Least Common Multiple: When Events Coincide
In the opening hour of a new store, a bell rang every 2 minutes and lights flashed every 3 minutes. If the store opened at 10:00, at what times did both events happen at the same time? Explain your strategy.
Already Got the Answer?
Setup: Let t = minutes after 10:00. Both events happen when t is a multiple of 2 and a multiple of 3, so t is a multiple of LCM(2, 3) = 6.
Answer: 10:06, 10:12, 10:18, 10:24, 10:30, 10:36, 10:42, 10:48, 10:54, and 11:00 (the very end of the hour).
One-line check:60 ÷ 6 = 10 coincidences, and every time listed is divisible by both 2 and 3.
What This Problem Teaches
- Recognizing a repeating-cycle problem. The words "every ... minutes" and "at the same time" are the signature of the least common multiple, even though the problem never uses those words.
- Translating clock language into arithmetic. Measuring from the start (
tminutes after 10:00) turns a story about bells into a statement about divisibility. - Seeing why LCM, not a product or a sum, answers "when do they line up?" You will see the reason in a table, not just a rule.
- Handling boundaries. Does "the opening hour" include 10:00? Does it include 11:00? Good problem solvers notice and state these choices.
- Describing a strategy in words. The prompt asks you to explain, so the page shows you how to put the reasoning into a clear sentence or two.
Picture This: One Hour, Minute by Minute
The most convincing way to understand this problem is to lay out all 60 minutes and mark what happens. The table below is arranged in rows of six minutes, which is a deliberate choice. Watch what happens in each column.
| Block | Col 1 | Col 2 | Col 3 | Col 4 | Col 5 | Col 6 |
|---|---|---|---|---|---|---|
| Min 1–6 | 1 | 2bell | 3light | 4bell | 5 | 610:06 |
| Min 7–12 | 7 | 8bell | 9light | 10bell | 11 | 1210:12 |
| Min 13–18 | 13 | 14bell | 15light | 16bell | 17 | 1810:18 |
| Min 19–24 | 19 | 20bell | 21light | 22bell | 23 | 2410:24 |
| Min 25–30 | 25 | 26bell | 27light | 28bell | 29 | 3010:30 |
| Min 31–36 | 31 | 32bell | 33light | 34bell | 35 | 3610:36 |
| Min 37–42 | 37 | 38bell | 39light | 40bell | 41 | 4210:42 |
| Min 43–48 | 43 | 44bell | 45light | 46bell | 47 | 4810:48 |
| Min 49–54 | 49 | 50bell | 51light | 52bell | 53 | 5410:54 |
| Min 55–60 | 55 | 56bell | 57light | 58bell | 59 | 6011:00 |
The pattern is striking. Column 2 and column 4 always hold a bell, column 3 always holds a light, columns 1 and 5 are silent, and column 6 is always both. The whole hour is the same six-minute pattern repeated ten times. That repetition is the entire problem.
Solution: Method 1 — The List-and-Match Approach
Before reaching for a formula, do what the problem literally describes: write down when each event happens and look for the overlap. This method needs nothing beyond skip-counting, and it builds the intuition that the formula later compresses.
Step 1 — Measure time from the opening
Clock times such as 10:06 are awkward to work with. Let t be the number of minutes after 10:00, so the opening hour covers t = 0 to t = 60. The first bell rings at t = 2, since it rings "every 2 minutes" after opening, and the first flash comes at t = 3.
Step 2 — List the bell times
The bell rings at multiples of 2:
Step 3 — List the light times
The lights flash at multiples of 3:
Step 4 — Find the numbers that appear in both lists
Scan for repeats. The first is 6, then 12, then 18, then 24. The gap between matches is always 6, so the pattern continues at 30, 36, 42, 48, 54, and 60. This shared list of multiples is exactly what "common multiple" means, and the first one is the least common multiple.
