Alex hangs a poster on his kitchen wall. The wall has white and grey tiles of the same size (see picture). How many grey tiles are completely covered by the poster?
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Answer: B — 21
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Hint 1 of 3
A tile counts only if the poster hides ALL of it; skip any tile the poster just clips at the edge.
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Hint 2 of 3
Mark the block of whole tiles that sit completely under the poster.
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Hint 3 of 3
Among only those fully-hidden tiles, count the grey ones (the white ones don't matter).
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Approach: find the tiles completely under the poster, then tally just the grey ones
First outline the tiles that the poster covers completely, ignoring tiles it only partly overlaps.
These fully-hidden tiles form a grey-and-white checkerboard block.
Now count just the grey tiles inside that block, one by one.
Leonie has one stamp for each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. She uses them to stamp the date of the kangaroo competition (see picture). How many of the stamps does Leonie use?
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Answer: B — 6
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Hint 1 of 2
Write out the whole date as it is stamped: 15 03 2018.
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Hint 2 of 2
She has only one stamp of each digit, so a digit that shows up twice still uses just one stamp — count the different digits.
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Approach: write the date and circle the different digits, counting each kind once
The date is stamped as 1 5 0 3 2 0 1 8.
Cross out repeats: the 0 and the 1 each appear twice, but she only owns one stamp of each.
The different digits are 0, 1, 2, 3, 5, 8 — that is 6 stamps, answer B.
Barbara wrote the word MATHEMATIC on a piece of paper. She used the same colour for letters that are the same, and a different colour for letters that are different. How many different colours did she use?
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Answer: A — 7
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Hint 1 of 2
Write out the letters of MATHEMATIC and circle the ones that repeat.
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Hint 2 of 2
Count how many different letters appear, not how many letters there are.
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Approach: count the distinct letters
MATHEMATIC uses the letters M, A, T, H, E, M, A, T, I, C.
The repeats are the second M, the second A and the second T.
The different letters are M, A, T, H, E, I, C — that is 7.
60 children stand in a row. Each child wears a high-visibility vest and a backpack. The vest colours always alternate: yellow, green, yellow, green, … The backpack colours follow the pattern red, brown, purple, red, brown, purple, … How many children have both a yellow vest and a purple backpack?
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Answer: E — 10
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Hint 1 of 2
Yellow vests are on odd positions; purple backpacks repeat every third child.
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Hint 2 of 2
Find positions that are both odd and a multiple of 3 — that's every 6th child starting at 3.
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Approach: combine the two repeating patterns
Vests go yellow, green, ..., so yellow is on positions 1, 3, 5, ... (odd).
Backpacks go red, brown, purple, ..., so purple is on positions 3, 6, 9, ...
Both happen at positions 3, 9, 15, ..., i.e. every 6th child starting at 3.
There are exactly 2 frogs in each row and in each column (see picture). At the same moment, two of the six frogs each hop to an empty neighbouring square. Afterwards there are again 2 frogs in each row and in each column. In how many ways can two frogs hop like this?
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Answer: D — 4
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Hint 1 of 3
There are only three empty squares, so each hopping frog must land on one of those empty squares.
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Hint 2 of 3
When a frog leaves a row (or column) and another frog must keep that row (or column) at two, the two moves have to balance each other.
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Hint 3 of 3
Carefully try every pair of frogs that can hop into empty neighbours, and keep only the pairs that still leave two frogs in every row and every column.
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Approach: try each pair of frogs hopping into empty neighbours and keep the ones that stay balanced
On the 3-by-3 grid the six frogs leave exactly three empty squares; a hopping frog can only move onto an empty neighbour.
If one frog hops out of a row, the row drops to one frog, so a second frog must hop back into that same row to keep it at two — and the same must hold for columns.
Go through the pairs of frogs that can both hop into empty neighbours and check which pairs keep every row and every column at two frogs.
Exactly four such pairs of hops work, so the answer is 4 (D).
Three boys enter a room one after the other. Hermann is not the first. Felix is not the second. Clemens is not the third. How many different orders are there for the boys to enter the room?
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Answer: B — 2
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Hint 1 of 2
List all six orders of the three boys and cross out the forbidden ones.
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Hint 2 of 2
Hermann cannot be 1st, Felix cannot be 2nd, Clemens cannot be 3rd.
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Approach: enumerate the orders and discard those breaking a rule
There are 6 possible orders of three boys.
Keeping only those with Hermann not first, Felix not second, and Clemens not third leaves Felix-Clemens-Hermann and Clemens-Hermann-Felix.
