MYP 3 · Maths

NUMBER

Lowest common multiple

QUESTION 1 [7 marks] — Criterion B Medium
Investigate the relationship between the HCF and LCM of two numbers, and their product.
a. For the pairs (12,18), (20,30), and (15,25), calculate HCF, LCM, and the product HCF$\times$LCM for each pair.
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Show Solution
(12,18): HCF$=6$, LCM$=36$, product$=216$. (20,30): HCF$=10$, LCM$=60$, product$=600$. (15,25): HCF$=5$, LCM$=75$, product$=375$.
b. Compare each product HCF$\times$LCM to the product of the original two numbers ($a \times b$). What do you notice?
[2]
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$12\times18=216$, $20\times30=600$, $15\times25=375$ — each matches the HCF$\times$LCM product exactly.
c. State a general rule connecting HCF, LCM, and the product of two numbers $a$ and $b$.
[1]
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$$\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b$$
QUESTION 2 [5 marks] — Criterion C Medium
Two buses leave a depot at the same time. Bus A returns to the depot every 18 minutes, and Bus B returns every 24 minutes.
a. Explain, in your own words, why the LCM of 18 and 24 gives the time until both buses are next at the depot together.
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The LCM is the smallest number that both 18 and 24 divide into exactly — meaning it's the first time that is simultaneously a whole multiple of Bus A's cycle AND Bus B's cycle, so both buses complete a whole number of trips and arrive together.
b. Calculate the LCM of 18 and 24, showing your full working using prime factorization.
[3]
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$18=2\times3^2$, $24=2^3\times3$. LCM (highest power of each prime): $2^3\times3^2=72$. So LCM $=72$ minutes.
QUESTION 3 [7 marks] — Criterion D Medium
Three warning lights at a construction site flash at regular intervals: the red light every 8 seconds, the yellow light every 12 seconds, and the blue light every 18 seconds. All three flash together at exactly 9:00:00 am.
a. Find the LCM of 8, 12, and 18 to determine how many seconds until all three lights flash together again.
[3]
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$8=2^3$, $12=2^2\times3$, $18=2\times3^2$. LCM $=2^3\times3^2=72$ seconds.
b. At what time will the lights next flash together?
[1]
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$9{:}00{:}00$ am $+$ 72 seconds $= 9{:}01{:}12$ am.
c. A site supervisor claims the lights will flash together again exactly 5 times within the next 6 minutes. Is this claim correct? Justify your answer.
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6 minutes $=360$ seconds. Number of 72-second intervals in 360 seconds: $360\div72=5$. So the lights DO flash together exactly 5 times in the next 6 minutes (at 72, 144, 216, 288, and 360 seconds) — the supervisor's claim is correct.
QUESTION 4 [6 marks] — Criterion A Hard
Three lighthouses flash at regular intervals: Lighthouse A every 18 seconds, Lighthouse B every 24 seconds, Lighthouse C every 30 seconds. All three flash together at exactly midnight.
a. Find the LCM of 18, 24, and 30 using prime factorization.
[3]
Show Solution
$18=2\times3^2$, $24=2^3\times3$, $30=2\times3\times5$. LCM (highest power of each prime): $2^3\times3^2\times5=360$.
b. Find how many times all three lighthouses flash together SIMULTANEOUSLY within the first hour (3600 seconds) after midnight, not counting the initial flash at midnight itself.
[3]
Show Solution
Number of 360-second intervals in 3600 seconds: $3600\div360=10$. So they flash together 10 times in the first hour (after midnight, not counting midnight itself).
QUESTION 5 [7 marks] — Criterion B Hard
Investigate the relationship between LCM$(a,b)$ when $a$ and $b$ share NO common factors (other than 1) versus when they do.
a. Find LCM$(8,9)$ (which share no common factor besides 1) and compare it to the simple product $8\times9$.
[2]
Show Solution
HCF$(8,9)=1$. LCM$=\frac{8\times9}{1}=72$, which equals the product $8\times9=72$ exactly.
b. Now find LCM$(8,12)$ (which DO share a common factor) and compare it to the product $8\times12$.
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HCF$(8,12)=4$. LCM$=\frac{8\times12}{4}=24$, which is LESS than the product $8\times12=96$.
c. State a general rule: when does LCM$(a,b)$ equal the simple product $a\times b$, and explain why, using the formula LCM$(a,b)=\frac{a\times b}{\text{HCF}(a,b)}$.
[3]
Show Solution
LCM$(a,b)$ equals $a\times b$ exactly when HCF$(a,b)=1$ (i.e. $a$ and $b$ are 'coprime', sharing no common factors). This follows directly from the formula: dividing by HCF$=1$ leaves the product unchanged. Whenever the numbers share a common factor greater than 1, the LCM will always be LESS than the simple product.
QUESTION 6 [5 marks] — Criterion C Hard
A student calculates LCM$(6,10)$ by simply multiplying them: '$6\times10=60$, so LCM=60' — without checking for a common factor.
a. Verify whether 60 is INDEED the smallest common multiple of 6 and 10, by listing the first several multiples of each and finding where they first coincide.
[3]
Show Solution
