Warm-Up Math Riddles You Can Solve in One Step
An easy math riddle is not a small hard one. It has a single hinge: one sentence that can be read two ways, or one quantity that is not what it appears to be. Once you spot the hinge the answer follows in a line of arithmetic, which is why these work as a first round with mixed ages. Nobody is left grinding through a page of working while everyone else waits.
Read each of these slowly, because most of them are won or lost on a single word. Watch for a phrase that looks like it describes part of a quantity when it describes the whole of it, and for a comparison that only means something once you have fixed the unit it is measured in. Riddles at this level are usually reading tests wearing a number as a disguise.
If you are using these with children, the honest test of an easy riddle is whether a wrong answer can be explained in one sentence. All eight below pass that test, and the working under each one is short enough to read aloud without losing the room.
Q1.An old potter was selling pots at the market. A woman bought one and paid its full price, which turned out to be one dollar plus half the price of the pot. What did the pot cost?Tap to reveal
Answer: $2.
Call the price x. She paid 1 + x/2, and that was the entire price, so x = 1 + x/2. Take x/2 from both sides: x/2 = 1, so x = 2. Check it: half of $2 is $1, and $1 plus $1 is $2. The riddle works because 'a dollar plus half the price' sounds like a description of part of the money when it is a description of all of it. The same shape turns up in every price riddle of this kind: a fast, confident, wrong answer arrives before you have finished reading the sentence, and the only reliable defence is to write the sentence down as an equation before answering it.
Q2.Jessie asked her grandfather how old he was. He answered: 'Divide my age by six, then subtract six, and you are left with six.' How old is he?Tap to reveal
Answer: 72.
As an equation, x/6 - 6 = 6, so x/6 = 12 and x = 72. Working backwards is quicker and needs no algebra at all: undo the subtraction (6 + 6 = 12), then undo the division (12 x 6 = 72). Any riddle that describes a chain of operations on a hidden number is easiest unwound from the end, because every arithmetic step has exactly one inverse.
Q3.A farmer keeps hens and roosters. He has 40 roosters, and that number is half of a quarter of the number of hens. How many hens are there?Tap to reveal
Answer: 320.
Half of a quarter is an eighth, so 40 = H/8 and H = 320. Combining the two fractions before you do anything else turns a two-step problem into a single multiplication, and it also removes the temptation to divide when you should multiply. Note that the roosters are not part of the hen count, so the farm holds 360 birds in total, which is what a follow-up question usually asks for.
Q4.Three cats eat three mice in an hour and a half. How long will ten cats take to eat twenty mice?Tap to reveal
Answer: Three hours.
Three cats and three mice is one mouse each, so a single cat takes 1.5 hours for a single mouse. Ten cats therefore finish ten mice in 1.5 hours, and twenty mice is two rounds of that: 3 hours. The trap is to treat the numbers as one big proportion and scale the time by 10/3 or 20/3. The rate per cat never changes, so it cancels out, and the only thing that matters is how many mice each cat has to get through.
Q5.The lilies on a lake double in number every day. On day 15 the lake is exactly half covered. On which day is it completely covered?Tap to reveal
Answer: Day 16.
One doubling takes half to whole, so the very next day finishes the job. The surprise comes from reading 'half covered' as 'halfway through in time', which is how doubling always fools people. Under this growth the lake was a quarter covered on day 14, an eighth on day 13, only about three percent on day 11, and under two percent on day 10, so almost the entire visible change happens in the last few days.
Q6.A relative of mine is 600 times younger than me. Can that be true?Tap to reveal
Answer: Yes.
Not in years, but nothing says years. Count in months: a newborn grandchild of one month against 600 months of my own, which is exactly 50 years. The riddle is a units trick rather than an arithmetic one, and 'times younger' is the tell - a ratio is only a ratio once you have fixed the unit it is measured in.
Q7.A farmer's geese are walking in single file. One goose leads three others, another follows three, and two walk in the middle. How many geese are there?Tap to reveal
Answer: Four.
