Arithmetic & Operations — Look for structure before you calculate.

An arithmetic expression is not read straight left to right like English. It's read in tiers, with the higher tiers handled before the lower ones. The rule is called PEMDAS in U.S. schools (or BODMAS in many other countries — same rule, different letters).

One example of each tier:

• Parentheses: 2 + (3 × 4) = 2 + 12 = 14. The (3×4) is one unit.
• Exponents: 2 + 3² = 2 + 9 = 11. The 3² happens before the +. (Don't compute 2+3 first and then square.)
• Multiplication before addition: 2 + 3 × 4 = 2 + 12 = 14 (not 20). The × belongs to a higher tier than +.
• Left-to-right within a tier: 12 ÷ 3 × 2 = 4 × 2 = 8 (not 12÷6 = 2). M and D are same tier; do them left-to-right.

Watch the order in action

One PEMDAS tier per row. Highlighted piece is what gets computed; the answer drops to the next row.

The big idea behind PEMDAS: multiplication is repeated addition, so it has to happen before any plain addition does. Exponents are repeated multiplication, so they happen before any plain multiplication. Parentheses are how we say 'pretend this is a single number'.

An arithmetic series is a list of numbers that go up (or down) by the same step each time: 1, 2, 3, … or 10, 13, 16, 19, … or 81, 83, 85, 87, … The amount added each step is called the common difference.

Arithmetic series have a beautiful property: the average of the whole list equals the average of just the first and the last number. The reason is balance — for every number above the middle, there's a number an equal distance below it. They cancel out.

The pair (4, 16) sums to 20. The pair (7, 13) sums to 20. The middle (10) is half of 20. Every pair averages 10.

Sum = 5 × (average) = 5 × 10 = 50.

The Gauss story makes this memorable. When Carl Friedrich Gauss was about 9 years old (around 1786), his teacher tried to keep the class busy by asking them to add 1+2+3+…+100. Gauss did it in about a minute by pairing the ends: 1+100=101, 2+99=101, …, 50+51=101. Fifty pairs of 101 = 5050.

'Pair from the ends' is just this formula seen with your eyes. Whenever a list is long but evenly spaced, reach for it — never add term by term.

The most powerful identity in arithmetic — and the one most underused on AMC 8 — is the distributive property:

It says: multiplying first then adding gives the same answer as adding first then multiplying. The reason is that a × (b + c) means 'a copies of (b + c)', which is a copies of b plus a copies of c.

The contest trick uses the reverse direction. When you see two or three multiplications that share a factor, factor it out, add the leftover pieces, multiply once at the end. One multiplication beats three.

A tiny example. 17 × 6 + 17 × 4 = 17 × (6+4) = 17 × 10 = 170. You never had to compute 17×6 = 102 or 17×4 = 68. The contest setters love hiding 17×6 + 17×4 inside a longer problem.

When a problem says closest to or approximately, look at the answer choices. They're usually 10s, 100s, or 1000s apart. That gap tells you how precise your answer needs to be.

The big idea: if the choices are 100 apart, you can afford an error of up to about 50 without hitting the wrong box. So you can round each factor to whatever's convenient, as long as the total rounding error stays within your budget.

Two rules that almost never fail:

• Round before multiplying, not after. 2.46 × 8.163 is 20.08, but 2.5 × 8 = 20 — exactly right for our purposes, and you didn't multiply two ugly decimals.
• Round in opposite directions to cancel. If you round 2.46 up to 2.5, try to round 8.163 down to 8 so the errors push against each other.

This works because multiplication is roughly linear for small changes. A 2% bump in one factor changes the product by ~2%. So even rounding aggressively (say, 5% per factor) keeps you within 10–15% of the truth — usually way inside the answer-choice gap.

When a problem says given a = −2, which of [these five expressions] is the largest?, the student instinct is to reason: 'well, a² is positive so it's bigger than the negative ones, and 24⁄a is more negative than 4a, so…'. This is slower than just plugging in and ranking.

The principle is simple: comparing five concrete numbers is faster than comparing five symbolic expressions, even for a fast reasoner. Lay them out in a column, evaluate each, pick the biggest.

Where this matters most is when the input is negative or between 0 and 1. Both of these break your usual instincts:

• For a negative number, a² is positive, a³ is negative, −a is positive, 1/a is negative.
• For a number between 0 and 1, squaring makes it smaller (½² = ¼), and the reciprocal is bigger (1/½ = 2). The size order of expressions flips compared to numbers bigger than 1.

Both traps appear everywhere on AMC. The shield is: plug in. Don't try to reason. Computation is faster than philosophy.

An average is a summary. The thing you can always do arithmetic with is the total:

This sounds obvious, but the AMC tests whether you treat averages as numbers in their own right (a beginner mistake) or as a compressed form of (total, count).