Multiples of 6: $6,12,18,24,30,36,42,48,54,60$. Multiples of 10: $10,20,30,40,50,60$. The multiples first coincide at 30, not 60 — so the TRUE LCM is 30, not 60.
b. Explain the student's error, and show the correct method using HCF, to confirm LCM$(6,10)=30$.
[2]
Show Solution
The student assumed LCM always equals the simple product, forgetting to check for a shared common factor. HCF$(6,10)=2$, so LCM$=\frac{6\times10}{2}=30$ — matching the true smallest common multiple found by listing.
QUESTION 7 [6 marks] — Criterion D Hard
A factory produces Product X every 15 minutes and Product Y every 40 minutes, starting simultaneously at the beginning of a shift. A quality inspector visits the factory floor every 25 minutes, ALSO starting at the beginning of the shift.
a. Find the LCM of 15, 40, and 25, to determine when all three events (Product X, Product Y, and an inspection) will next coincide exactly.
[3]
Show Solution
$15=3\times5$, $40=2^3\times5$, $25=5^2$. LCM$=2^3\times3\times5^2=600$ minutes.
b. Convert 600 minutes into hours and minutes, and comment on whether this coincidence is likely to be OBSERVED in practice, given that a typical factory shift is 8 hours long.
[3]
Show Solution
$600$ minutes $=10$ hours $0$ minutes. Since a typical 8-hour shift is SHORTER than 10 hours, this exact triple-coincidence would NOT actually occur within a single shift — it would only happen if the same schedule continued into a second shift, or if shifts were unusually long.
QUESTION 8 [6 marks] — Criterion A Hard
Three lighthouses flash at regular intervals: Lighthouse A every 18 seconds, Lighthouse B every 24 seconds, Lighthouse C every 30 seconds. All three flash together at exactly midnight.
a. Find the LCM of 18, 24, and 30 using prime factorization.
[3]
Show Solution
$18=2\times3^2$, $24=2^3\times3$, $30=2\times3\times5$. LCM (highest power of each prime): $2^3\times3^2\times5=360$.
b. Find how many times all three lighthouses flash together SIMULTANEOUSLY within the first hour (3600 seconds) after midnight, not counting the initial flash at midnight itself.
[3]
Show Solution
Number of 360-second intervals in 3600 seconds: $3600\div360=10$. So they flash together 10 times in the first hour (after midnight, not counting midnight itself).
QUESTION 9 [7 marks] — Criterion B Hard
Investigate the relationship between LCM$(a,b)$ when $a$ and $b$ share NO common factors (other than 1) versus when they do.
a. Find LCM$(8,9)$ (which share no common factor besides 1) and compare it to the simple product $8\times9$.
[2]
Show Solution
HCF$(8,9)=1$. LCM$=\frac{8\times9}{1}=72$, which equals the product $8\times9=72$ exactly.
b. Now find LCM$(8,12)$ (which DO share a common factor) and compare it to the product $8\times12$.
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Show Solution
HCF$(8,12)=4$. LCM$=\frac{8\times12}{4}=24$, which is LESS than the product $8\times12=96$.
c. State a general rule: when does LCM$(a,b)$ equal the simple product $a\times b$, and explain why, using the formula LCM$(a,b)=\frac{a\times b}{\text{HCF}(a,b)}$.
[3]
Show Solution
LCM$(a,b)$ equals $a\times b$ exactly when HCF$(a,b)=1$ (i.e. $a$ and $b$ are 'coprime', sharing no common factors). This follows directly from the formula: dividing by HCF$=1$ leaves the product unchanged. Whenever the numbers share a common factor greater than 1, the LCM will always be LESS than the simple product.
QUESTION 10 [5 marks] — Criterion C Hard
A student calculates LCM$(6,10)$ by simply multiplying them: '$6\times10=60$, so LCM=60' — without checking for a common factor.
a. Verify whether 60 is INDEED the smallest common multiple of 6 and 10, by listing the first several multiples of each and finding where they first coincide.
[3]
Show Solution
Multiples of 6: $6,12,18,24,30,36,42,48,54,60$. Multiples of 10: $10,20,30,40,50,60$. The multiples first coincide at 30, not 60 — so the TRUE LCM is 30, not 60.
b. Explain the student's error, and show the correct method using HCF, to confirm LCM$(6,10)=30$.
[2]
Show Solution
The student assumed LCM always equals the simple product, forgetting to check for a shared common factor. HCF$(6,10)=2$, so LCM$=\frac{6\times10}{2}=30$ — matching the true smallest common multiple found by listing.
QUESTION 11 [6 marks] — Criterion D Hard
A factory produces Product X every 15 minutes and Product Y every 40 minutes, starting simultaneously at the beginning of a shift. A quality inspector visits the factory floor every 25 minutes, ALSO starting at the beginning of the shift.
a. Find the LCM of 15, 40, and 25, to determine when all three events (Product X, Product Y, and an inspection) will next coincide exactly.
[3]
Show Solution
$15=3\times5$, $40=2^3\times5$, $25=5^2$. LCM$=2^3\times3\times5^2=600$ minutes.
b. Convert 600 minutes into hours and minutes, and comment on whether this coincidence is likely to be OBSERVED in practice, given that a typical factory shift is 8 hours long.
[3]
Show Solution
$600$ minutes $=10$ hours $0$ minutes. Since a typical 8-hour shift is SHORTER than 10 hours, this exact triple-coincidence would NOT actually occur within a single shift — it would only happen if the same schedule continued into a second shift, or if shifts were unusually long.