JMC 2015：全部卡片

360°

提示One full circuit brings you back facing the way you started.

为什么This is true for a triangle, a hexagon or a fifty-sided shape: the exterior angles sum to 360° whatever the number of sides. It is often quicker than the interior-angle formula — find the missing exterior angle first, then use the straight line (180°) to reach the interior angle beside it.

Exterior angle = sum of the two opposite interior angles

提示Extend one side past a corner: the turn made there matches the pair of corners far away.

为什么It follows from two facts you already know: the angles of a triangle add to 180°, and the angles on a straight line add to 180°. Competition setters love it because it lets you leap straight to an answer without finding every angle in the diagram.

When either is 1

提示Try a few pairs where one number is tiny.

为什么1 + 7 = 8 but 1 × 7 = 7, so a 1 is the one case where the plus sign wins. With 2 and 2 the two results are equal (4 and 4); from 2 and 3 upwards, multiplying is always bigger. So when you may choose + or × to make an expression as large as possible, multiply everything — except that a 1 should be added, not multiplied.

8

提示Four of them are single digits; the rest sit in the teens.

为什么The primes below 20 are 2, 3, 5, 7, 11, 13, 17 and 19. Because 2 is the only even prime, every other prime is odd. That drives a family of parity arguments: odd + odd is even, so the sum of two odd primes can never be prime, and any question about sums of primes usually turns on whether a 2 is involved.

2520

提示Not 10 factorial — that repeats factors. One highest power per prime.

为什么2520 = 8 × 9 × 5 × 7. Any number divisible by everything from 1 to 10 must be a multiple of 2520, so the fast check is: does it have three 2s, two 3s, a 5 and a 7 among its prime factors? Being divisible by 5 needs a last digit of 0 or 5, which alone eliminates most options.

60; 90; 120

提示The three shapes that tile a flat floor on their own: think how many meet at a point.

为什么These three appear together constantly: a square sitting on a triangle, a hexagon with a square on one side. Whenever shapes meet at a point, add the known corners and subtract from 360° to find the gap; whenever they meet along a line, subtract from 180°.

Only their units digits

提示Think about what the last column of the working depends on.

为什么The last digit of a product depends only on the last digits of the factors: 7 × 6 = 42, so 4837 × 2596 ends in 2. Two products with different final digits cannot be equal, which lets you reject answer options without doing any long multiplication.

Total of areas − area covered

提示Where two tiles sit on the same spot, that spot has been counted twice.

为什么Adding the tiles' areas counts every overlapping patch once per tile, while the covered region counts it once. The difference is exactly the overlap (as long as no spot is covered three times over). It is the same idea as a Venn diagram: add the parts, then take away what was counted twice.

A number that reads the same forwards and backwards

提示Like the words noon and level, but made of digits.

为什么A multi-digit palindromic number can never end in 0, because it would then have to begin with 0. So whatever a last-digit test tells you about the units digit — even, or a multiple of 5 — it tells you about the leading digit as well, before you have done any arithmetic.

4

提示Neither 0 nor 1 is prime, and every even digit but one is out.

为什么The prime digits are 2, 3, 5 and 7. Knowing them by heart speeds up any question about primes built from digits.

equilateral

提示The angle at the centre is a sixth of a full turn, and two sides of each piece are the same length.

为什么Each central angle is 360° ÷ 6 = 60°, and the two sides from the centre are equal, so every piece is an equilateral triangle with the hexagon's own side length. Shaded-fraction questions on hexagons are usually solved by counting these six triangles — or the twenty-four smaller ones you get by quartering each.

Brick above minus neighbour

提示Sums going up mean the reverse operation coming down.

为什么Working downwards is the inverse of working upwards, so every step down is a subtraction. Fill in only the bricks you actually need: a missing bottom corner often needs just a diagonal chain of subtractions, not the whole pyramid.

It's the one where the amount being subtracted is smallest

提示Total up only the minus terms in each version, then compare those totals.

为什么If several calculations all use the same set of numbers with + and − mixed in different ways, every calculation subtracts some of the numbers and adds the rest. The result is biggest when the numbers being subtracted add up to the least — you don't need to work out every total from scratch, just compare what's being taken away in each version. This saves real time when checking several similar-looking multiple-choice options.

Count on to the next full hour, then add the remaining whole hours to midnight

提示Break the gap at a round point first — that keeps the numbers small and easy to add.

为什么Jumping to a round time first turns an awkward subtraction into two easy steps: how far to the next full hour, then how many full hours after that until the target time. This trick works for any 'how long until a deadline' question, and generalises to any calculation where a detour through a friendly round figure is quicker than one messy subtraction.

