IMC 2025：全部卡片

6

提示What does 2 × 8 end in?

为什么Powers of 2, 3, 7 and 8 all have units-digit cycles of length 4; 4 and 9 have length 2; 0, 1, 5, 6 never change. The remainder on dividing by 5 (or 10) is read straight from that units digit — 8¹⁰ ends in 4, so it leaves remainder 4 on division by 5.

The sum of the two opposite interior angles

提示Combine the straight line with the triangle's own total.

为什么If the base angles are b, the exterior angle at the base is 180 − b and also equals apex + b. Isosceles triangles hand you two equal base angles for free — say so before you write anything else.

The number of items

提示Think of paying whole pounds and getting change.

为什么Prices like 99p and £2.99 are whole pounds minus a penny, so 25 such items cost 25p less than a whole number of pounds. Read the count off the pence, then a second equation on the pounds finds how many of each.

Ninths

提示0.1 recurring is 1 over what?

为什么A single repeating digit d is d/9: 0.1 recurring = 1/9, 0.6 recurring = 2/3, 0.9 recurring = 1. So a sum of them is a fraction over 9, which may itself be a recurring decimal.

sum

提示Not the numbers themselves.

为什么Expand the brackets, then substitute the given product rather than solving for a and b — most 'find the sum' questions never need the numbers themselves.

Opposite corners: equal sums of squares

提示Pythagoras four times, then pair up.

为什么Dropping perpendiculars from P to the sides makes four right-angled triangles sharing legs; adding the two 'opposite' Pythagoras equations gives PA² + PC² = PB² + PD². Distances 6, 8 and 10 to three corners in order force the fourth to be √(36 + 100 − 64) = √72.

r + R

提示The touching point lies on the line of centres.

为什么Centres and touching point are collinear, so the centre distance is r + R when the circles touch from outside and R − r when one sits inside the other.

A multiple of 4

提示How many equal shares make the whole?

为什么A ratio a : b in whole parts makes the total a multiple of a + b: 3 : 2 → multiple of 5, 5 : 1 → multiple of 6, 7 : 3 → multiple of 10. Matching totals to ratios is then a divisibility check, not an equation.

3n and 5n

提示One shared multiplier scales both parts.

为什么Introduce one letter for the common multiplier: 3n red and 5n blue counters; add 4 red and the ratio is 1 : 1, so 3n + 4 = 5n and n = 2.

even; odd; 2

提示Which multipliers keep the parity and which force it?

为什么So 5n + 3 even forces n odd; then 7n + 2 is odd and 4n + 1 is odd for every n. Parity questions are settled by checking n = 1 and n = 2 — nothing else changes the pattern.

Factorise each bracket

提示Both brackets share a common power.

为什么5⁴(5² − 1) ÷ 2⁶(2² − 1) = 625 × 24 ÷ (64 × 3) = 625 × 8 ÷ 64 = 625/8. Never expand a power you can factorise.

2

提示Each pair got counted in both orders.

为什么Choosing 2 from 6 gives 6 × 5 ÷ 2 = 15, from 7 it is 21. Constraints such as 'not next to each other' are handled by listing by the first chosen item and counting what remains legal.

The number with the smallest absolute value

提示Ignore the sign and compare sizes.

为什么Two negative numbers can look far apart yet one sits nearer zero — always strip the sign and compare what remains. The same idea also settles "which is bigger" once every option is negative: the smaller absolute value is the larger number.

Split it into place-value chunks that are already multiples of that divisor

提示Think about grouping digits, not checking one at a time.

为什么Spot that many multiples of the divisor already sit inside the number's place-value blocks — millions, thousands, hundreds — so only a small remainder needs checking directly, saving a full long division.

Divide the total surplus by the surplus per item

提示Set up one equation using the count as the unknown.

为什么Write (surplus per item) × n = total surplus, then solve for n. This turns a wordy description into a one-line equation before any numbers are plugged in — the same move works for any "each one always contributes X extra" scenario.

multiplying

提示What's the opposite operation to dividing?

为什么So a ÷ (b ÷ c) becomes a × c ÷ b — useful when a division chain is nested in brackets, because the innermost fraction flips to a multiplication as you unwrap it outward.

1

提示One of the two claims must be false, and the other true.

为什么This settles the first contradicting pair straight away; anyone else's statement is then checked against what that pair's truth-values force — a chain of only-one-can-be-right deductions, not full case-by-case guessing.

Equals 2 — no fraction arithmetic needed

提示Check whether one mixed number is simply twice the other first.

为什么Spotting that relationship first (comparing whole-number parts and denominators) is faster than converting both to improper fractions and cancelling — the full method still works, but the shortcut skips it entirely when it applies.

Factor out a common term to leave a perfect square inside the root

提示Look for what multiplies both parts before you try to simplify directly.

为什么Rewriting the expression so a shared factor comes out front often turns a messy subtraction under a root into (factor) × (perfect square), which then square-roots cleanly instead of needing an estimate.

Cancel the matching factors from both sides first

提示Spot which numbers repeat on each side of the equation.

为什么Spotting a shared factor on both sides and dividing it away first turns a big multiplication into a much smaller one — the two remaining products must still be equal, so solve that smaller equation instead.

Multiply each blend's share of the mix by its own percentage, then add the results

提示Think of it as a weighted average of the two mixtures.

为什么Each blend contributes to the final mix in proportion to how much of it is used, and each blend itself is a percentage split — multiplying those two percentages together and summing across blends gives the true overall percentage, avoiding a false halfway average.

Add all three equations, then eliminate to find the one unknown whose coefficient differs

提示Combine every relationship at once rather than solving one at a time.

为什么Summing every equation is still the right first move - it collapses three relationships into one - but the coefficients rarely land equal, so what comes out is a multiple of the total plus one leftover unknown. Pin that unknown down first by eliminating between two of the original equations, then substitute it into the summed equation and the total drops out. The saving is real: you find one unknown, never all three.

Combine the area equations so the hidden lengths cancel, leaving an equation in the total area alone

提示Try adding or multiplying the region relationships together and see what disappears.

为什么This often produces a quadratic in the total once you substitute back in — solve it, discard any root too small to contain the known regions, and the genuine total remains. No individual side length needs to be found at all.

Use a rough bound from the total cost to rule out the too-large options

提示Check which candidate count actually stays under the given maximum.

为什么Because the shortfall repeats every fixed number of items, several counts satisfy the pence condition alone; only one keeps the overall spend under the actual amount stated, so a quick size check (not more algebra) picks it out.

hypotenuse; leg

提示Picture the two centres, the surface, and a right angle between them.

为什么The line joining the two centres has length r + R (sum of radii, the hypotenuse); the vertical offset between where each touches the surface equals the difference of the radii (one leg); the horizontal offset is the third leg, found with Pythagoras.