Step 5 — Convert minutes back to clock times
Add each value of t to 10:00:
t = 24 → 10:24 t = 30 → 10:30 t = 36 → 10:36
t = 42 → 10:42 t = 48 → 10:48 t = 54 → 10:54
t = 60 → 11:00
Step 6 — State the strategy in words
Because the problem asks you to explain, here is a model answer: "The bell rings at multiples of 2 minutes and the lights flash at multiples of 3 minutes, so both happen together at common multiples of 2 and 3. The smallest is 6, so the events coincide every 6 minutes after opening, at 10:06, 10:12, and so on up to 11:00."
Solution: Method 2 — Prime Factorization and the LCM Formula
Listing works well for small numbers, but it becomes tedious when the intervals are 14 and 21 instead of 2 and 3. Prime factorization finds the first coincidence directly, without any list.
Step 1 — Factor each interval into primes
3 = 3
Both numbers are already prime, and they share no prime factors.
Step 2 — Build the LCM
The LCM uses every prime that appears in either number, each raised to the highest power it reaches. Here that is one 2 and one 3:
Step 3 — Generate every coincidence from the LCM
Every common multiple of two numbers is a multiple of their LCM. So the coincidence times are 6×1, 6×2, 6×3, ... up to the end of the hour:
That gives ten times: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60 minutes after 10:00, the same list Method 1 produced.
Solution: Method 3 — The Alternating-Flash Argument (Remainders)
Here is a third way to see it, using even and odd numbers as the idea behind modular arithmetic. It explains why the answer is 6 instead of just confirming that it is.
The lights flash at 3, 6, 9, 12, 15, ..., which is 3k for k = 1, 2, 3, .... The bell rings at every even minute. Is 3k even? Since 3 is odd, 3k is even exactly when k is even. So the flashes alternate: odd minute (no bell), even minute (bell), odd, even, and so on. Every second flash lines up with a bell, and the flashes are 3 minutes apart, so the coincidences are 6 minutes apart.
In remainder language, we need t ≡ 0 (mod 2) and t ≡ 0 (mod 3). Because 2 and 3 share no common factor, these two conditions together are equivalent to the single condition t ≡ 0 (mod 6).
The bell and the lights happened together every 6 minutes after opening: at 10:06, 10:12, 10:18, 10:24, 10:30, 10:36, 10:42, 10:48, 10:54, and 11:00 (the closing instant of the opening hour).
Strategy: the events coincide at common multiples of 2 and 3, and the least common multiple is LCM(2, 3) = 6.
Verification
Check 1 — Substitute each time into both conditions. For t = 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, divide by 2 to get 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 and divide by 3 to get 2, 4, 6, 8, 10, 12, 14, 16, 18, 20. Every quotient is a whole number, so the bell rings and the lights flash at each of these minutes.
Check 2 — Make sure none were missed. A coincidence requires t to be divisible by both 2 and 3. Between multiples of 6, the minutes are 1, 2, 3, 4, 5 (plus 6 more each block). Of these, 2 and 4 are even but not multiples of 3, 3 is a multiple of 3 but odd, and 1 and 5 are neither. So no non-multiple of 6 qualifies, which matches column 6 being the only gold column.
Check 3 — Count with an independent method. The bell rings 60 ÷ 2 = 30 times and the lights flash 60 ÷ 3 = 20 times. The overlap should be 60 ÷ 6 = 10, which equals the number of gold cells in the table. Also, the bell alone rings 20 times and the lights alone flash 10 times, and 20 + 10 + 10 = 40 distinct event minutes, which matches the table's count of non-blank cells (60 minutes minus 20 silent ones in columns 1 and 5).
Two Judgment Calls Hidden in the Wording
The arithmetic here is easy. The wording is where careful solvers earn their points. Two small decisions change the length of your list, and a good answer states them out loud.
Does 10:00 count? We assumed the first bell is 2 minutes after opening and the first flash is 3 minutes after opening, so nothing happens at the exact moment of 10:00. If your teacher's version has both events fire at the opening instant, then 10:00 is also a coincidence, because 0 is a multiple of every number. In that case add 10:00 to the front of the list.
