Decoding a divisibility trick with algebra

What to assume they know

• Writing a two-digit number as ten times its tens digit plus its units digit, and a three-digit number in the corresponding way
• Adding and subtracting expressions, and collecting like parts
• Taking a common factor out of an expression
• What "divisible by" means, and that a factored expression displays its factors
• Turning a stated word problem into an equation with one or two unknowns, and solving it
• Splitting a whole number into its prime factors

What they should be able to do

• Write a two-digit number and its reversal as expressions in the same two digits
• Subtract those expressions, simplify, and take out the common factor to display the 9
• Read the quotient off the factored form and identify it as the gap between the digits
• Carry out the other case, when the tens digit is the larger, and say why the conclusion is unchanged
• Add the two expressions instead, and account for the 11 that appears
• Extend the method to three digits cycled around, factor the result, and name both divisors it displays
• Explain why repeating a three-digit block gives a number divisible by 7, by 11 and by 13, and what three successive divisions return
• Set up and solve each of the chapter's end-of-section word problems by naming the unknown and writing what the situation asserts
• Say, in general, why algebra is what justifies a claim about all numbers where examples cannot

Where it usually goes wrong

• "It worked because 47 has something special about it." Nothing about 47 enters the argument. The expression covers every two-digit number with unequal digits at once, which is a different kind of statement from a hundred successful examples.
• "9 must be special." 9 is 10 − 1. The nine is manufactured by the two place weights the reversal swaps, and the addition version manufactures 11 = 10 + 1 in exactly the same way from exactly the same weights.
• "The digits have to be far apart." They only have to differ, so that the difference is not zero. A gap of 1 gives 9 itself, which the trick handles fine.
• "The quotient is unpredictable." It is the digit gap, and once you know that the trick has a second half: name the gap as well.
• "If the tens digit is the larger, the trick breaks." The difference is nine times the gap the other way round, which is still a multiple of 9. The chapter leaves this case to the reader precisely because it changes nothing.
• "Adding should also give a multiple of 9." It gives a multiple of 11. Students who have internalised "reversal means nines" will predict wrongly here, which is what makes item 2 worth doing before item 3.
• "Divisibility by 7, 11 and 13 for a repeated block is a coincidence." It is one factorisation, 1001 = 7 × 11 × 13, applied once. The chapter's own hint says as much.
• "Justifying means checking enough cases." Three instances of the addition rule are printed, and the item still asks for a justification. The chapter's SUMMARY states outright that this distinction is what algebra is for.
• "The word problems at the end belong to a different chapter." Every one is the same act: name what you do not know, write down what the situation asserts about it, solve. That is why they are here and not in a chapter on equations.
• "Karim was unlucky." Karim was arithmetically doomed. The break-even charge equals the holding, and 8 exceeded his 7 from the outset; the genie, knowing the holding, chose accordingly.

Questions to check understanding

• Justify, using letters, that the difference between a two-digit number and its reversal is a multiple of 9, and state the quotient
• Do the same for the sum and the divisor 11
• Given a three-digit number, justify that the total of its three cyclic versions is a multiple of 37, and of 3
• Explain why a six-digit number made by repeating a three-digit block is divisible by 7, by 11 and by 13
• Given a stated quotient from the nines trick, list the two-digit numbers that could have produced it
• Set up and solve a two-unknown word problem of the heads-and-legs type, and then solve it again without letters
• Solve an age problem stated as two ratios at two different times
• Work out a break-even condition for a repeated doubling-and-charging process
• Short-answer reasoning: why do three worked instances not justify a claim about all numbers

Examples worth working on the board

Values marked verified are worked out here on the chapter's stated inputs. No answer to any exercise the chapter sets the reader is printed anywhere in Part II pp.135–147, and Part II has no answer-key appendix. This section's Figure it Out block is the longest in the chapter. (The chapter does print worked answers to its own demonstrations — the 291 decode resolved to 25th December on Part II p.137, the apex-10 and apex-60 pyramids shown completed on Part II pp.138–139, the sum-36 block given as 5, 6, 12, 13 on Part II p.141, and the worked algebra grid resolving to 9 and 5 on Part II p.142 — so do not say the chapter prints no answers at all.)