Number them 1 to 4 along the line. Goose 1 has three behind it, goose 4 has three in front of it, and geese 2 and 3 are in the middle. Every clause in the riddle is satisfied at once by four birds. The answer of eight comes from assuming the three behind the leader must be different birds from the three ahead of the last one, which is the commonest error in counting riddles: describing the same objects twice and then adding the descriptions together.
Q8.You have three lemons. Two weigh the same and one is lighter. Using a pair of balance scales, what is the smallest number of weighings guaranteed to find the light one?Tap to reveal
Answer: One.
Put any two lemons on the scales. If one side rises, that is the light lemon. If they balance, the light one is the lemon you left off the scales. One weighing has three possible outcomes, left up, right up or level, and there are only three candidates, so a single weighing can carry enough information to settle it. Counting outcomes against candidates like this is the standard way to work out the floor for any coin-weighing puzzle before you go looking for a method.
Age and Family Riddles
Age riddles have one property that makes them unusually tractable: the gap between two people's ages never changes. If a father is 30 years older than his son today, he was 30 years older a decade ago and will be 30 years older a decade hence. Almost every age riddle is solved by writing that unchanging gap down and letting the changing ratio do the rest.
Family riddles work on a different mechanism. Words like mother, daughter and mother-in-law name relationships, not people, and one person can occupy two of them at once, which is why these puzzles so often produce an answer one smaller than expected.
The five below run from a one-line sum to a relationship you have to unpick from the inside out. If you get stuck on the last one, resolve the innermost phrase first and write down who that person is before reading the rest of the sentence.
Q1.Five years ago, a brother and sister's ages added up to 8. What will their two ages add up to five years from now?Tap to reveal
Answer: 28.
Each of them gains ten years over that decade, so the sum gains twenty: 8 + 20 = 28. You never learn either individual age, and you do not need to. The sum of two ages grows by two years for every year that passes, which is the shortcut. Anyone who tries to solve for the brother and sister separately ends up with two unknowns and one equation and wrongly concludes the riddle is broken.
Q2.My son is a third of my age. Five years ago he was a quarter of my age. How old am I?Tap to reveal
Answer: 45.
Let my age be x, so my son is x/3. Five years ago the statement reads x/3 - 5 = (x - 5)/4. Multiply everything by 12: 4x - 60 = 3x - 15, so x = 45. I am 45 and my son is 15; five years ago we were 40 and 10, which is the required quarter. The 30-year gap between us is the fixed quantity, and the changing ratio of 3 to 4 is what pins down where in our lives we are.
Q3.A father has seven sons, and each of the sons has one sister. How many children are in the family?Tap to reveal
Answer: Eight.
The seven boys are brothers, so they all share the same sister. One girl satisfies all seven statements, giving 7 + 1 = 8 children. The wrong answer of 14 treats 'each son has a sister' as seven separate girls, which is the double-counting error again, this time hidden inside the word each.
Q4.A room contains two mothers and two daughters, but only three people. How?Tap to reveal
Answer: Three women: a grandmother, her daughter, and her granddaughter.
The woman in the middle is counted twice, once as a mother and once as a daughter. Since family words describe relationships rather than people, one person can fill two of the four roles. The trick scales up in a tidy way: three mothers and three daughters can be four people, and n of each can be n + 1, as long as the family runs in a single line of descent.
Q5.A woman says of a man: 'I am not his wife, but his mother is my mother's mother-in-law.' What are the two of them to each other?Tap to reveal
Answer: He is her father, so they are father and daughter.
Work from the inside out. Her mother's mother-in-law is the mother of her mother's husband, which means her own father's mother. The man's mother is that same woman, so the man and her father have the same mother. The intended answer is that the man is her father. Strictly speaking a paternal uncle also fits the clue, which is why the original version of this riddle makes you choose from a fixed list, and the opening 'I am not his wife' is there to block the reading where the two are husband and wife.
