How to read a hard problem without giving up

You rarely give up on a hard problem after failing to solve it. You give up before you start. You read the statement, nothing happens; you read it again more slowly, still nothing happens; and somewhere in the third pass you conclude that this one is beyond you. Notice what did not occur in that sequence. Nothing was attempted. The problem was stared at.

That failure has a cause and the cause is fixable. You were reading for meaning when you should have been reading for material. Comprehension is what you do to a paragraph of a novel. A competition problem wants something else from a reader: it wants to be taken apart, and taking it apart happens on paper, with a pen moving, long before you have any idea how the solution goes.

What follows is a reading procedure, not a solving method. It is what you do to a statement in the first five minutes so that there is something in front of you to solve.

Separate the givens from the goal

Two columns. On the left, every condition the problem hands you, one per line. On the right, the single thing you are asked to produce.

Take the number that quadruples when its last digit moves to the front, IMO 1962 Problem 1. The left column has three entries: it is a natural number, its last digit is a six, and erasing that six and writing it in front produces four times the original. The right column has one: the smallest such number.

Splitting like this catches a specific and expensive mistake. "Smallest" is a property of the answer, not a condition on the object, and filing it on the left makes the problem feel much harder than it is. On the right it is only bookkeeping you do at the end.

The reason this column is worth ruling is that every given is load-bearing. Nobody puts a hypothesis into a competition problem for decoration. That gives you a test you can run on your own work at any moment: if you have an argument that never used one of the lines on the left, the argument is wrong, and you now know exactly where to look. A hypothesis you never spent is a piece of the problem you never read.

Name the objects

Three-digit numbers divisible by eleven whose quotient is the sum of the squares of their digits — IMO 1960 Problem 1 — is unreadable as English. As English it is a search through nine hundred candidates, and a search is not something you can think about, only something you can grind.

Write $N = 100a + 10b + c$ and the sentence turns into machinery. Divisibility by eleven becomes a condition on $a - b + c$. The digit-square condition becomes an equation in three bounded integers. Both can now be manipulated, bounded and split into cases, which is what "solving" will consist of. The answers are $550$ and $803$, and you can see the second one work: $803 = 11 \times 73$, and $73 = 8^2 + 0^2 + 3^2$.

Nothing was discovered in that substitution. That is exactly why you should do it immediately and without deliberation. Naming costs nothing, forecloses nothing, and commits you to no plan, so there is never a reason to wait until you understand the problem better. You will not understand it better until you do.

Compute the smallest case you are allowed to compute

Back to the 1962 problem, now with names. Let $n = 10m + 6$ and suppose $m$ has $k$ digits. Moving the six to the front produces $6 \cdot 10^k + m$, so the requirement reads $6 \cdot 10^k + m = 4(10m + 6)$, which rearranges to $$m = \frac{2 \cdot 10^k - 8}{13}.$$

Now you can simply try values. For $k$ up to four the numerator is never divisible by thirteen, so no such number has five digits or fewer. At $k = 5$ it works: $m = 15384$, hence $n = 153846$, and indeed $615384 = 4 \times 153846$. Because you climbed through the smaller cases rather than around them, "smallest" was answered on the way past.

The move worth stealing is what the substitution did to the word smallest. An open-ended superlative over infinitely many integers became "the least $k$ for which a specific fraction is an integer", and that is a question with a shape. Vague questions are unanswerable in a way that has nothing to do with difficulty.

Read the goal backwards

Some problems refuse to be entered from the front. Proving that $\tfrac{21n+4}{14n+3}$ is irreducible for every natural number is one: there is no first step, because "irreducible" is not a thing you can start doing.

So start from the end instead. Irreducible means the numerator and denominator share no factor, which means their greatest common divisor is one. And a standard way to prove that two integers have greatest common divisor one is to exhibit an integer combination of them equal to one. The goal has now changed shape. You are no longer proving a property; you are hunting for two coefficients $x$ and $y$ with $x(21n + 4) + y(14n + 3) = 1$.

That is a search you can actually run. The $n$ terms have to cancel, so $21x + 14y = 0$, so $x$ and $y$ are $-2$ and $3$ up to scale, and the identity $3(14n + 3) - 2(21n + 4) = 1$ holds for every $n$ at once. The problem is finished, and the only insight was refusing to start at the beginning.

This is the whole content of working backwards: assume the ending, and let it tell you what the middle has to contain. It is the oldest habit in geometry for the same reason. The end of a problem carries more information than the start, because the start is deliberately vague and the end never is.

Ask what the problem does not care about

The last pass over the statement is a question about what is absent. If swapping two letters leaves every given unchanged, the problem does not distinguish them, and neither should your work: you may assume an order between them for free.

Absence is also a hint about where the answer lives. Fix $a + b = 10$ and ask for the largest possible product. The set-up is unchanged by swapping $a$ and $b$, and sure enough the product peaks at $a = b = 5$, where it equals $25$. That is symmetry doing the work an optimisation would otherwise cost you.

One caution, because it is where the habit goes wrong. Symmetric problems put their extremes at the symmetric point or at the boundary of the region, and you have to check both. What symmetry buys you is a shortlist of two, not an answer.

The rule that decides when to stop

Everything above takes five to ten minutes and produces no solution. That is the interval where giving up actually happens, so it needs a rule rather than a mood.

Give up on a budget, not on a feeling. Decide before you start how long the problem gets — a few minutes for a mid-paper AMC question, the better part of an hour for an olympiad problem — and hold to it in both directions. The budget stops you from grinding a lost cause for an afternoon, and, far more importantly, it stops you from quitting at minute four, when the feeling arrives and the evidence has not.

Two conditions make the budget honest. The clock only runs while the pen moves; time spent rereading the statement is not time spent on the problem. And when the budget expires, take one hint, not the solution, and then set a fresh budget. A hint returns you to the problem. A full solution ends it, and a solution read after a genuine attempt teaches something, while the same solution read after no attempt merely feels as though it did.

What to do with this

Practise the reading separately from the solving. Take a page of problems you have not seen and, for each one, write only the two columns, the names for the objects, and one sentence on what the final line would have to say. Do not solve any of them. You will get through the page in the time a single solution takes, and the thing you are short of is not solutions.

Lemma builds the same habit into the ladder, teaching working backwards and symmetry as named moves you reach for on purpose rather than tricks you happen to remember. The technique index lists them all, the problem archive carries the statements above in full with hints staged one at a time, and the daily problem posts a fresh one every day in three tiers with no account needed, which is the cheapest way to keep the reading muscle warm.