Two Bells, 12 and 15 Seconds Apart, Meet at a Full Minute

One bell rings five times a minute, evenly spaced. Another rings four times a minute, also evenly spaced. They ring together on the first beat. When do they next ring at the same instant?

The reflex is to multiply: four times five is twenty, so twenty seconds. The multiplication is correct and the answer is not, and the reason is visible before any number theory. Check what the units are doing.

Rates do not multiply into times

Bell one rings at $5\ \text{min}^{-1}$. Bell two rings at $4\ \text{min}^{-1}$. Their product is $20\ \text{min}^{-2}$, which is not a duration and is not anything else recognisable either. Nothing in the problem asks for a quantity with those units, so the number twenty was never a candidate. A dimension check kills the trap before any argument about bells.

Convert to the quantity that does have units of time. Five evenly spaced rings a minute is one ring every

and now the question is about two sets of instants rather than two rates.

Two arithmetic progressions, one intersection

A time in both sets is a common multiple of $g_1$ and $g_2$, and every common multiple is a multiple of the least one, so the intersection is itself a progression:

so the answer is a least common multiple and nothing more:

One full minute. The two bells ring together only on the beats that begin each minute, and nothing happens jointly in between.

Multiplying the gaps instead of the rates is a better mistake and still a mistake: $12 \times 15 = 180$ is a genuine common multiple, so the bells really do ring together at three minutes. It is three times too late, and the factor of three is exactly what the two gaps share:

The general pair of bells

Something tidier is true, and it explains why this particular problem lands on a round number.

The proof is two applications of one identity. Since $a$ and $b$ divide $N = 60$, we have $\gcd(N/a, N/b) = N/\operatorname{lcm}(a,b)$, and therefore

using $\gcd(a,b)\operatorname{lcm}(a,b) = ab$ at the last step. For $a = 5$ and $b = 4$, the greatest common divisor is one and the answer is the whole minute. That is not a coincidence of five and four: any two coprime rates take a full minute to line up again, which is a much less interesting-looking fact than the original question and a much more useful one.

Read the diagonal for the degenerate case, and read the sixes and the fours for the interesting one: six and four rings a minute give gaps of ten and fifteen seconds, which coincide at thirty. Six and three coincide at twenty. The trap answer of twenty seconds is the correct answer to a problem one digit away from this one.

Three bells, or thirty

Nothing above depended on there being two of them. Any number of bells starting together next coincide at the least common multiple of all their gaps, since a shared instant has to be a common multiple of every one of them, and the rule in (5) extends by the same argument to

for rates $a_i$ all dividing $N$. Add a third bell at three rings a minute to the original pair and the gaps become twelve, fifteen and twenty seconds, whose least common multiple is still sixty, because five, four and three share no factor. A tower of bells at six, four and two rings a minute lands on thirty instead, since the greatest common divisor is two.

Reading (7) the other way is the useful direction. A carillon coincides less often the more its rates have in common, so if you want the bells to line up rarely, choose rates that share factors. If you want them to line up on every minute mark and never in between, choose rates with no common factor at all. It is the same rule that decides how often two gears return to their starting alignment, which is the mechanical version of the same question.

The wording has two readings and they agree

"Five rings a minute" can mean five gaps of twelve seconds, which is what (1) assumes, or five rings spanning the minute with four gaps of fifteen seconds each. The second reading changes both gaps: fifteen and twenty seconds instead of twelve and fifteen. It does not change the answer.

which is why the problem can say "a minute" without disambiguating itself. It is worth noticing that under this second reading the trap answer of twenty seconds is exactly the slower bell's own gap, so the wrong number is a real ring time for one of the two bells. That is part of why it survives a second look.

Drop the shared start and the answer can vanish

The one hypothesis doing the most work is that both bells ring at time zero. Take it away and the problem changes shape entirely. Suppose the second bell is late by $\delta$ seconds, so it rings at $\delta, \delta + 15, \delta + 30, \dots$. A coincidence needs integers $k, m$ with

and the left side is always a multiple of $\gcd(12,15) = 3$. So the bells coincide at some point if and only if $\delta$ is a multiple of three seconds, and when they do the coincidences recur every sixty seconds as before. A one-second offset means the two bells never ring together, not once, ever. The bottom timeline in Fig. 1 is that case.

This is the part of the problem I would push on if I were asking it, because it separates someone who computed a least common multiple from someone who understands why the least common multiple was the right object. The first answer is a number. The second survives a change to the setup.

What else the model assumes

The rings have to be evenly spaced. A bell that rings five times a minute at irregular intervals makes the question unanswerable, and "five times a minute" is a rate rather than a schedule unless evenness is granted.

A ring has to be an instant rather than an interval. Two bells whose sounds last two seconds each overlap at plenty of moments that are not shared ring times, and if the question were about hearing them together the answer would be a set of intervals rather than a point.

Sources and further reading

• The object in (3) and the identity in (4) — Least common multiple
• The divisor in (5) and (6) — Greatest common divisor
• Why (8) is solvable exactly when three divides the offset — Bézout's identity
• The check that killed the trap before any arithmetic — Dimensional analysis

Everything above was checked before publication by symbolic arithmetic and by exhaustive second-by-second enumeration over a full hour, which confirmed that no instant in the first fifty-nine seconds is a ring time for both bells, and that the shared ring set is exactly the set of common multiples. The rule in (5) was then verified against enumeration for all three thousand six hundred gap pairs up to sixty seconds.