A small zoo has a giraffe, an elephant, a lion and a turtle. Susi wants to visit exactly two of the animals today but does not want to start with the lion. How many different possibilities does she have, to visit the two animals one after the other?
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Answer: D — 9
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Hint 1 of 2
Count ordered visits of two different animals, then remove the forbidden starts.
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Hint 2 of 2
There are 4 x 3 ordered pairs; throw out the ones beginning with the lion.
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Approach: count ordered pairs, then subtract those starting with the lion
Choosing two animals in order gives 4 x 3 = 12 possibilities.
Of these, the ones starting with the lion number 1 x 3 = 3.
Thomas drew a pig and a shark. He cuts each animal into three pieces. Then he takes one of the two heads, one of the two middle sections and one of the two tails and lays them together to make another animal. How many different animals can he make in this way?
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Answer: E — 8
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Hint 1 of 2
A new animal needs a head, a middle and a tail, and there are two choices for each part.
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Hint 2 of 2
Multiply the number of choices for the three parts.
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Approach: multiply the choices for each of the three parts
There are 2 heads to choose from, 2 middle sections, and 2 tails.
Each new animal is one choice for each part, so 2 × 2 × 2 = 8 animals can be built.
The black diamonds ◆ and white diamonds ◇ follow a fixed pattern. The first 3 levels are shown on the right. Each level (from the 2nd level on) has one more row than the level before. In every level, the two outermost diamonds of the last row are white, and all the other diamonds are black. How many black diamonds are there in level 6?
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Answer: C — 26
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Hint 1 of 3
Each level is a little triangle of rows: 1 diamond, then 2, then 3, and so on down.
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Hint 2 of 3
Count ALL the diamonds in level 6 first, then take away the white ones.
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Hint 3 of 3
In every level only the two ends of the very bottom row are white.
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Approach: count all the diamonds, then take away the two white ones
Level 6 has 6 + 1 = 7 rows, with 1, 2, 3, 4, 5, 6, 7 diamonds in them.
Anna, Laura, Lisa and Katharina wanted to take a photo together. Anna and Katharina are best friends and wanted to stand next to each other. Lisa also wanted to stand next to Anna. In how many different ways can the photo be taken, if all their wishes are met?
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Answer: B — 4
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Hint 1 of 2
Anna must stand next to both Katharina and Lisa, so Anna is in the middle of those two.
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Hint 2 of 2
Treat Katharina-Anna-Lisa as one block and place Laura at either end.
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Approach: bundle the friends who must be adjacent
Anna is next to Katharina and next to Lisa, so the order around Anna is Katharina-Anna-Lisa (or its reverse): 2 ways.
This block of three can have Laura at the left end or the right end: 2 ways.
You can place together the cards pictured to make different three‑digit numbers, for instance 989 or 986. How many different three‑digit numbers can you make with these cards?
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Answer: E — 12
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Hint 1 of 3
Two of the cards are the kind that show a 6 one way and a 9 when you turn them upside down; the 8 looks the same either way.
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Hint 2 of 3
First decide which of the three spots the 8 sits in, then fill the other two spots.
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Hint 3 of 3
For each place the 8 can go, the two leftover spots can each be a 6 or a 9.
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Approach: place the 8, then fill the other two spots with 6 or 9
There are three cards: two flip-cards that each show a 6 or a 9, and one 8-card that looks the same whichever way up it is.
Pick where the 8 goes: it can be the first, middle, or last digit — that is 3 choices.
For each of those, the two remaining spots can each be a 6 or a 9, giving 4 numbers per choice: for example with 8 in front you get 866, 869, 896, 899.
So the count is 3 groups of 4, which is 3 × 4 = 12 different numbers, answer E.
Lucas has these five puzzle pieces (shown on the right): a smiling head, a banana-tail, and three middle pieces. They snap together only where a bump fits into a notch. He wants to make a caterpillar with a head, a tail, and 1, 2 or 3 pieces in between. How many different caterpillars can Lucas build?
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Answer: B — 4
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Hint 1 of 3
The head only joins on one side and the tail only joins on one side; the pieces fit only where a bump (tab) meets a notch (socket).
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Hint 2 of 3
A caterpillar is head, then 1, 2, or 3 middle pieces, then tail — build the chains by matching bumps to notches.
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Hint 3 of 3
Go case by case: count the legal chains with exactly 1 middle piece, then 2, then 3, and add them up.
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Approach: match the connectors (bump-to-notch) and count the legal chains of 1, 2, or 3 middle pieces
A piece can join another only where a bump fits into a notch, and the head and tail each connect on just one side.
