Here is a puzzle that a 12-year-old can understand and that took the world's mathematicians 55 years to settle. Take a handful of numbers on a clock. Can you always put them in an order so that, as you add them up one at a time, you never land on the same hour twice? In February 2026 the final piece of the proof was published — and the key tool was not cleverness about any particular order, but randomness.
The puzzle
GCSE core Clock arithmetic is just arithmetic where you wrap round. On an 11-hour clock, 8 + 5 is not 13 but 2, because 13 goes past 11 and wraps round. Mathematicians call this working “modulo 11”. Now take some different numbers that are not 0 — say 1, 2, 3, 5, 9 and 10 — and add them in some order, writing down each running total. Try it:
Try it yourself — distinct partial sums on an 11-hour clock
The numbers 1, 2, 3, 5, 9, 10, added up one at a time on a clock with 11 hours (so 12 o'clock is 0, and 13 wraps round to 2). Can you order them so every running total lands on a different hour?
| Order | 1 | 10 | 2 | 9 | 3 | 5 |
|---|---|---|---|---|---|---|
| Running total | 1 | 0 | 2 | 0 | 3 | 8 |
Repeated total: 0 — try again.
Ronald Graham — a celebrated mathematician who was also an expert juggler (Quanta notes the question echoes juggling patterns, where two balls must never land on the same beat) — asked in 1971 whether a good order always exists, for every prime p and every set of non-zero numbers.
Illustration — the running totals as hops round an 11-hour clock
Start at 0, then add 1, 2, 3, 9, 5, 10 in turn:
- +1 → hour 1
- +2 → hour 3
- +3 → hour 6
- +9 → hour 4
- +5 → hour 9
- +10 → hour 8
The distinction students often miss
Checking examples is not a proof
You can find a good order for this six-number example in seconds, and computers can check millions of small cases. But there are infinitely many primes and, for each, astronomically many sets. Only a proof covers them all — the same gap between evidence and proof that runs through our Unresolved Mathematical Mysteries hub.
Why it was hard: three different sizes of problem
Undergraduate extension What makes the problem difficult depends on how many numbers you have compared with the size of the clock.
- Very small sets (say 100 numbers on a clock with a billion hours): there is lots of room, but you need to rule out every unlucky coincidence. Kravitz and Bedert settled this case in 2024.
- Nearly full sets (almost every hour is used): there is almost no room to manoeuvre. Müyesser and Pokrovskiy found a way: set aside a few specially chosen numbers, shuffle the rest randomly, then repair any trouble spots by inserting the spares.
- The middle (about half the hours): too crowded for the small-set methods and too sparse for the large-set ones. This was the last gap.
The final piece: randomness and waves
Postgraduate extension Sauermann and Pham started with a random order of the numbers. In a random order, a repeated running total is a “bad event”. They then needed to show two things: that bad events are rare enough, and that the few that occur can be repaired by rearranging. To show bad events are rare, they used Fourier analysis — the same idea that breaks a sound into pure sine waves — to prove that random combinations of the numbers do not favour any particular total. Their repair procedure fixes the ordering at least 90% of the time, which is all you need: if a random process succeeds with positive probability, a good order must exist.
As the researchers explained to Quanta Magazine, the striking thing about the method is that if you do it randomly it is likely to work, even though it is hard to describe explicitly what the solution looks like. This “probabilistic method” — proving something exists without ever writing it down — is one of the most powerful ideas in modern combinatorics.
Quick check
Sauermann and Pham showed a random order can be repaired at least 90% of the time. Why does that prove a good order always exists?
The timeline
Timeline
- 1971Ronald Graham poses the rearrangement question for numbers on a prime-sized clock.
- 2024Noah Kravitz and Benjamin Bedert settle the case of very small sets — say 100 numbers on a clock with about a billion hours.
- 2022–2025Alp Müyesser and Alexey Pokrovskiy handle sets containing almost every possible number; a joint team extends the methods to more cases.
- Feb 2026Lisa Sauermann and Huy Tuan Pham close the last gap — medium-sized sets — in a 27-page paper, completing the proof.
- Sep 2026Quanta Magazine reports the four-paper resolution of the 55-year-old conjecture.
Solved by people, in an age of AI
2026 has been full of headlines about AI solving maths problems — some genuine, some not, as our article Solved by Computer explains. Graham's conjecture is a useful reminder that the deepest progress still often comes from people: four teams, building on each other's ideas over several years, turning randomness into certainty.
Sources and further reading
- Mathematicians Harness Randomness To Crack a 55-Year-Old Conjecture — Quanta Magazine (28 Sep 2026) — the four papers, the case split, and Sauermann & Pham's method
- Probabilistic method — Wikipedia
- Modular arithmetic — Wikipedia
Last reviewed: 3 October 2026. Figures are typical published values and rules can change. Exam wording varies by specification, so check yours.
Frequently asked questions
What exactly does Graham's rearrangement conjecture say?
Take a prime number p and any collection of different non-zero numbers on a clock with p hours (arithmetic 'modulo p'). Then you can always put them in some order so that the running totals — first number, first two added, first three added, and so on — all land on different hours. It is now a theorem: proved across four papers, the last in February 2026.
Who solved Graham's conjecture?
No single person. Noah Kravitz and Benjamin Bedert handled very small sets; Alp Müyesser and Alexey Pokrovskiy handled sets containing almost every number, with a joint team extending their methods; and Lisa Sauermann (University of Bonn) and Huy Tuan Pham (University of Chicago) closed the final, medium-sized case in February 2026.
Was the proof found by a computer or AI?
No. This is a human proof, built from probability (random orderings) and Fourier analysis. Computers can check small cases quickly — like the six-number example on this page — but no amount of checking covers every prime and every set, which is why a proof was needed.
Why does it matter if the clock size is prime?
On a prime-sized clock, adding any non-zero number repeatedly visits every hour before coming back to the start, which gives the arithmetic very regular structure. The conjecture, and its proof, are about that prime case.
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Sudershan Soni
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