Why the HCF-times-LCM shortcut works for two numbers but breaks for three

What to assume they know

• Reading HCF and LCM straight off two factorisations — reading HCF and LCM off aligned prime columns as the lower and the higher index
• Why a composite number has one prime factorisation and no other — that a factorisation is unique, which is what makes the columns well defined at all
• Adding indices when multiplying prime powers
• Division with large numbers, and comfort with a five-digit product

What they should be able to do

• Verify the two-number identity on a worked pair and state what it lets you compute
• Explain the identity by showing that the lower and higher indices of a prime reassemble its two indices in the product
• Use the identity in both directions: HCF and product to LCM, and HCF and LCM to the product
• Show, on the chapter's own three numbers, that the identity fails, and measure the size of the failure
• Attribute that failure to the middle index, prime by prime, and predict the discrepancy before dividing
• Give three numbers for which the identity nevertheless holds, and say what is special about them
• Apply the corrected three-number formulas from the chapter's closing note and check them against a known answer

Where it usually goes wrong

• "HCF × LCM = product is a law of numbers." It is a consequence of there being exactly two indices in each column. State it that way once and the three-number case stops being a surprise rule to memorise.
• "So for three numbers the identity is always wrong." No — 17, 23, 29 and 8, 9, 25 both satisfy it. The chapter's remark is written about the example in front of it; the defensible general statement is that the identity is not guaranteed for three numbers.
• "You can find the HCF of three numbers by pairing them and using the identity twice." The closing note exists precisely because that does not work; it needs all three pairwise HCFs and the three-way one together.
• "The shortfall is random." It is 24 for 6, 72, 120 and 3 for 12, 15, 21, and both are read off the columns before dividing. Predict it, then divide, and let the prediction be confirmed.
• "If the HCF is 1 the LCM is the product, for any count of numbers." True only when no two of them share a prime. Three numbers can have HCF 1 while two of them still share a factor — 6, 10, 15 is the standard trap, and their LCM is 30, not 900.
• "Q4 wants the factorisations of 306 and 657." It hands over the HCF so that the identity does the work. Reaching for factor trees there is a sign the identity has not landed.

Questions to check understanding

• Given the HCF of two numbers and the numbers themselves, find the LCM
• Given the LCM and one number and the HCF, find the other number
• Verify the identity on a stated pair by computing all four quantities
• Show by a single counterexample that the identity does not extend to three numbers, and separately give three numbers for which it happens to hold
• Apply the corrected three-number formula to compute an LCM from pairwise HCFs

Examples worth working on the board

Values marked verified are worked out here on the chapter's printed data; the chapter prints no answers to its exercises, and no answer key was consulted.

• 6 and 20 (§1.2, p. 4). Printed: HCF 2, LCM 60, and the chapter's own observation that these multiply to 6 × 20. Verified: 2 × 60 = 120 and 6 × 20 = 120. Column check when explaining it: the prime 2 has indices 1 and 2, and the lower and higher of those are 1 and 2 — the same two numbers, reused.
• 96 and 404 (§1.2, pp. 4–5). Printed inputs: 96 = 2⁵ × 3, 404 = 2² × 101, HCF 4, and the chapter obtains the LCM by dividing the product by the HCF. Verified: 96 × 404 = 38784, and 38784 ÷ 4 = 9696. This is the identity being used as a labour-saving device rather than admired.
• Exercise 1.1 Q4 (§1.2, p. 5). Given data: HCF(306, 657) = 9; the LCM is asked for. Verified: 306 × 657 = 201042, so the LCM is 201042 ÷ 9 = 22338. Independent check from factorisations, which the question does not require: 306 = 2 × 3² × 17 and 657 = 3² × 73, so the HCF really is 3² = 9 and the LCM is 2 × 3² × 17 × 73 = 22338.
• Example 4 and the Remark — 6, 72 and 120 (§1.2, p. 5). Printed inputs: 6 = 2 × 3, 72 = 2³ × 3², 120 = 2³ × 3 × 5, with HCF 6 and LCM 360. The chapter then remarks that multiplying all three together does not agree with HCF times LCM. Verified: 6 × 72 × 120 = 51840, while 6 × 360 = 2160. The two differ by a factor of 24.
• Locating the shortfall (not in the book). The 2 column holds indices 1, 3, 3: lowest and highest are 1 and 3, so 4 of the 7 available are used and 3 are dropped. The 3 column holds 1, 2, 1: 1 and 2 are used out of 4, so 1 is dropped. The 5 column holds 0, 0, 1 and nothing is dropped. Verified: the dropped indices amount to 2³ × 3¹ = 24, which is exactly the factor measured above. This is the strongest moment in the topic — the discrepancy is predicted from the columns before any division is done.
• Exercise 1.1 Q3(i) as a second test (§1.2, p. 5): 12, 15 and 21. Verified: HCF 3, LCM 420, product 3780, and 3 × 420 = 1260, a factor of 3 short. The 3 column holds indices 1, 1, 1, and dropping the middle one loses exactly a factor of 3; no other column loses anything.
• Exercise 1.1 Q3(ii) and Q3(iii) as the counter-case (§1.2, p. 5): 17, 23 and 29; then 8, 9 and 25. Verified: the first triple has HCF 1 and LCM 11339, and 1 × 11339 is exactly 17 × 23 × 29. The second has HCF 1 and LCM 1800, and 1 × 1800 is exactly 8 × 9 × 25. In both, every prime appears in only one of the three numbers, so every middle index is 0 and there is nothing to drop. These two items keep the explanation honest.
• The corrected formulas (closing note, p. 9). The chapter's note gives the LCM of three numbers as their product times their three-way HCF, divided by the three pairwise HCFs; and the HCF as their product times their three-way LCM, divided by the three pairwise LCMs. Verified on 6, 72, 120: HCF(6, 72) = 6, HCF(72, 120) = 24, HCF(6, 120) = 6, so the first formula gives 51840 × 6 ÷ (6 × 24 × 6) = 311040 ÷ 864 = 360, the LCM. And LCM(6, 72) = 72, LCM(72, 120) = 360, LCM(6, 120) = 120, so the second gives 51840 × 360 ÷ (72 × 360 × 120) = 18662400 ÷ 3110400 = 6, the HCF. Both land on the values Example 4 already produced.

Figures to have open

• A prime-column table that can show two rows and then three, with individual index cells that can be highlighted or greyed. Standard schematic, and the same table used in Reading HCF and LCM straight off two factorisations — reuse it so the student sees one picture across the module.
• A balance-scale or two-pan comparison holding 51840 against 2160, with the 24 shown as what is missing from one pan. Standard schematic.
• The two corrected formulas set as displayed fractions. They are printed in the chapter's closing note on p. 9; set them fresh rather than reproducing the printed box.

Where this sits in the book

• NCERT Mathematics, Textbook for Class X, printed Chapter 1 "Real Numbers", §1.2, p. 4 for the two-number identity as first noticed, pp. 4–5 for Example 3, p. 5 for Example 4 and the Remark that follows it.
• Exercise 1.1 Q3 and Q4, p. 5.
• The closing note printed after §1.4 on p. 9, which states the inequality for three numbers and gives the two corrected formulas.