Counting Riddles About Money, Animals and Weights
This is the oldest family of math riddle there is. Problem 79 of the Rhind Mathematical Papyrus, copied in Egypt around 1550 BC, is a list of houses, cats, mice and measures of grain that exists purely to be counted up, and the English nursery rhyme about meeting a man with seven wives on the way to St Ives is the same puzzle wearing different clothes. Fibonacci's Liber Abaci of 1202 is largely made of merchants, coins and livestock.
The reason they survive is that each sentence in them converts directly into a line of arithmetic. 'Four glasses and a jug weigh as much as 17 balls' is an equals sign with words around it. Two sentences give two equations and settle two unknowns, and when there are more unknowns than equations, the requirement that animals and coins come in whole numbers usually does the remaining work.
Three techniques cover almost everything in this round. First, add rates rather than times when several workers share a job. Second, assume everything is the cheap kind and count the surplus, which kills any legs-and-heads problem without algebra. Third, total the whole thing and check it divides before you try to split it.
Q1.Four glasses and a jug together weigh the same as 17 iron balls. The jug on its own weighs the same as one glass plus seven balls. How many balls does the jug weigh?Tap to reveal
Answer: Nine balls.
Write it as 4g + j = 17b and j = g + 7b, then substitute: 4g + g + 7b = 17b, so 5g = 10b and one glass weighs two balls. That gives j = 2b + 7b = 9b. Check: four glasses are 8 balls, plus a jug at 9, which is 17. This is a pair of simultaneous equations with the equals signs written out as 'weighs as much as', and substitution is easier than elimination here because the second sentence already has the jug on its own.
Q2.A lion can finish a sheep in 2 hours, a wolf in 3 hours and a dog in 6 hours. How long do the three of them take together?Tap to reveal
Answer: One hour.
Add rates, never times. In one hour the lion gets through 1/2 of the sheep, the wolf 1/3 and the dog 1/6, and 1/2 + 1/3 + 1/6 = 1, which is exactly the whole job, so the answer is one hour. Adding the times, or averaging them to about 3.7 hours, is the standard mistake. There is a sanity check that catches it every time: a group is always faster than its fastest member, so the answer had to come out below 2 hours.
Q3.Can seven identical loaves of bread be divided equally among twelve people, with no weighing and no leftovers?Tap to reveal
Answer: Yes. Each person gets 7/12 of a loaf.
Cut three loaves into quarters, which gives twelve quarters, and the other four into thirds, which gives twelve thirds. Hand everyone one quarter and one third: 1/4 + 1/3 = 3/12 + 4/12 = 7/12, and twelve shares of 7/12 make exactly 7 loaves. The move is to split the awkward fraction 7/12 into pieces whose denominators divide 12 cleanly. Egyptian arithmetic did precisely this as a matter of course, writing fractions as sums of unit fractions.
Q4.A peasant takes two baskets of 30 eggs each to market. He means to sell the small eggs at three for a dollar and the large ones at two for a dollar. At the market he changes his mind, mixes all 60 eggs and sells them five for two dollars. Did he do better?Tap to reveal
Answer: No. He lost a dollar.
Sold separately: 30 divided by 3 is 10 sales at a dollar, plus 30 divided by 2 is 15 sales at a dollar, which is $25. Mixed: 60 divided by 5 is 12 lots at two dollars, which is $24. The missing dollar has an exact source. Five for two dollars is 40 cents an egg, but his real mixture is half the eggs at 33.3 cents and half at 50 cents, which averages 41.7 cents. The gap of about 1.7 cents across 60 eggs is precisely one dollar. Bundling three-for-a-dollar with two-for-a-dollar looks like adding the two prices, when a fair blend has to average the price per egg.
Q5.Three men visit a barber. He tells each of them in turn to put into the drawer as much money as he finds in it, then take two dollars out. After the third man the drawer is empty. How much was in it at the start?Tap to reveal
Answer: $1.75.