Does 11:00 count? The opening hour runs from 10:00 up to 11:00. A coincidence at exactly 11:00 sits on the boundary. We included it because the bell and lights are both "due" at that moment, but it is the one time you could reasonably leave off. If your class treats the hour as ending just before 11:00, your list has 9 times, ending at 10:54.
Either way, the structure is identical: coincidences every 6 minutes. Writing one sentence that states your assumption is what turns a correct list into a complete answer.
What Trips People Up
Adding the intervals describes nothing in this story. Neither event happens at minute 5 (the bell is at 4 and 6, the lights at 3 and 6). Matching requires a time that is a multiple of each interval, which is a multiplication idea, not an addition one.
The question asks "at what times," so it wants clock times, and all of them in the hour. Finding the first coincidence is only the halfway point. Add the LCM repeatedly (or use 6k) until you reach the end of the hour, then convert to clock times.
For 2 and 3 the product happens to equal the LCM, because they share no factors. That makes the shortcut feel universal. It is not: for 4 and 6 the product is 24, but the events actually line up at 12. The shortcut a × b only works when the numbers have no common factor. Try the first What-If below to see this firsthand.
Minutes only go up to 59. Sixty minutes after 10:00 is 11:00. Whenever you convert t into a clock time, check whether you passed the top of the hour.
If You See These Words...
This problem is a member of a large family, and the surface story changes constantly. Look for these signals:
- "Every ... minutes / days / laps": two or more things repeat on their own schedule.
- "At the same time," "together again," "next time they meet": you are being asked about a shared moment, which means a common multiple.
- "How often" asks for the LCM itself. "At what times" asks for the list of multiples of the LCM.
- "Smallest number that is divisible by both" or "buy equal amounts": same structure with no clock in sight.
Hot dogs come in packs of 10 and buns in packs of 8. What is the smallest number of hot dogs you can buy so that nothing is left over? There is no clock and nothing repeats in time, but you are still looking for the smallest number that is a multiple of both 10 and 8, which is LCM(10, 8) = 40.
Contrast this with GCF problems, which ask you to split something into equal groups (the greatest group size that divides evenly). "Meets again" and "lines up again" mean LCM. "Cut into equal pieces" means GCF.
The Pattern Behind This
For two repeating events with intervals a and b, starting together at time 0, the coincidence times are exactly the multiples of the LCM:
LCM(a, b) = (a × b) ÷ GCF(a, b)
Number of coincidences in T minutes: ⌊ T ÷ LCM(a, b) ⌋
For our problem, GCF(2, 3) = 1, so LCM = (2 × 3) ÷ 1 = 6, and the count is ⌊60 ÷ 6⌋ = 10. For the 4-and-6 case, GCF = 2, so LCM = 24 ÷ 2 = 12, which is why the plain product overshoots by exactly the shared factor.
Limit of the shortcut: The formula (a × b) ÷ GCF is for two numbers. For three events, take the LCM in stages, LCM(LCM(a, b), c), or use prime factorization on all three at once. Do not multiply all three and divide by one GCF.
Where This Shows Up in Real Life
- Music: A "two against three" polyrhythm, where one hand plays 2 beats while the other plays 3, lines up every 6 subdivisions. It is this very problem played on a piano.
- Traffic engineering: Signals with different cycle lengths drift in and out of sync. Planners use LCMs to work out how often a corridor of lights will line up.
- Biology: Periodical cicadas emerge on 13- or 17-year cycles. Because these are prime, they share almost no common multiples with predator cycles of 2, 3, 4, 5, or 6 years, so overlaps are rare.
- Scheduling: Two staff members on 4-day and 6-day rotations are both off on the same day every 12 days.
Where This Leads
The next step is to let the two events start at different moments, such as a bell that starts at 10:01. The coincidences are then no longer multiples of the LCM but numbers that satisfy two remainder conditions at once, a system of congruences. Number theorists solve these with the Chinese Remainder Theorem, and the fourth What-If below is a small, friendly version of it.
