LCM • Topic 3 of 3

LCM Applications

Two CAT staples live here. First, LCM of fractions: the rule is LCM = LCM(of numerators) / HCF(of denominators), and symmetrically HCF of fractions = HCF(numerators)/LCM(denominators). Reduce each fraction first. Second, the "events recurring together" family: if bells ring at intervals of a, b, c seconds and start together, they next ring together after LCM(a,b,c) seconds, and the count of times in T seconds is T/LCM (plus the start instant if asked). Third, the "least number leaving remainder r" problem: the smallest number that leaves the SAME remainder r when divided by several divisors is LCM(divisors) + r, and the general form is LCM·k + r. If instead each divisor leaves a remainder that is a constant LESS than the divisor (e.g. always 2 short), the number is LCM − (common deficit). Set the unknown as LCM·k + r and solve for k to fit any extra condition.

✅ Solved examples

1. Find the LCM of the fractions 2/3, 4/9 and 5/6.
LCM(numerators 2,4,5) = 20; HCF(denominators 3,9,6) = 3. LCM = 20/3.
2. Three bells toll at 6, 8 and 12 second intervals, starting together. After how long do they next toll together, and how many times in 6 minutes (excluding the start)?
LCM(6,8,12) = 24 s, so they next toll together after 24 s. In 360 s: 360/24 = 15 times.
3. Find the least number which leaves remainder 3 when divided by 4, 6 and 8.
LCM(4,6,8) = 24. Least such number = 24 + 3 = 27.
4. Find the least number which when divided by 5, 6 and 7 leaves remainders 3, 4 and 5 respectively.
Each remainder is 2 less than its divisor (5−3 = 6−4 = 7−5 = 2). So N = LCM(5,6,7) − 2 = 210 − 2 = 208.

✏️ Practice — try these, take hints as needed

1. Find the LCM of 3/4, 6/7 and 9/8.
LCM of numerators 3,6,9.
HCF of denominators 4,7,8.
LCM(num) = 18, HCF(den) = 1.
18
2. Four lights flash every 12, 15, 20 and 30 seconds, starting together. After how many seconds do they next flash together?
Take the LCM of the intervals.
12 = 2²×3, 15 = 3×5, 20 = 2²×5, 30 = 2×3×5.
Highest powers 2²×3×5.
60 seconds
3. Find the least number that leaves remainder 5 when divided by 8, 12 and 16.
Find LCM(8,12,16).
LCM = 48.
Add the common remainder 5.
53
4. Find the least number which when divided by 9, 10 and 15 leaves remainders 4, 5 and 10 respectively.
Check the gap divisor − remainder.
9−4 = 10−5 = 15−10 = 5.
N = LCM(9,10,15) − 5.
85
5. Three runners complete a circular lap in 30, 40 and 60 seconds. If they start together, after how long are they together at the start again, and how many times do they meet there in 1 hour (excluding the start)?
Meeting interval = LCM of lap times.
LCM(30,40,60) = 120 s.
3600 ÷ 120.
Every 120 s; 30 times in an hour

📝 Topic test — 8 questions

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