Each visit turns x into 2x - 2. Three visits give 8x - 14, and that is zero, so x = 1.75. Running it backwards is safer and needs no expansion: before the last man took his two dollars the drawer held $2, which was double what he found, so he found $1; before him the drawer held $1 + $2 = $3, double $1.50; before him $1.50 + $2 = $3.50, double $1.75. Each backwards step is one addition and one halving, which is much harder to get wrong than multiplying out three nested brackets.
Q6.A cowhand looks after 100 animals made up of oxen, cows and calves, and is paid exactly $100 in total: $10 for each ox, $5 for each cow and 50 cents for each calf. How many of each are there?Tap to reveal
Answer: 1 ox, 9 cows and 90 calves.
There are two equations and three unknowns, so whole numbers have to finish the job. From o + c + f = 100 and 10o + 5c + 0.5f = 100, double the second and subtract the first: 19o + 9c = 100. That means c = (100 - 19o)/9, which is a whole number only when o = 1, giving c = 9 and f = 90. Check the money: $10 + $45 + $45 = $100. Puzzles of this shape are called Diophantine, after the Greek mathematician Diophantus, and the pattern is always the same: too few equations, rescued by insisting that the answers cannot be fractions.
Q7.On a meadow there are horses and children. Count all the legs and you get 74. Count all the heads and you get 22. How many of each?Tap to reveal
Answer: 7 children and 15 horses.
Pretend all 22 are children: that would be 44 legs. There are 74, an excess of 30, and every horse you swap in adds two legs beyond a child's two, so 30 divided by 2 is 15 horses, leaving 7 children. Check: 15 x 4 + 7 x 2 = 60 + 14 = 74. The assume-and-adjust method solves every version of this puzzle, including the ones with spiders and beetles, without writing down a single equation.
Q8.Some donkeys are resting in a meadow. A group of boys arrives and each boy climbs onto a donkey, but one boy is left standing. They try again with two boys to a donkey, and this time one donkey is left empty. How many boys and how many donkeys?Tap to reveal
Answer: 4 boys and 3 donkeys.
The first arrangement says boys = donkeys + 1. The second says the boys fill half their number of donkeys and one is spare, so donkeys = boys/2 + 1. Setting the two together: b = (b/2 + 1) + 1, so b/2 = 2 and b = 4, which leaves 3 donkeys. Check both scenes: four boys on three donkeys leaves one boy on foot, and in pairs they occupy two donkeys and leave one empty.
Q9.A club has two dance floors. If one person moves from the first floor to the second, both floors hold the same number. If instead one person moves from the second floor to the first, the first holds twice as many as the second. How many are on each floor?Tap to reveal
Answer: 7 on the first floor and 5 on the second.
The two scenes give a - 1 = b + 1 and a + 1 = 2(b - 1). The first says a = b + 2; substituting into the second, b + 3 = 2b - 2, so b = 5 and a = 7. Check both moves: 6 and 6, then 8 and 4. Every sentence about moving one person is worth one equation, so two such sentences are exactly enough to fix two unknowns, and a riddle that gives three is either redundant or inconsistent.
Q10.A farmer has nine cows yielding 1, 2, 3, 4, 5, 6, 7, 8 and 9 gallons of milk. He wants to give each of his three sons the same number of cows and the same total yield. Can it be done?Tap to reveal
Answer: Yes: 1+5+9, 2+6+7 and 3+4+8 give three cows and 15 gallons each.
Total the milk first: 1 + 2 + ... + 9 = 45, and 45 divided by 3 is 15, so the target is 15 gallons per son. If the total had not divided by three the answer would be no before you tried a single arrangement, which is why totalling first is the habit worth taking from this riddle. Then it is quick: the middle cow, 5, needs 1 and 9; the 2 needs 6 and 7; and what is left, 3, 4 and 8, also makes 15.
Number Sequence Riddles
A sequence riddle asks you to find the rule, not the number, and the number follows once the rule is right. Two habits find almost every rule going. Write out the differences between consecutive terms, and if that shows nothing, write out the ratios. Steady differences mean addition, steady ratios mean multiplication, and a pattern that alternates between the two means the sequence is running two rules in turn.