• The trick as performed (Part II §6.6, p.144, a two-character comic strip; Mukta gives the instructions this time and Shubham computes). The instructions: pick a two-digit number whose two digits are not the same and keep it secret; write its digits the other way round; take the smaller from the larger; divide what you get by 9. Mukta then asserts the division will come out exactly. Shubham's worked column is printed: he chooses 47, reverses it to 74, computes 74 − 47 = 27, and divides to get 3.
• The algebra (Part II §6.6, p.145). The chapter names the two digits and writes the number as ten times the first plus the second, and the reversal the other way about. Taking the case where the units digit is the larger, it prints the subtraction, the collecting step, and the factored result: nine times the gap between the two digits. It then states that this is divisible by 9 and hands the other case — first digit larger — to the reader.
• The other case. Verified: the subtraction gives nine times the gap the other way round, so the difference is again a multiple of 9. Nothing in the argument depends on which digit is larger; only the sign of the gap changes, and the trick takes the smaller from the larger so the gap is positive either way.
• Figure it Out item 1 (Part II §6.6, p.145) asks what the quotient is and whether it relates to the two numbers. Verified: the quotient is the gap between the digits. Against the printed example, 7 − 4 = 3, which is the quotient printed in the comic strip. This is the moment the trick stops being a trick — the performer can also announce the digit gap.
• Figure it Out item 2 (Part II §6.6, p.145) replaces the subtraction with an addition, and prints three instances: 31 with 13 gives 44; 28 with 82 gives 110; 12 with 21 gives 33. The item observes that all three are multiples of 11 and asks for a justification. Verified: adding the two expressions gives eleven times the total of the two digits, so the quotient on dividing by 11 is that digit total — 4, 10 and 3 for the three printed instances.
• Figure it Out item 3 (Part II §6.6, p.145). A three-digit number and the two numbers made by cycling its digits round, all three added. The item asks for a justification that the total is always a multiple of 37, and whether it is always a multiple of 3; a printed hint points at multiples of 37. Verified: each digit lands once in each of the three positions, so the total is 111 times the sum of the three digits, and 111 is 3 times 37. Both divisibility claims follow at once, from the same factorisation.
• Figure it Out item 4 (Part II §6.6, p.145, carrying a Math Talk badge). A three-digit number written twice in a row to make a six-digit number, then divided by 7, by 11 and by 13 in turn; a printed hint says to multiply those three together. Verified: the six-digit number is the three-digit one multiplied by 1001, and 1001 is 7 times 11 times 13, so the three divisions return the original three-digit number and leave nothing over at any stage.
• Figure it Out item 5 (Part II §6.6, pp.145–146). Three shrines, each fronted by a pond that doubles whatever flowers are dipped in it. The flowers are dipped in the first pond and some are left at the first shrine; what remains is dipped in the second and some left at the second; what remains is dipped in the third and all of it left at the third. The same number was left at each shrine. Verified: writing the starting count and the offering as two unknowns gives eight times the start equal to seven times the offering, so the smallest whole-number answer is 7 flowers to start and 8 left at each shrine. Trace: 7 doubles to 14, leave 8, carry 6; 6 doubles to 12, leave 8, carry 4; 4 doubles to 8, leave 8.
• Figure it Out item 6 (Part II §6.6, p.146, with a Math Talk badge and an illustration of horses and hens behind a fence). Heads number 55 and legs 150. The item also asks for a solution without letters, and prints a hint: suppose all 55 were hens, then work from the shortfall in legs. Verified: 20 horses and 35 hens. The hint's route: 55 hens would give 110 legs, the shortfall is 40, and each horse accounts for 2 extra legs.
• Figure it Out item 7 (Part II §6.6, p.146). A mother's age is at present five times her daughter's; in six years it will be three times. Verified: the daughter is 6 and the mother 30; in six years they are 12 and 36.
• Figure it Out item 8 (Part II §6.6, p.146). Gauri and Naina are cowherds; Naina's herd is twice Gauri's, and Naina says that handing over three cows would leave them equal. Verified: Gauri has 6 and Naina 12; after the transfer both have 9.
• Figure it Out item 9 (Part II §6.6, p.146). A dosa cart: ₹5000 a day in rent, ₹10 to make one dosa. (i) At 100 dosas a day, what price gives a profit of ₹2000? (ii) At a price of ₹50, how many dosas a day give a profit of ₹2000? Verified: ₹80; and 175 dosas.
• Figure it Out item 10 (Part II §6.6, p.146). Three fractions to evaluate: 1 over 3; then 1 + 3 over 5 + 7; then 1 + 3 + 5 over 7 + 9 + 11. A printed hint points at what an opening run of odd numbers adds up to. Verified: all three equal one third, because the nth fraction has as its numerator the odd numbers from 1 up to the nth of them, totalling n squared, and as its denominator the next n odd numbers after those, totalling 3 times n squared.
• Figure it Out item 11, "Karim and the Genie" (Part II §6.6, p.147, with an illustration of the genie above the banyan tree). Each circuit of the tree doubles the coins in Karim's pocket; he pays the genie 8 coins per circuit. After the third doubling he holds exactly 8 coins, which is exactly what he owes. Three questions, the last two with a Math Talk badge: how many coins he began with; what charge per circuit he should have insisted on if he wanted to gain; and how the genie should set the charge to take everything. Verified: (i) 7 coins — 7 doubles to 14 and 8 paid leaves 6; 6 doubles to 12 and 8 paid leaves 4; 4 doubles to 8, which is the 8 he owes. (ii) The charge must be strictly less than what he holds when he sets out: doubling then paying leaves twice the holding less the charge, which beats the holding exactly when the charge is smaller than it. At a charge equal to the holding the deal is a treadmill, and 8 was more than his 7, so he was losing from the first circuit. (iii) A charge of twice his current holding empties him in a single circuit.
• The SUMMARY (Part II p.147). Three bullets: that algebra is useful for modelling and understanding numerical situations and therefore turns up almost everywhere; that it is indispensable for justifying mathematical claims; and a list of what this chapter applied it to. The middle bullet is the one to close the explanation on, because it is the thesis of the whole chapter.

