Given several candidate expressions, decide which describe a stated family, and justify each rejection
• Show that two expressions describe the same family, stating any restriction on the letter
• Given the remainders of two numbers under one divisor, state the remainders of their sum and difference without dividing
• Find numbers satisfying two remainder conditions at once, and write an expression for all of them
• Find the smallest number satisfying several remainder conditions, and explain why it is the smallest
• Examine a classmate's claim about sums of numbers drawn from one remainder family

Examples worth working on the board

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

• The opening prompt (Part I, §5.1, p.121, subheading "What Remains?"). Find a number leaving 3 under division by 5, then write more such numbers. The chapter supplies none — the student's own list is the raw material.
• The six candidate expressions, printed as options (Part I, §5.1, p.121): (i) 3k + 5, (ii) 3k − 5, (iii) 3k over 5 written as a stacked fraction, (iv) 5k + 3, (v) 5k − 2, (vi) 5k − 3. The chapter goes on to develop only two of the six — 5k + 3 in its own paragraph and 5k − 2 on the next page — and never says in so many words that the other four fail. That rejection is the student's work.
• The first table, printed (Part I, §5.1, p.121). Values of k running 0, 1, 2, 3, 4 across the top; the row beneath gives 5k + 3 as 3, 8, 13, 18, 23. Note the step of 5 between consecutive entries — that is the visual the section turns on.
• The row picture, printed (Part I, §5.1, p.121). Rows of five yellow tokens, the block labelled k rows with the width marked 5, and a short row of three tokens beneath it. This is the same drawing convention used all through "Pairs to Make Fours", and reusing it deliberately is worth one sentence saying so.
• The second table, printed (Part I, §5.1, p.122). Values of k running 1, 2, 3, 4, 5; the row beneath gives 5k − 2 as 3, 8, 13, 18, 23 — the same five numbers as the first table. The page states the condition that k is at least 1 here, and says these numbers can equally be seen as two short of a multiple of 5. The page then asks whether still other expressions generate the same family, and leaves it open.
• Exercise inputs, §5.1 "Figure it Out" (Part I pp.122–123).
  • No. 4: numbers leaving 2 under division by 3 and 2 under division by 4; find several, then write one expression covering all of them.
  • No. 5: a verse puzzle about a handful of pebbles. Its conditions, in order: grouped in threes one is left; paired up one is left, so the count is odd; grouped in fives one is left; grouped in sevens nothing is left; and the count is not more than one hundred. Hand the conditions over as a list — the puzzle is the exercise, and the number is not printed anywhere in the chapter.
  • No. 6: Tathagat's claim. He collects numbers leaving 2 under division by 6 and says any three of them add to a multiple of 6.
  • No. 7: 661 leaves 3 under division by 7, and 4779 leaves 5. Find the remainders of 4779 + 661 and of 4779 − 661 without dividing, and show the reasoning both algebraically and with a picture. This is the item that forces the reduction step, since the two remainders together exceed 7.
  • No. 8: a number leaving 2 under division by 3, 3 under division by 4 and 4 under division by 5; find the smallest such, and explain simply why it is the smallest. Note the shape of the three conditions — each remainder is one short of its divisor — because that shape is the simple explanation the item is fishing for.
• The chapter prints no answers to any of these.

Figures to have open

• A number line with one remainder family marked as evenly spaced ticks, and a second family marked on the same line in another colour, so the common members can be seen. Standard schematic. The chapter draws no number line here and the section needs one.
• The rows-of-five block with a short row of three (Part I, §5.1, p.121), redrawn as a schematic and reused at other divisors.
• The two substitution tables, drawn to the same grid so the identical value rows can be laid over each other (Part I, §5.1, pp.121–122).
• A row-block picture for the sum in exercise 7: two blocks of sevens with short rows of 3 and 5, the two short rows combining into one further complete row plus a leftover. Standard schematic; the exercise asks for a visual and the chapter provides none.
• No photograph is needed. The pebble puzzle's pot illustration on Part I p.122 is decorative and carries no data.

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 "What Remains?", Part I pp.121–122.
• §5.1 "Figure it Out", Part I pp.122–123, items 4, 5, 6, 7 and 8.
• Backward pointer inside the same section: the two-class split of even numbers by remainder under division by 4 is set up at Part I p.116 under "Pairs to Make Fours", and this subheading generalises it.
• Forward pointer inside the same chapter: negative intermediate remainders reappear in the test for divisibility by 11, Part I pp.128–129.