Hard Number Theory Problems: A Method When You Are Stuck

Every student who has attempted a serious competition knows the feeling. You read the question twice. You understand every word. And you have absolutely no idea what to do first. Hard number theory problems produce that silence more reliably than any other topic on the paper.

Most students interpret that silence as a verdict on their ability. It is not. Strong problem-solvers feel it too — the difference is that they do not treat it as a signal to stop. They have a short list of moves to try when nothing is obvious, and they start working through it. Hard number theory problems yield to method far more often than to inspiration.

What follows is that list, in the order it is usually worth trying.

Move one: make the problem smaller

If a problem involves a large or general number, replace it with a small one. Not because the small case is the answer, but because patterns are visible at small scale and invisible at large scale.

Suppose a question asks for the last two digits of 7²⁰²⁷. There is no sensible way to begin on that directly. So do not. Compute the last two digits of 7¹, 7², 7³, 7⁴: you get 07, 49, 43, 01. And at 7⁴ ending in 01, the whole pattern must restart, because multiplying by something ending in 01 reproduces the cycle. The period is 4, and a problem that looked impossible now needs one division.

The instinct to protect is this: when you cannot start, compute something. Almost any concrete calculation is better than staring. Five small cases take ninety seconds and they very often expose the structure that the general argument then confirms.

One honest caution. A pattern in five cases is a conjecture, not a proof. On a multiple-choice paper like the AMC 8 that distinction rarely costs you anything, and acting on a well-tested pattern is entirely sensible. In an olympiad, where you must justify the claim, the pattern is where the real work begins rather than where it ends.

Move two: ask what cannot change

Many problems that look like they require tracking a complicated process are actually asking about something that the process never affects. Those quantities are called invariants, and spotting one often collapses the entire question.

The classic example: the numbers 1 to 10 are written on a board. You repeatedly erase two of them and write their sum. After nine operations one number remains. What is it?

You could try to track the choices, and there are a great many. But notice that each operation removes two numbers and adds back exactly their sum — so the total never changes. It starts at 55 and it ends at 55, whatever choices you make. The answer is 55, and the branching was irrelevant.

The invariants worth checking first in number theory are small in number and repay the habit:

Parity. Is the quantity odd or even, and does the operation preserve that? Parity arguments settle a surprising number of "is this possible?" questions in a single line.

Remainders. Work modulo a small number — often 3, 4, 9 or 10 — and see whether something is fixed. If a target value is impossible modulo 4, it is impossible, and no amount of searching will produce it.

Digit sums. These are preserved modulo 9, which is why they answer divisibility questions and questions about rearranged digits.

When a question asks whether something can happen, and you suspect it cannot, an invariant is almost always the reason.

Move three: work backwards from the options

On a multiple-choice competition this is not cheating. It is using the information the paper gives you, and on some questions it is plainly the intended route.

If a question asks for the smallest integer with some property and offers five candidates, testing candidates is often faster than deriving the answer. Start with the middle option: if it fails in a way that tells you the answer must be larger, you have eliminated three options with one test.

The same reasoning applies partially. Even when you cannot finish a problem, you may be able to show that the answer must be even, or must be divisible by 3, or must exceed 100. Each such fact removes options. Because the AMC 8 has no penalty for a wrong answer, narrowing five choices to two and then guessing is a genuinely good outcome — it is worth far more than an elegant half-solution left unwritten.

Students trained only in school mathematics often resist this, because school rewards the full derivation. A competition rewards the correct letter. Those are different games and it is worth being explicit with students about which one they are playing.

Move four: change what the problem is about

The hardest questions usually respond to a reframing rather than to more effort. Three reframings do most of the work.

Rewrite in prime factors. A question about divisors, multiples or squares is a question about exponents. Converting it makes the structure visible, as with the perfect-square questions where "is it a square?" becomes "are the exponents even?".

Rewrite as remainders. A question about last digits, cycles, days of the week or repeating patterns is a modular question. Once stated that way, the tools are standard.

Count the complement. If counting the things you want is awkward, count the things you do not want and subtract. "How many numbers under 100 are divisible by neither 2 nor 3?" is unpleasant head-on and straightforward as 99 minus those divisible by 2 or 3.

The common thread is that you are not working harder on the original phrasing. You are trading it for an equivalent phrasing that a known method already handles.

A worked example of the method

What is the smallest positive integer with exactly 15 divisors?

Read it and there is no obvious opening. So apply the moves.

Reframe into primes. The divisor count of 2^a × 3^b × … is (a+1)(b+1)…, so we need those factors to multiply to 15.

Small cases — the factorisations of 15. Either 15 = 15, or 15 = 3 × 5. That is the complete list, so there are only two shapes to consider.

The first shape means a single prime with exponent 14, and the smallest such number is 2¹⁴ = 16384.

The second means exponents 2 and 4, on two different primes. To keep the number small, put the larger exponent on the smaller prime: 2⁴ × 3² = 16 × 9 = 144. The other assignment, 2² × 3⁴ = 324, is bigger — which is worth checking rather than assuming.

So the answer is 144. Notice that no step required insight. Reframing gave the structure, enumerating small cases gave the two possibilities, and a comparison finished it.

Knowing when to stop

This is the part most preparation ignores, and it costs students real marks.

Forty minutes for 25 questions is a little over 90 seconds each. Later questions are harder, so the early ones must be faster than that to leave room. A student who spends eight minutes on question 21 has not been brave; they have surrendered three or four easier questions later in the paper.

A workable rule: if you have made no progress in about two minutes, mark your best guess, flag the question, and move on. Return only if time remains. Because a blank and a wrong answer score identically, never leave anything unanswered — record something before you move, not at the end when you may be rushing.

Students find this genuinely difficult, because abandoning a problem feels like failing at it. Reframe it for them: the goal is the highest total score, and a question you cannot do in two minutes is worth exactly as much as one you can — which means time spent on it is time borrowed from questions you would have got right.

Building a routine for hard number theory problems

These moves only help if they are automatic under pressure, and that requires practising them specifically rather than just doing more problems.

When a student gets stuck, resist giving the hint. Ask instead: have you tried small cases? what stays the same? what do the options tell you? The aim is for the student to run that checklist themselves, eventually without being prompted.

It is also worth normalising the difficulty. A student who believes strong solvers see answers immediately concludes, on getting stuck, that they are not one. A student who knows that everyone starts most hard problems with no idea keeps working — and working is what solves them.

Registering for the AMC 8

The AMC 8 is 25 multiple-choice questions in 40 minutes, open to students in grade 8 and below, with no penalty for a wrong answer. It is a good first competition precisely because the later questions demand the kind of thinking described here rather than more advanced content.

Students cannot enter individually. Registration runs through a participating school, university, math circle or learning centre, whose Competition Manager orders and administers the competition. If your child's school does not take part, find a local centre that does — and begin looking early, because registration closes well ahead of the competition week.

The current cycle's dates, deadlines and fees are published at maa.org/amcreg. If you want structured practice at exactly this kind of problem, our AMC 8 preparation course works through the harder end of the paper under timed conditions, and the AMC 8 masterclass focuses on strategy and pacing. Students moving up after Class 8 continue with AMC 10 preparation, where these four moves matter even more.