Test-day Habits — Save the points you already know how to win.

Here's the part most kids miss: most of the points you'll lose on a contest aren't from problems you can't solve. They're from problems you CAN solve but where you misread one word, mixed up cents and dollars, or bubbled the wrong letter at the end.

You don't have to learn new math to win those points back. You just have to build a few habits. We call them test-day habits. Practice them on every problem until they feel automatic — then on test day you don't even think about them.

Tiny words that flip answers

Here are real traps from past contests — one word changes the answer:

The unit-slip trap

The problem says: “A pencil costs 25 cents. How many pencils can you buy with $5?” If you write 5 ÷ 25, you get 0.2 — nonsense. Translate first: $5 = 500 cents. Then 500 ÷ 25 = 20. The units have to match before you divide.

The two-minute rule

If you've been on one problem for more than two minutes without making progress, mark it and move on. Come back later if there's time. A contest isn't graded on which problems you tried — it's graded on how many you got right. A long battle with one hard problem that ends in a wrong guess just cost you all the easy points after it.

You have the habits. But what about a problem where you read it and your mind just goes blank — you don't even know how to start? That blank is normal, even for strong solvers. The difference is they have a short list of moves to try, so a blank turns into “let me try this.” Here are five of the most useful, each with our own tiny example.

Strategy framing inspired by Posamentier, The Art of Problem Solving (teacher resource).

Move 1 — Work backward from the end

When the finish is clearer than the start, run the problem in reverse. Whatever was done going forward, undo it going back: add becomes subtract, times becomes divide.

Our numbers. “I picked a number, doubled it, then added 6, and got 20. What was my number?” Don't guess — undo. Last thing done was “add 6,” so first undo that: \(20 - 6 = 14\). Before that we doubled, so undo with halving: \(14 \div 2 = 7\). The number was 7. Check forward: \(7 \times 2 = 14\), \(14 + 6 = 20\). It works.

Move 2 — Try the extreme case

If a problem gives you freedom (“some number,” “any rectangle,” sizes not given) yet expects ONE answer, push a quantity to its limit — make it as big, as small, or as simple as the rules allow. The extreme is often easy to see, and the promised answer can't change.

Our numbers. “Two friends split into a circle and shake hands once with every other person. With \(n\) people there are \(\tfrac{n(n-1)}{2}\) handshakes — is that formula right?” Test the smallest real case, \(n = 2\): two people, exactly 1 handshake. The formula gives \(\tfrac{2 \times 1}{2} = 1\). Try \(n = 3\): a triangle of \(3\) handshakes, and \(\tfrac{3 \times 2}{2} = 3\). The extreme small cases confirm it fast.

Move 3 — Organize the data into a table

If a problem hands you a pile of facts, stop juggling them in your head. Make a grid and fill in what you know; the missing entry often becomes obvious.

Our numbers. “Anna, Ben, and Cara each own one pet — a cat, a dog, or a fish. Anna does not own the dog or the fish. Ben does not own the fish. Who owns the fish?” Build a yes/no grid: Anna can only be the cat. That frees the dog and fish for Ben and Cara; since Ben isn't the fish, Ben is the dog, so Cara owns the fish. The grid does the remembering for you.

Fill in the forced “no”s; the “yes”s appear on their own.

Move 4 — Account for every possibility (casework)

When a problem asks “how many ways,” split it into a few tidy cases that don't overlap and don't leave anything out, then count each case. The trick is choosing one thing to organize the cases around.

Our numbers. “How many ways can you make 15 cents using only nickels (5¢) and dimes (10¢)?” Organize by number of dimes. Zero dimes: need three nickels (\(15\)) — works. One dime: need one nickel (\(10 + 5 = 15\)) — works. Two dimes: \(20\)¢, too much — stop. So there are 2 ways. By marching through the dime count we're sure we missed nothing.

Move 5 — Always CHECK before you bubble

The last move is the cheapest and the most skipped. Once you have an answer, run three quick tests:

One pass through these three saves more points than any single technique on the test.

Sometimes the toolbox doesn't crack it and time runs short. That's fine — on these contests there is no penalty for a wrong answer, so you should put a letter on every problem. But a blind 1-in-5 guess is a waste — you can almost always do better.

The secret: you often don't need the whole answer. One small, cheap-to-compute property of the answer can knock out most of the choices.

Find one property of the answer, and four choices fall.

By now you have the everyday habits, a toolbox to start a blank problem, and a way to guess smart. The last chapter is the ceiling: the hardest problems on the test — the last handful, where the difficulty spikes. Those are where the kid going for a top score actually battles. Two specific moves help here, and neither one requires more math knowledge. They’re ways of using the problem against itself.

Trick 1 — engineering induction

When a problem has a variable like “there are n people in a row” and you can’t compute the answer directly, try n = 1, then n = 2, then n = 3. Write the answers in a column. Look for the pattern. Assume it continues.

This isn’t a formal proof — but in competition math, it’s almost always right. The contest setters write problems where the pattern is honest.

Example. “How many regions does n straight lines divide the plane into, if no two are parallel and no three meet at a point?” Compute small cases: n=0 → 1 region. n=1 → 2. n=2 → 4. n=3 → 7. n=4 → 11. Differences: 1, 2, 3, 4 — each new line adds one more region than the previous. Pattern locked. For n=10, regions = 1 + (1+2+…+10) = 56.

Trick 2 — use the freedom in the problem

If the problem says “in any triangle” or “for any rectangle,” then ANY example you pick will give the right answer.

Pick the easiest one. Equilateral triangle. Square. Numbers like 1 or 10. The problem PROMISED the answer doesn’t depend on which one you pick — so use the simplest.

Example. “In an equilateral triangle ABC, point P is inside. The sum of distances from P to the three sides is constant. What is it?” The problem says “constant” — so the answer doesn’t depend on where P is. Pick P at the center. Now everything is symmetric and you can read off the answer in one line.