That alternating kind is the backbone of this round, and it is what makes these harder than school sequences. A list like 7, 12, 24, 29, 58 looks random because no single rule fits it: add five, then double, then add five again. The giveaway is a sequence that lurches, with small steps and sudden jumps taking turns. When two rules will not fit, try three, and always test the cycle all the way to the end of the list before you trust it.
One caveat worth stating plainly, because most sequence puzzles quietly pretend otherwise: no finite list of numbers has a single mathematically forced continuation. There is always some more elaborate rule that fits the terms you were given and then does something else. The answer being asked for is the simplest rule that fits every gap, and that is the standard these follow.
Q1.Which number is missing? 3, 6, 7, 14, 15, ___, 31, 62, 63, 126, 127Tap to reveal
Answer: 30.
Two rules take turns: double, then add one. 3 x 2 = 6, 6 + 1 = 7, 7 x 2 = 14, 14 + 1 = 15, 15 x 2 = 30, 30 + 1 = 31, and the pattern runs on to 127. There is a second way to see it: every other term, 3, 7, 15, 31, 63, 127, is one less than a power of two, so the gap has to sit between 15 and 31 as 15 doubled.
Q2.What is the missing number? 4, 20, 24, ___, 44, 60, 64Tap to reveal
Answer: 40.
Add 16, then add 4, alternately: 4 + 16 = 20, 20 + 4 = 24, 24 + 16 = 40, 40 + 4 = 44, 44 + 16 = 60, 60 + 4 = 64. Everything here is addition, so the differences are where to look. Write them out as 16, 4, then a gap, then 16, 4, and the missing difference is obviously 16, which fills the gap without any thought about the numbers themselves.
Q3.Find the missing number: 6, 18, 36, ___, 216, 648, 1296Tap to reveal
Answer: 108.
Multiply by 3, then by 2, in turn: 6 x 3 = 18, 18 x 2 = 36, 36 x 3 = 108, 108 x 2 = 216, 216 x 3 = 648, 648 x 2 = 1296. Since it is all multiplication, divide each term by the one before to get the ratios 3, 2, 3, 2, 3, 2. Each pair of steps multiplies by 6, which gives a fast check: 6, 36, 216, 1296 are the first four powers of six.
Q4.What number is missing? 11, 16, 32, ___, 74, 79, 158Tap to reveal
Answer: 37.
Add 5, then double: 11 + 5 = 16, 16 x 2 = 32, 32 + 5 = 37, 37 x 2 = 74, 74 + 5 = 79, 79 x 2 = 158. There is a shortcut that skips the rule entirely. The term after the gap, 74, is even and looks like a double, and 32 is not half of it while 37 is. That single observation pins the answer, and the rest of the sequence only confirms it.
Q5.Find the missing number: 10, 20, 100, ___, 280, 560, 640Tap to reveal
Answer: 200.
Double, then add 80: 10 x 2 = 20, 20 + 80 = 100, 100 x 2 = 200, 200 + 80 = 280, 280 x 2 = 560, 560 + 80 = 640. The clean way in is from the right-hand end, undoing each rule: 640 - 80 = 560, then 560 halved is 280, then 280 - 80 = 200. Two steps backwards land on the gap, and backwards is usually easier because the large numbers at the end make the rule obvious.
Q6.What is the missing number? 8, 16, 28, 44, ___, 88, 116, 148, 184Tap to reveal
Answer: 64.
This one has a single rule, but a growing one. The steps are 8, 12, 16, 20, 24, 28, 32, 36, each four more than the last, so 44 + 20 = 64 and then 64 + 24 = 88. When the differences form their own arithmetic run, the sequence is quadratic: every term here is 2n squared + 6n + 8, counting n from zero. That is also why no pair of alternating rules will fit it, and recognising a quadratic early saves a lot of wasted guessing.
Q7.Guess the missing number: 3, 6, 18, 36, ___, 216, 234, 468, 1404Tap to reveal
Answer: 72.