Common trap: never average two averages. If a class of 20 kids scored an average of 80 and a class of 10 kids scored an average of 70, the combined average is NOT 75. It's:

combined total = 20 × 80 + 10 × 70 = 1600 + 700 = 2300

combined count = 20 + 10 = 30

average = 2300 ÷ 30 ≈ 76.7

The combined average leans toward the bigger group (the 20 kids at 80). That's weighted averaging. Every weighted-average problem on AMC is just this — go to totals, do plain arithmetic, divide at the end.

Same idea for 'one extra item changed the average': convert each average to a total, subtract, and you've isolated the new item.

The deviations trick — ups and downs must cancel

Here’s a beautiful fact: the amounts numbers go ABOVE the mean must equal the amounts they go BELOW.

Picture a seesaw, with the mean as the balance point. Every number is a kid sitting at some distance from the balance. The seesaw is level — so the total “weight” on each side must match. Add up all the “+ how much above” values; add up all the “− how much below” values; they cancel out to zero.

Below: −10 + (−5) = −15. Above: +5 + 10 = +15. They cancel.

Why it’s useful. When the AMC says “average of 5 scores is 80,” you know the deviations from 80 sum to zero. So if four scores are 70, 75, 88, 92 (with deviations −10, −5, +8, +12 totaling +5), the fifth must contribute −5 — meaning it’s 75.

Often this is faster than “total = average × count; missing = total − sum so far.” Just track distances from the mean.

Stories on the AMC love a sequence of give-aways: “Jack had X apples. He sold 25% to Jill, then 25% of what was left to June, then gave 1 to the teacher.” The slow way is to compute each step separately. The fast way is to stack the keep-fractions.

Picture the bar shrinking at each step:

The trap: 25% twice is NOT 50%

The second discount is taken from the already-reduced amount, not from the original. That’s why the losses don’t add.

Three different ways to describe the 'middle' of a list of numbers:

They tell you different things. Look at the list {1, 2, 3, 4, 100} on a number line:

The 100 yanks the mean far from the other values. The median sits right in the middle of the sorted list.

The mean is pulled around by extreme values; the median isn't. That's why news reports use median income (not mean income) — a few billionaires would skew the mean.

When does the median matter more than the mean? AMC often asks 'find n so the mean equals the median' or 'add a value so the average changes by 2'. These force the value into a specific range depending on where it falls in the sorted list.

To find the median of n values: sort them and take position (n+1)/2 if n is odd, or average positions n/2 and n/2+1 if n is even. Don't compute the others — just sort.

About 1 in every 3 AMC contests has a 'read the table' or 'read the graph' problem. Despite being arithmetic at heart, these problems cause more errors than they should — because students rush the reading.

Habits for tables:

• Identify what each row means and what each column means.
• Read the question to identify which cells to extract.
• If you need a sum or difference, mark the relevant cells and add.
• For 'percent of total' on a row, compute the row total first.

Habits for bar graphs and line graphs:

• Identify the axes and units (one tick = ?). A '50' tick can mean 50 dollars or 50 millions — always check.
• Read each relevant data point exactly.
• For cumulative graphs (where each new point adds to the previous), answers are differences of heights, not direct reads.

Habits for pie charts: each slice's angle ÷ 360° = its fraction of the whole. Multiply that fraction by the given total to get the slice's value.

Some AMC problems give you ugly numbers and answer choices that are orders of magnitude apart. The question isn't really "what's the exact value?" — it's "what's the size of the answer?" Exact computation is overkill; smart estimation wins.

Estimation by rounding

The cleanest tool is rounding: round each number to one or two significant digits, do the easy multiplication, count the zeros, and compare to the choices.

Example. 0.000315 × 7,928,564. Looks awful. But:

• 0.000315 ≈ 0.0003 = 3 × 10⁻⁴
• 7,928,564 ≈ 8,000,000 = 8 × 10⁶
• Product ≈ 3 × 8 × 10⁻⁴⁺⁶ = 24 × 10² = 2400.

Done. Order of magnitude is 10³, leading digits are about 24. If the answer choices are 210 / 240 / 2100 / 2400 / 24000, you can confidently pick 2400.

Powers of 10 — the secret weapon

Whenever you see scary numbers, rewrite as (small number) × (power of 10). This separates the shape of the number from its size.

'Closest to' and 'nearest' problems

For 'nearest dollar' or 'nearest hundred' problems, you have two options:

• Round first, then add (faster but loses precision). $1.98 + $5.04 + $9.89 → $2 + $5 + $10 = $17.
• Add exact, then round (safer when the rounding errors might cancel or compound). $1.98 + $5.04 + $9.89 = $16.91 → $17.

Both give $17 here. But beware: rounding three numbers and adding can produce a different answer than adding-then-rounding (rounding errors accumulate). When choices are very close together, do the exact computation and round only at the end.