Repeating events sync on the LCM. Splitting into the biggest equal pieces is the GCF. The whole battle is recognising which one the story is asking for.
The problem
Two bells ring every 12 minutes and every 18 minutes. They ring together at exactly 12:00 noon. When do they next ring together?
They coincide every LCM(12, 18) = 36 minutes — so next at 12:36 pm. A uses 12 + 18 = 30, but the bells do not wait for each other. C multiplies, 12 × 18 = 216 minutes = 3 h 36 min — a time they do ring together, but not the first. D, 54 minutes, is a multiple of 18 only. Sync questions are LCM questions, every time.
12 = 2² · 3, 18 = 2 · 3²
prime factorise both
LCM = 2² · 3² = 36
highest power of every prime
Why LCM: bell one rings at 12, 24, 36, 48 …; bell two at 18, 36, 54 …. The first number on both lists is the least common multiple — that is literally its definition doing the work.
Ribbons of 84 cm and 126 cm are cut into the longest possible equal pieces with nothing wasted.
84 = 2² · 3 · 7, 126 = 2 · 3² · 7
factorise again
GCF = 2 · 3 · 7 = 42 cm
lowest power of the SHARED primes
| The story… | Use | Typical words |
|---|---|---|
| combines repeating cycles | LCM | "together again", "at the same time", "next coincide" |
| splits quantities into equal parts | GCF | "largest equal pieces", "biggest groups", "no leftover" |
Sanity anchor: the LCM is at least as big as the larger number; the GCF is no bigger than the smaller one. If your "largest piece" came out bigger than a ribbon, the tools got swapped.
The LCM of 12 and 18 is 36 — but the answer is 12:36 pm, a clock time, and the ribbon answer might be "how many pieces" (8442 + 12642 = 5) rather than the length. After the LCM or GCF drops out, reread the last line of the problem and convert.
| The last line asks | Convert the LCM or GCF like this |
|---|---|
| how many square tiles | a 90 by 150 floor, GCF 30: (90 ÷ 30)(150 ÷ 30) = 3 × 5 = 15 tiles — multiply for an area |
| how many laps | runners lapping in 40 s and 50 s meet after LCM 200 s; the faster has run 200 ÷ 40 = 5 laps |
| how many turns of a gear | 18 and 30 teeth: the same teeth meet after LCM 90 teeth, which is 90 ÷ 18 = 5 turns of the smaller gear |
Two different units: one light flashes every 1.5 minutes, another every 40 seconds. They flash together now. When next?
1.5 min = 90 s
one unit, and whole numbers
90 = 2 · 3² · 5, 40 = 2³ · 5
factorise
LCM = 2³ · 3² · 5 = 360 s = 6 minutes
convert back for the answer
What is different: the LCM of 1.5 and 40 means nothing. Convert to the smaller unit first so both cycles are whole numbers.
How many meetings: buses leave every 15 and every 25 minutes, together at 6:00 a.m. How many times do they leave together from 6:00 a.m. to 12:00 noon, both included?
LCM(15, 25) = 75 min
one meeting every 75 minutes
360 ÷ 75 = 4 r 60 → 4 more after the start
6 hours = 360 min
6:00, 7:15, 8:30, 9:45, 11:00 → 5 times
count the 6:00 departure too
What is different: the question wants a count, not a time. Dividing gives the gaps; the starting meeting adds one more.
Unequal leftovers: the largest number that divides 70 leaving 4, and divides 100 leaving 1.
70 − 4 = 66 and 100 − 1 = 99
each loses its own leftover
66 = 2 · 3 · 11, 99 = 3² · 11
factorise
GCF = 3 · 11 = 33
check: 70 = 2 · 33 + 4, 100 = 3 · 33 + 1 ✓
What is different: the leftovers differ, so each number is reduced by its own. The divisor must also beat both leftovers — 33 > 4 ✓.
“With 1 left over” means the remainder must be set aside before the LCM or GCF can work.
Eggs packed in 4s, 5s or 6s always leave 1 egg over. What is the smallest possible number of eggs (more than 1)?
eggs − 1 is a common multiple of 4, 5 and 6
set the spare egg aside
LCM(4, 5, 6) = 2² · 3 · 5 = 60
the packs line up at 60
eggs = 60 + 1 = 61
put the spare back
100 pencils and 145 erasers are shared equally among as many students as possible, with exactly 1 pencil and 1 eraser left over. How many students?
shared exactly: 100 − 1 = 99 and 145 − 1 = 144
take the leftovers away
99 = 3² · 11, 144 = 2⁴ · 3²
factor
GCF = 3² = 9 students
check: 100 = 9 · 11 + 1, 145 = 9 · 16 + 1 ✓
60 forgets to add the spare egg back; GCF(100, 145) = 5 ignores the leftovers and shares everything out. With a leftover, the LCM answer is plus r and the GCF works on the numbers minus r.
These stories come as bells, buses, runners and gears (LCM) or tiles, ribbons and groups (GCF), often with a remainder or a “how many” twist at the end. Choose by whether the answer should be bigger or smaller than the numbers, shift by any remainder, then reread the last line — a clock time, a lap count or a number of pieces.
The problem
Ropes of 60 m and 84 m are cut into the longest possible equal pieces with nothing left over. How long is each piece?
Splitting into biggest equal pieces is GCF: 60 = 2² · 3 · 5 and 84 = 2² · 3 · 7 share 2² · 3 = 12 m. A and B divide both ropes cleanly but are not the LONGEST such length; D does not divide 60 at all — always check the "biggest" claim divides both.