Three rules in rotation: double, triple, add 18. So 3 x 2 = 6, 6 x 3 = 18, 18 + 18 = 36, 36 x 2 = 72, 72 x 3 = 216, 216 + 18 = 234, 234 x 2 = 468, 468 x 3 = 1404. Two alternating rules cannot fit both 18 to 36 and 36 to 216, and that failure is the signal to try a cycle of three. Always run a three-rule cycle to the end of the list, since a longer cycle fits more easily by accident.
Hard Math Riddles Where the Trick Matters More Than the Arithmetic
What makes a math riddle hard for an adult is not more arithmetic. Extra steps make a puzzle longer, not harder. Hard riddles work by producing a wrong answer instantly and making it feel obviously right, so you have to notice you were caught and reverse. Two of the twelve below are built that way: one buries a decoy quantity that turns out to lie behind you, and one invites you to buy back lost time with speed.
The other reliable source of difficulty is a shift of level. The tribe whose chief's headcount refuses to add up is not an arithmetic question at all, it is a question about which base the digits are written in. The city with no two people sharing a hair count is a question about how many labels exist rather than about hair. Once you notice which level the question is really on, the sums are trivial.
The twelve below are the hard round, and several are worth reading even if you already know the answer, because the reason is better than the answer. The wine-drop puzzle in particular can be settled in one sentence of bookkeeping, and the algebra underneath it is only there to reassure the sceptical.
Q1.Your train leaves in 2 minutes and the station is 2 miles away. You drive the first mile at 30 miles per hour. How fast do you have to drive the second mile to catch the train?Tap to reveal
Answer: No speed will do it. The train is already gone.
At 30 miles per hour a mile takes exactly two minutes, so the whole allowance is spent at the halfway point and the second mile would have to be covered in no time at all. In general, to average 60 miles per hour over two miles you need two minutes in total, and any speed below 60 on the first mile uses up more than half of that. The underlying rule is the one people forget under pressure: average speed is total distance divided by total time, and it is never the average of the two speeds.
Q2.A worm chews through 1 mm of paper a day. Two volumes stand side by side on a shelf, each with 4 cm of pages between covers 2 mm thick. How long does the worm take to get from the first page of volume one to the last page of volume two?Tap to reveal
Answer: Four days.
Shelve the two volumes in reading order with the spines facing out, and volume one's front cover ends up facing volume two's back cover. The first page of volume one sits just inside that front cover, and the last page of volume two sits just inside that back cover, so the worm crosses two covers and nothing else: 2 mm + 2 mm = 4 mm at 1 mm a day. The 4 cm of paper in each volume is a decoy, because all of it lies on the far side of the starting point. The answer of 84 days comes from adding both volumes' pages and the two covers between them without ever asking which way the books face.
Q3.A driver sees 15951 miles on the odometer and notices it reads the same in both directions. Two hours later the odometer again shows a number with that property. What was the car's average speed?Tap to reveal
Answer: 55 miles per hour.
The next palindrome after 15951 is 16061. The middle digit is already a 9, so it has to roll over, and that forces the second and fourth digits to move up together. That is 110 miles in two hours, so 55 miles per hour. It is worth checking nothing sits in between: a five-digit palindrome starting 159 has to end 951, and 15951 is the only one, so the next candidate must start 160.
Q4.Find a number that leaves no remainder when divided by 5, but leaves a remainder of 1 when divided by 2, 3 or 4.Tap to reveal
Answer: 25, and after that 85, 145, and every 60 more.
Leaving remainder 1 on division by 2, 3 and 4 means the number is one more than a multiple of 12, since 12 is the least common multiple of those three: 13, 25, 37, 49, 61 and so on. Of those, the ones divisible by 5 must end in 0 or 5, and 25 is the first. The whole family repeats every 60, because 60 is the least common multiple of 12 and 5. This is the Chinese remainder theorem in miniature, and the practical lesson is to start from the most restrictive condition and step through it rather than testing numbers one by one.
Q5.In a certain city no two people have the same number of hairs on their head, and nobody has more than 517 hairs. What is the largest number of people the city can hold?Tap to reveal
Answer: 518.
