Count the beads three ways

The idea

For positive integers p, q, r: LCM(p,q,r) = p × q × r × HCF(p,q,r) ÷ [HCF(p,q) × HCF(q,r) × HCF(p,r)], and HCF(p,q,r) = p × q × r × LCM(p,q,r) ÷ [LCM(p,q) × LCM(q,r) × LCM(p,r)].

Three numbers

6, 72, 120: HCF = 6, LCM = 360. But 6 × 72 × 120 = 51840, and HCF × LCM = 2160. They are not equal.

Can we still get the LCM of three numbers from their HCFs?

• Yes, with a clever formula
• No, never
• Not sure

Three steps on the beads

Run the formula one step at a time and watch the beads of each prime.

Two formulas

For positive integers p, q, r: LCM(p, q, r) = p × q × r × HCF(p, q, r) ÷ [HCF(p, q) × HCF(q, r) × HCF(p, r)], and HCF(p, q, r) = p × q × r × LCM(p, q, r) ÷ [LCM(p, q) × LCM(q, r) × LCM(p, r)].

Why? For one prime, call its powers x ≤ y ≤ z. The product has x + y + z beads. The HCF of all three adds x. The three pairwise HCFs take away x, y and x. What is left is z, the greatest power, which is the LCM.

For 6, 72, 120: LCM = 6 × 72 × 120 × 6 ÷ (6 × 24 × 6) = 311040 ÷ 864 = 360.

Notes

LCM(p,q,r) = p·q·r·HCF(p,q,r) ÷ [HCF(p,q)·HCF(q,r)·HCF(p,r)] and HCF(p,q,r) = p·q·r·LCM(p,q,r) ÷ [LCM(p,q)·LCM(q,r)·LCM(p,r)].

Check yourself

Use the LCM formula for 4, 6, 8. Here HCF(4, 6, 8) = 2, HCF(4, 6) = 2, HCF(6, 8) = 2, HCF(4, 8) = 4. Find LCM(4, 6, 8).

Answer: 24

4 × 6 × 8 = 192. Times 2 is 384. Divide by 2 × 2 × 4 = 16: 384 ÷ 16 = 24.

Use the HCF formula for 4, 6, 8. Here LCM(4, 6, 8) = 24, LCM(4, 6) = 12, LCM(6, 8) = 24, LCM(4, 8) = 8. Find HCF(4, 6, 8).

Answer: 2

4 × 6 × 8 × 24 = 4608. Divide by 12 × 24 × 8 = 2304: 4608 ÷ 2304 = 2.

For one prime the powers in p, q, r are 1, 3, 3. The product has 7 beads. After adding the HCF of all three (1 bead) and removing the pairwise HCFs (1 + 3 + 1 beads), how many beads are left?

• 1. Count again: 7 + 1 − 5 = 3.
• 3 — correct. Yes! 7 + 1 − 5 = 3, the greatest power, which is the LCM’s.
• 7. That is before any step. Add 1 and take away 5.
• 8. That is after adding. Now take away 1 + 3 + 1 = 5.