Extend Your Thinking
Try each problem on your own first, then open the solution to compare. They get harder as you go.
Suppose the bell rings every 4 minutes and the lights flash every 6 minutes, starting from the 10:00 opening. At what times during the first hour (up to and including 11:00) do both happen together?
Bell: 4, 8, 12, 16, 20, 24, ... Lights: 6, 12, 18, 24, 30, ...
The first number in both lists is 12. Notice this is not 4 × 6 = 24, which is a match but not the first one.
4 = 2² and 6 = 2 × 3. The LCM takes the highest power of each prime: 2² × 3 = 12.
12, 24, 36, 48, 60 minutes after opening, which are 10:12, 10:24, 10:36, 10:48, and 11:00.
12 ÷ 4 = 3 and 12 ÷ 6 = 2, both whole. Also 60 ÷ 12 = 5 coincidences, which matches the five times listed. The shared factor 2 is why the answer is half of the product 24.
The bell rings every 2 minutes and the lights flash every 3 minutes, as before. Now a door chime also sounds every 5 minutes. At what times in the first hour (up to and including 11:00) do all three happen together?
Bell and lights already coincide at multiples of 6. The chime must also be at that moment, so we need multiples of 6 that are also multiples of 5.
LCM(6, 5): since 6 and 5 share no common factor, the LCM is 6 × 5 = 30. Equivalently, LCM(2, 3, 5) = 2 × 3 × 5 = 30.
30, 60 minutes after opening.
All three events happen together at 10:30 and 11:00.
At t = 30: 30 ÷ 2 = 15, 30 ÷ 3 = 10, 30 ÷ 5 = 6, all whole. Any smaller candidate fails: the multiples of 6 below 30 are 6, 12, 18, 24, and none is divisible by 5.
The bell rings every 2 minutes. The lights flash every m minutes, where m is a whole number. If the bell and lights happen together every 10 minutes exactly, what could m be?
"Together every 10 minutes" means LCM(2, m) = 10.
The LCM is always a multiple of m, so m must divide 10. The only options are m = 1, 2, 5, 10.
LCM(2, 1) = 2 ✗, LCM(2, 2) = 2 ✗, LCM(2, 5) = 10 ✓, LCM(2, 10) = 10 ✓.
m = 5 or m = 10.
For m = 5, the lights flash at 5, 10, 15, 20, ..., and the even ones (10, 20, 30, ...) match the bell, so the gap is 10. For m = 10, every flash (10, 20, 30, ...) is at an even minute, so every flash matches the bell and the gap is again 10. Both answers work, and the problem has two valid solutions.
The bell rings every 2 minutes starting at 10:01 (so at 10:01, 10:03, 10:05, ...). The lights flash every 3 minutes starting at 10:02 (so at 10:02, 10:05, 10:08, ...). At what times during the first hour (before 11:00) do both happen at the same time?
The bell rings at t = 1, 3, 5, 7, ... (the odd minutes). The lights flash at t = 2, 5, 8, 11, 14, 17, ....
Compare the lists: 2 is even (no bell), but 5 is odd, so t = 5 is the first coincidence.
The flashes alternate even, odd, even, odd (2, 5, 8, 11, 14, 17, ...). Only the odd ones match the bell, so every second flash coincides. Flashes are 3 minutes apart, so matches are 6 minutes apart: 5, 11, 17, 23, ... (that is, t = 6k − 1).
5, 11, 17, 23, 29, 35, 41, 47, 53, 59. The next value, 65, is past the hour. So both events happen together at 10:05, 10:11, 10:17, 10:23, 10:29, 10:35, 10:41, 10:47, 10:53, and 10:59.
Check the last one: t = 59 is odd, so the bell rings, and 59 − 2 = 57 = 3 × 19, so it is a flash time. Check a near miss: t = 8 is a flash but even, so no bell. In remainder language, t ≡ 1 (mod 2) and t ≡ 2 (mod 3) combine into t ≡ 5 (mod 6). The offset shifts the pattern but does not change the 6-minute spacing.
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2026-11-07