The available hair counts are 0, 1, 2 and so on up to 517, which is 518 different values, and each can be used by only one person. Forgetting that zero is allowed is the usual slip and produces 517. Read the other way round, this is the pigeonhole principle, often called Dirichlet's drawer principle: in any such city of 519 people, two of them would have to match.
Q6.A tribe counts using the hands of its deity. The chief says there are 16 women and 13 men, 32 people in all. How many hands does the deity have?Tap to reveal
Answer: Seven.
The sum is wrong in base ten, since 16 + 13 = 29, so the digits cannot be base ten. In base b, '16' means b + 6, '13' means b + 3 and '32' means 3b + 2. So (b + 6) + (b + 3) = 3b + 2, which gives b = 7, and the digit 6 confirms it since a base-7 system uses digits 0 to 6. In our numerals the tribe has 13 women and 10 men, 23 people. Counting bases really do follow anatomy, from base ten for fingers to the base-twenty system the Maya used, so a seven-handed deity plausibly yields a seven-based system.
Q7.In one of these pairs, subtracting one number from the other gives the same result as dividing one by the other. Which pair: 24/6 and 24/8; 3 and 2/12; 30/6 and 36/6; or 25/4 and 5/4?Tap to reveal
Answer: 25/4 and 5/4.
25/4 - 5/4 = 20/4 = 5, and 25/4 divided by 5/4 = 25/5 = 5. Pairs like this are not scarce. If a - b = a/b then a = b squared divided by (b - 1), so any b other than 1 generates a partner. Put b = 5/4 and you get a = 25/4, which is where this pair comes from; put b = 2 and you get the tidy pair 4 and 2, where the difference and the quotient are both 2.
Q8.A pan browns two slices of bread at a time and each side takes half a minute. What is the shortest time to brown three slices on both sides?Tap to reveal
Answer: One and a half minutes.
Six sides need browning and the pan handles two at a time, so three half-minute rounds is the theoretical floor, and the only question is whether a schedule exists that never wastes a space. It does. Round one: slices A and B, first sides. Round two: turn A over, take B out, put C in. Round three: the second sides of B and C. The pan is full in all three rounds. Finding the floor first and then hunting for a schedule that hits it is how these scheduling puzzles are meant to be attacked.
Q9.A rectangular room contains ten chairs. Can all ten be stood against the walls so that each of the four walls has the same number of chairs along it?Tap to reveal
Answer: Yes, three against every wall.
Put one chair in each of two opposite corners, since a corner chair touches two walls and counts for both. Then stand two more against each wall, which uses the other eight. Every wall now has its own two plus exactly one corner chair, because opposite corners share themselves out one per wall, giving 2 + 1 = 3. The arithmetic explains why it works: ten chairs give twelve wall-slots when two of them are doubled, and 12 divides by 4. Using all four corners would give fourteen slots, which does not divide by four, so counting the slots first tells you which arrangement can possibly succeed.
Q10.A square yard has an apple tree at each corner. The owner doubles the yard's area and keeps it square. Can all four trees still stand on the fence line?Tap to reveal
Answer: Yes.
Turn the new square through 45 degrees so that each of the four old corners sits at the midpoint of one of its sides. A square built that way has sides equal to the diagonal of the old square, so its area is exactly double. That is not a coincidence, it is the fact that the square drawn on a diagonal has twice the area of the original, which is the construction Socrates draws out of a slave boy in Plato's Meno. The new yard covers every bit of the old one, meeting the old boundary only at the four trees, which now sit at the midpoints of the fence.
Q11.There is a bottle of white wine and a bottle of red. One drop of red is put into the white bottle and mixed in, then one drop of the new mixture is put back into the red bottle. Which is larger: the red now in the white bottle, or the white now in the red bottle?Tap to reveal
Answer: They are exactly equal.