Odd

提示Try 15 ÷ 3 and 21 ÷ 7, and look at what kind of number both answers turn out to be.

为什么This is a fast filter on a multiple-choice paper: whenever a division is expected to come out as a whole number, and both the number being divided and the number dividing it are odd, the answer itself must be odd too — so any listed option that is even can be ruled out without doing the division at all. The reverse also holds: odd × odd always equals odd.

Combine the two fractions into one single fraction of the number

提示Find a common denominator for the two amounts before doing anything else.

为什么Say the difference between two different fractions of a number is known. Combining those two fractions using a common denominator collapses the puzzle into 'a single simple fraction of the number equals this difference' — from there, divide the given difference by that fraction to find the whole number in one step, without ever writing an algebraic equation.

Whether it also leaves remainder 1 when divided by 3

提示Odd numbers already score 1 on the 2s check — now put each option through the 3s check too.

为什么Every odd number leaves remainder 1 when divided by 2, so for an odd candidate that half of the check is already settled — the real work is testing each one against division by 3. Among odd numbers, two in every three leave 0 or 2 on the 3s check, and only those leaving 1 match. The shortcut is deliberately scoped to odd candidates: an even number can match too, but only by leaving 0 on both checks, which means it is a multiple of 6. So every number with matching remainders has the form 6k or 6k + 1, and the odd-only check covers just the 6k + 1 half — enough when every option in front of you is already odd.

friendly

提示Think nice, round numbers you could divide in your head without a written sum.

为什么Exact division isn't needed when a question only asks 'roughly how many' — round the total and the count involved to nearby clean values, then divide those tidier numbers. The estimate lands close enough to the true rate to pick the correct multiple-choice option, using far less working than a precise calculation would need.

One minus the stated fraction, of the whole quantity

提示If a fifth of the total is left over, the fixed number itself must be the other four-fifths of it.

为什么Picture a quantity built from two parts: a fixed lump, plus a fraction of the whole quantity added on top. Since the fraction-of-itself part is already 'inside' the total, the fixed lump must be exactly the complementary portion — subtract the stated fraction from 1 to find what share the fixed number represents, then use that to work backward to the total quantity.

List each achievable multiple of 3, then count the pairs that reach it

提示Group your search by target total instead of checking every combination in turn.

为什么Rather than testing every possible pair of numbers one at a time, work out which totals within your range are actually multiples of 3, then for each of those target totals list the pairs of numbers that add up to it. Adding the counts for every target total gives the answer far faster than a pair-by-pair check, especially as the set of numbers grows larger.

Test both starting assumptions and check they give the same total

提示Follow the chain of accusations twice, once from each possible opening statement, and compare.

为什么When each person in a chain says the one before them is lying, the truth values have to alternate: truthful, lying, truthful, lying. So you often cannot pin down whether the very first person was honest. Work through the whole chain assuming they were, then again assuming they weren't. If the chain holds an even number of people, the alternating pattern splits them exactly half and half either way, so both routes land on the same total and the count is fixed even though the starting fact stays unknown. With an odd number of people the two routes disagree, which is why you check that they match rather than assume it.

Leaves the fraction's value exactly unchanged

提示Multiplying top and bottom by two over two changes nothing at all.

为什么If a fraction's numerator and denominator are both exactly doubled, that is the same as multiplying the whole fraction by two over two, which equals one — so the value cannot change. Spotting this factor-of-two relationship between two fractions built from the same digits saves you from simplifying the second fraction from scratch.

Triple the largest cube smaller than the target and see whether it reaches the target

提示Multiply the single biggest building-block value by three, then see if it beats your candidate total.

为什么Before searching for three positive cubes that add to a target cube, check whether the target is reachable at all: multiply the largest cube below it by three, since the same cube may be reused. If that ceiling falls short of the target, no combination of three cubes below it can possibly reach it, so the target is ruled out without testing a single combination. Clearing the ceiling does not prove a solution exists.

above; below; sides

提示Photograph the shape from the top, the underneath, and each vertical face, then add those totals together.

为什么Rather than counting individual exposed cube faces (many of which are hidden inside the stack), work out the area you would see looking straight down, straight up from underneath, and straight at each of the four vertical faces. Because two of those views usually match the footprint exactly, this cuts the counting down from dozens of small squares to a handful of view areas.

There is always one fewer gap than the number of objects

提示Try lining up just two items, then three, and count the spaces between them each time.

为什么Picture two objects: there is exactly one gap between them. Add a third object and you add one more gap, not two. This pattern continues for any row of equally spaced objects with one sitting at each end of the fixed length: the count of gaps is always one less than the count of objects. Once you know the total length and the objects' combined width, sharing the leftover length across that number of gaps gives the size of each gap.