Figures to have open

• A place-value pair diagram: two digit tiles with the weights 10 and 1 above them, and the same two tiles with the weights exchanged. This is the explanation's central figure and carries sections 3, 4 and 7 at once. Not in the chapter.
• The subtraction and the addition written out as annotated line-by-line derivations. Standard notation panels; the chapter prints the subtraction (Part II p.145) and leaves the addition to the reader.
• A factor tree for 111 and another for 1001. Standard schematic; not in the chapter, which only hints at them.
• A four-panel reduction of the story problems: the story in one line, the unknown named, the equation, and nothing else. Standard schematic; deliberately withholds the answers.
• The genie and the horses-and-hens illustrations (Part II pp.146–147) are decorative and need not be reproduced. A coin counter that halves visibly across three circuits does more work than either.

Where this sits in the book

• NCERT Ganita Prakash Class 8, Part II, printed Chapter 6, "Algebra Play", §6.6 "Decoding Divisibility Tricks", Part II pp.144–147. The section opens in the lower half of Part II p.144 with the comic strip, the algebra fills the top third of Part II p.145, and the eleven-item Figure it Out block runs from the middle of Part II p.145 to the middle of Part II p.147.
• Item 11 is printed with its own bold title, "Karim and the Genie", and runs as a short narrative down most of Part II p.147.
• Math Talk badges sit in the outer margin against item 4 (Part II p.145), item 6 (Part II p.146) and item 11 (Part II p.147). The badge lettering is artwork and does not appear in extracted text.
• Printed hints are attached to items 3, 4, 6 and 10.
• The chapter's SUMMARY occupies the foot of Part II p.147 and is the last thing in the chapter file.
• §6.5 "The Largest Product", which ends higher on Part II p.144, is covered by Arranging given digits to make the largest product.