Each bottle ends with the volume it started with, one drop out and one drop in. So whatever red is missing from the red bottle has been replaced by exactly that much white, and the red that is missing is precisely the red now sitting in the white bottle. The two quantities are equal by bookkeeping alone, and no amount of stirring, or any drop size, changes it. The algebra agrees: if the white bottle holds q drops, the drop returned is 1/(q+1) red and q/(q+1) white, so the red left behind is 1 - 1/(q+1) = q/(q+1), which is exactly the white that crossed over.
Q12.Three men of different heights are stuck in a pit. They stand on each other's shoulders, but the shortest man, who is on top, is still 5 cm below the edge. With no help and no tools, can they get out?Tap to reveal
Answer: Yes. Restack with the tallest man on top.
Stand three men on each other's shoulders and the crown of the top man lands at the sum of their three heights minus two shoulder-to-crown gaps. That gap is much the same for all three, so the pile is the same height whatever the order - what changes is the reach above the crown, and the tallest man has the longest arms. Put him on top: the pile is no higher, the hands at the top are, and 5 cm is covered. The only thing left to choose is which man's reach ends up on the outside.
Frequently Asked Questions
What is the hardest math riddle on this page?
The wine-drop puzzle, on the reasoning rather than the arithmetic. Almost everyone assumes the answer depends on how well the first drop was mixed in, and it does not: because each bottle finishes with the volume it started with, the red missing from one bottle must have been replaced by exactly that much white. The train riddle is the most reliably annoying one, since the correct answer is that no speed works at all, and people keep looking for a number. Neither needs anything past school arithmetic, which is what makes them hard rather than long.
How do you solve a math riddle without using algebra?
Three methods cover most of them. Work backwards from the end, undoing each step, which is how the barber's drawer and the grandfather's age fall over in two lines. Assume everything is the cheaper or simpler kind and count the surplus, which solves any legs-and-heads problem instantly. And total everything before you try to split it, since a total that does not divide tells you the answer is no before you attempt an arrangement. Algebra is a fallback for when a riddle has two genuinely independent unknowns.
Which of these math riddles work for children?
The warm-up round is the one to use with primary-age children, along with most of the age and family riddles: each has a single hinge and a wrong answer you can explain in one sentence. The counting round suits roughly ages ten and up, because it needs fractions or a little bookkeeping. The first five number sequences work as soon as a child can double and add confidently, and they are the best of the lot for a classroom, since the rule can be checked out loud by the whole room; the last two in that round, where the steps grow and then three rules rotate, are a step up. Save the hard round for adults or for a strong older group.
What is the difference between a math riddle and a word problem?
A word problem is honest: it tells you what it wants and the difficulty lives in the calculation. A riddle is built so that a wrong answer arrives first and feels right, and the difficulty lives in noticing. The pot that costs a dollar plus half its price, the lake that is half covered on day fifteen, and the two-mile drive to the station are all trivial to compute once you have read them correctly, which is why the working matters more than the number.
How do I solve number sequence riddles?
Write out the differences between consecutive terms first. If those are constant, the rule is addition; if they form their own steady run, the sequence is quadratic. If the differences look random, write out the ratios instead. When neither gives a single rule, assume two rules are alternating, which is what the sequences in the fourth round do, and if two will not fit, try a cycle of three. One honest caveat: no finite list has a single forced continuation, so what a sequence puzzle actually asks for is the simplest rule that fits every gap.
Why does every answer come with the working?
Because the working is the part that gets asked about and the part most lists leave out. Knowing that the eggs lose a dollar because 40 cents an egg is not the average of 33.3 and 50, or that the bookworm only meets two covers once you notice which way books face on a shelf, is more useful than the answer itself. Writing out the reason also keeps a riddle honest. Several riddles were dropped from this page while the explanations were being written, because the stated answer turned out not to follow.
How many riddles should I use in one round?
Fewer than the length of this page suggests. Eight to twelve is right for a session of twenty-five minutes or so if you read the working aloud, and about six is plenty at a dinner table where people are also talking. Pick one round rather than skimming across all five, because mixing warm-ups with the hard set tends to leave half the group idle. Read the riddle, wait through the silence, and only then read the explanation.
