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ESO Y2 · Mathematics
Numbers · Natural numbers

Lowest common multiple

One alarm sounds every $12$ minutes, another every $15$ and another every $20$. All three have just gone off together. When will they next sound together? That is exactly the question that the lowest common multiple answers.

Before you start, make sure you know: Multiples, divisors and divisibility tests (1st ESO)
Before you start, make sure you know: Prime numbers and prime factor decomposition (1st ESO)

What the l.c.m. is

Number lines comparing the multiples of 4 and 6, with their first coincidences at 12 and 24 multiples of 4 multiples of 6 4 8 12 16 20 24 6 12 18 24 12 = m.c.m.(4, 6) (24 is also a common multiple, but not the lowest)
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In a platform video game, your character gets a protective shield every $4$ levels and a speed boost every $6$ levels. If you have just started the game at level $0$, at…

In a platform video game, your character gets a protective shield every $4$ levels and a speed boost every $6$ levels. If you have just started the game at level $0$, at which level will both rewards coincide for the first time? Calculate the result by finding the first multiples of each number.

The number line shows the first jump of each reward. Continue the jumps yourself until you find the first level where they coincide.

Number line of levels 0 to 14 with the first jump of 4 levels (shield) and the first jump of 6 levels (speed) from level 001234567891011121314+6 Speed+4 Shield
Solve
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A messaging application sends a "news digest" notification every $10$ minutes, while another social media application does so every $15$ minutes. If both sent a…

A messaging application sends a "news digest" notification every $10$ minutes, while another social media application does so every $15$ minutes. If both sent a notification at $10:00$ in the morning, how many minutes will pass until they coincide again for the first time? Write out the sets of multiples of $10$ and $15$ to find the lowest common multiple.

Solve

Calculating the l.c.m. by prime factor decomposition

Listing multiples is slow for large numbers. The quick method: break each number down into prime factors and take all the factors, common and non-common, raised to the highest power.

Calculation of the l.c.m. of 12 and 18 by taking the highest power of each prime factor 12 = 2² · 3 18 = 2 · 3² m.c.m. = 2² · 3² = 36 highest power of each factor
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Calculate the lowest common multiple of the numbers $12$ and $18$ in your exercise book using prime factorisation. To do this, follow these steps: 1. Find the prime…

Calculate the lowest common multiple of the numbers $12$ and $18$ in your exercise book using prime factorisation. To do this, follow these steps:

  1. Find the prime factorisation of both numbers.

  2. Identify the common and non-common factors raised to the highest power.

  3. Multiply them to obtain the $m.c.m.(12, 18)$.

12 2 6 2 3 3 1 12 = 2² · 3 18 2 9 3 3 3 1 18 = 2 · 3² m.c.m. = ? · ? = ?
Solve
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In a logistics centre, one maintenance alarm sounds every $45$ minutes and another every $60$ minutes. If at $08:00$ in the morning both alarms coincided, at what time…

In a logistics centre, one maintenance alarm sounds every $45$ minutes and another every $60$ minutes. If at $08:00$ in the morning both alarms coincided, at what time will they sound together for the second time after $08:00$?

Solve

The l.c.m. of three or more numbers

The method is the same, except that you compare more factorisations at once.

Diagram of prime factorisations for calculating the l.c.m. of three numbers 12 = 2² · 3 15 = 3 · 5 20 = 2² · 5 m.c.m. = 2² · 3 · 5 = 60
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Two professional cyclists train on a circular velodrome. The first completes a lap every $108$ seconds, while the second, who goes a bit slower, takes $120$ seconds. If…

Two professional cyclists train on a circular velodrome. The first completes a lap every $108$ seconds, while the second, who goes a bit slower, takes $120$ seconds. If both pass the finish line together right now, how many minutes will it take for them to coincide at the same point again for the first time?

FINISH LINE t₁ t₂ Lap progression (seconds) 0 108 216 324 ... 0 120 240 360 ... t = ? min lcm(108, 120)
Solve
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The product of two natural numbers is $6\,750$ and their highest common factor ($M.C.D.$) is $15$. Using the relationship between the product of two numbers, their…

The product of two natural numbers is $6\,750$ and their highest common factor ($M.C.D.$) is $15$. Using the relationship between the product of two numbers, their $M.C.D.$ and their lowest common multiple ($m.c.m.$), first calculate the value of the $m.c.m.$ of these numbers. Next, find out what those two numbers are, knowing that their difference is exactly $15$.

Solve

Applications of the l.c.m.

The l.c.m. appears whenever something repeats with a period and you ask when the repetitions coincide.

Diagram of two ships with periodic sailings that coincide again at the l.c.m. of their periods Ship A every 24 days Ship B every 35 days m.c.m.(24, 35) = 840 days (they coincide again after 840 days)
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Find the lowest common multiple of the following groups of numbers using prime factorisation: a) $m.c.m.(45, 75)$ b) $m.c.m.(12, 18, 24)$ c) $m.c.m.(80, 120, 200)$

Find the lowest common multiple of the following groups of numbers using prime factorisation:

  1. a) $m.c.m.(45, 75)$
  2. b) $m.c.m.(12, 18, 24)$
  3. c) $m.c.m.(80, 120, 200)$
Solve
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In a space station, four research satellites orbit an asteroid. The first completes its orbit every $12$ hours, the second every $18$ hours, the third every $20$ hours…

In a space station, four research satellites orbit an asteroid. The first completes its orbit every $12$ hours, the second every $18$ hours, the third every $20$ hours and the fourth every $30$ hours. If today at $12:00$ noon they are aligned for the first time, how many full days must pass for them to coincide in exactly the same position again? At what time of day will this meeting take place?

t₀=12:00 PERIODS T₁ = 12h T₂ = 18h T₃ = 20h T₄ = 30h
Solve

Back to the start: Since l.c.m.$(12, 15, 20) = 60$, the three alarms will sound together again in $60$ minutes (exactly one hour later).

Summary

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