How to check an answer when the clock is against you

The AMC 10 and AMC 12 give you 75 minutes for 25 questions. That is three minutes each, and the ones you can actually finish take less, which is where the checking time comes from. There is no separate checking phase at the end. There is only the twenty or thirty seconds you can afford on the question you just did, and the question is what to spend it on.

Almost everyone spends it wrong. They re-read their working.

Why re-reading catches nothing

Reading your own derivation again runs the same reasoning through the same brain that just produced it, under the same time pressure and holding the same wrong assumption. If you dropped a factor of two because you thought the shape had two halves rather than four, you will read the line about halves and it will still look correct, because the belief that caused the error is still installed.

A check is only worth seconds if it is independent of the derivation: a different route to the same number, or a property the true answer must have that your answer can be tested against without redoing anything. Everything below is one of those two.

Run them in this order. They are sorted by cost, and the cheap ones catch the embarrassing errors.

1. Did you answer the question that was asked

Free, and it catches the single most common lost mark. The problem asked for $x + y$ and you found $x$. It asked for the perimeter and you produced the area. It asked how many were red and you counted the blue.

Say the question out loud in your head, in five words, and look at what you wrote down. Under pressure this fails far more often than the algebra does.

2. Trap it between two walls

Before you check whether the answer is right, check whether it is possible. This is bounding used as a two-second instrument rather than as a proof technique: a probability lives in $[0, 1]$, a count of subsets of a ten-element set cannot exceed $1024$, an angle in a triangle is under $180^\circ$, and a length in a figure is less than the longest segment drawn.

The stronger version is a crude estimate you trust for a different reason than the exact computation. Suppose you are counting three-element subsets of $\{1, \dots, 10\}$ whose sum is divisible by 3. There are $\binom{10}{3} = 120$ subsets in total, and residues mod 3 are spread fairly evenly, so about a third of them, roughly 40, should work. The true count is 42. That estimate would not have proved 42 correct, but it kills 14, and it kills 126, instantly.

3. Cast out nines

For anything arithmetic-heavy, work the whole computation again modulo 9. It costs a few seconds because digit sums are fast, and it rests on the fact that a number and its digit sum leave the same remainder on division by 9, which is the content of the divisibility rule for 9.

Claim $37 \times 43 = 1591$. Then $37 \equiv 1$ and $43 \equiv 7$, so the product must be $\equiv 7 \pmod 9$; and $1 + 5 + 9 + 1 = 16 \equiv 7$. Consistent. Write 1691 by mistake and the digit sum is 17, which is $\equiv 8$, so the check fires.

Know what it misses, or you will over-trust it. Transpose two digits and the digit sum is unchanged: 1519 passes the mod 9 test cheerfully. Mod 9 catches slips in the values, not in the order. Parity is the same kind of instrument for a different failure: if the answer must be even because it counts pairs, an odd answer is dead without any further work.

4. Substitute one number

For any answer that is a formula rather than a number, test it at the smallest case you can compute by hand. Suppose you derived $\sum_{k=1}^{n} k(k+1) = \dfrac{n(n+1)(2n+1)}{6}$, having half-remembered the sum-of-squares formula. Set $n = 1$: the left side is $1 \cdot 2 = 2$, the right side is $\dfrac{1 \cdot 2 \cdot 3}{6} = 1$. Dead in four seconds. The correct closed form, $\dfrac{n(n+1)(n+2)}{3}$, gives 2 as it must.

Substitution is also the fastest way to kill a plausible-looking inequality. Whether $a + b + c \ge \tfrac1a + \tfrac1b + \tfrac1c$ holds for positive reals with $abc = 1$ looks like a standard exercise until you try $a = b = 2$, $c = \tfrac14$: the left side is $4.25$ and the right side is $5$. One substitution, claim destroyed.

Note the asymmetry, because it is the whole discipline. A test case can refute, but it cannot confirm. Plenty of triples with $abc = 1$ do satisfy that inequality, which is exactly why someone believed it. Passing your test case means you failed to find a problem, not that there is none.

5. Check the equality case

For an inequality you have proved, ask where equality holds and confirm it survives. $x^2 + 1 \ge 2x$ is tight exactly at $x = 1$, where both sides equal 2. If your proof produces a bound nothing achieves, you have proved something weaker than the problem wanted, and often that is because a step threw away more than it needed to.

The reverse failure is worth a warning. Some true statements have almost no margin, so sampling a couple of convenient points tells you nothing. Take $x^4 - x + \tfrac12 > 0$, true for every real $x$. At $x = 0$ and $x = 1$ you get $\tfrac12$ both times, which feels like comfortable room. The actual minimum sits a little under $0.03$, near $x \approx 0.63$. Two lucky samples looked safe while the real margin was thirty times smaller. When an inequality is a competition problem, assume it is nearly tight and check near where it would fail, not where it is convenient.

What this costs, and when to build it

Checks 1, 2 and 3 fit in about twenty seconds combined. Check 4 costs about ten seconds when a small case exists. Check 5 is slower and belongs to the proof-based papers rather than to a multiple-choice clock.

None of them are habits you can install on the day of a contest. Under real pressure you do what you have rehearsed, and if you have never once written the mod-9 line during practice, you will not write it in the exam either. Pick one check and use it on every single problem for a week, until it stops being a decision.

That means practising with the check attached, which is easier when problems arrive with a known answer and a time budget. The problem archive states an expected time on every entry, so you can see whether you had checking time or only thought you did. The daily problem is three tiers a day, no account needed, and one problem a day is enough to keep the habit alive. And the bounding and squeezing page treats the estimate above as what it really is: half of a proof, run in isolation because half is all the clock will pay for.