Try 1 middle piece between head and tail: only the pieces whose bumps and notches line up both ways work.
Now try 2 middle pieces, then 3 middle pieces, keeping every join a bump-into-notch fit.
Adding up all the chains that connect properly gives 4 (B) different caterpillars.
Anna has four discs of different sizes. She wants to build a tower using 3 discs. A smaller disc always has to lie on top of a bigger disc. How many ways are there for Anna to build this tower?
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Answer: C — 4
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Hint 1 of 2
A tower uses 3 of the 4 discs, and the sizes force their order.
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Hint 2 of 2
So really you are just choosing which one disc to leave out.
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Approach: each choice of 3 discs has exactly one legal stacking
Once three discs are chosen, they must go largest-on-bottom, so the order is fixed.
Thus the number of towers equals the number of ways to pick 3 discs from 4.
That is the same as choosing which disc to omit: 4 ways, so the answer is 4.
Aladdin’s carpet is a square. Along each edge there are two rows of dots (see diagram), and each edge has the same number of dots. How many dots does the carpet have in total?
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Answer: A — 32
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Hint 1 of 3
The dots make two square loops — a big loop on the outside and a smaller loop just inside it.
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Hint 2 of 3
Count the dots on one side of a loop, but be careful: the corner dots belong to two sides, so don't count them twice.
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Hint 3 of 3
Add up the big loop and the small loop.
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Approach: count the big square loop and the small square loop of dots
The dots make two square loops, one just inside the other, with the same number on every side.
The big loop has 6 dots along each side; counting around it (corners only once) gives 4 × 6 − 4 = 20 dots.
The small loop has 4 dots along each side, giving 4 × 4 − 4 = 12 dots.
Cynthia paints each region of the figure in a single colour: red, blue or yellow. Regions that touch each other must be painted different colours. In how many different ways can Cynthia paint the figure?
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Answer: E — 6
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Hint 1 of 3
Find which regions actually share a border, since only touching regions are forced to be different colours.
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Hint 2 of 3
Colour one region first, then count how many free colour choices are left for each region next to it.
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Hint 3 of 3
Multiply the number of choices region by region as you work inward.
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Approach: colour the regions one at a time and multiply the free choices
Colour the outermost region first: there are 3 colours to pick from.
Each region just inside it touches the one outside, so it must use a different colour, which usually leaves only a small number of choices.
Multiplying the choices region by region as you move inward gives 6 different ways to paint the whole figure, choice E.
A mouse wants to escape a labyrinth (see picture). On her way out she is only allowed to go through each opening once at most. How many different ways can the mouse choose to get outside?
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Answer: D — 6
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Hint 1 of 2
Trace each route from the mouse to the outside, never reusing an opening.
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Hint 2 of 2
Count the separate paths one by one and don't repeat any.
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Approach: trace and count the distinct escape routes
Start at the mouse and follow the openings outward, using each opening at most once.
List every different route that reaches the outside.
Joseph has got a toy car, a teddy bear, a ball and a ship. He wants to put them in a new order on the shelf. The ship must be next to the car, and the teddy bear should also be next to the car. In how many different orders can he put the toys on the shelf?
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Answer: B — 4
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Hint 1 of 2
The car must touch both the ship and the teddy, so the car sits between them.
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Hint 2 of 2
Treat ship-car-teddy as one block and place the ball at either end.
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Approach: form the forced block, then place the remaining toy
The ship and the teddy both must be next to the car, so the car is in the middle of a block: ship–car–teddy.
That block can be ordered 2 ways (ship–car–teddy or teddy–car–ship).
The ball goes at the left end or the right end of the block: 2 choices.
Two of the 5 ladybirds in the picture are always friends with each other if the difference between their number of dots is exactly 1. Today every ladybird has sent an SMS to each of their friends. How many SMS messages were sent?
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Answer: C — 6
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Hint 1 of 2
First count the spots on each ladybird, then pair up those that differ by exactly one spot.
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Hint 2 of 2
Each friendship means two messages, since each friend texts the other.
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Approach: find friend pairs, then count messages both ways
The ladybirds carry 2, 3, 3, 5 and 6 spots.
Friends differ by exactly 1 spot: the 2-spot is friends with each 3-spot, and the 5-spot is friends with the 6-spot — 3 friendships in all.
In each friendship both friends send a message, so each pair accounts for 2 messages.
Katja throws darts at the target shown on the right. If she does not hit the target she gets no points. She throws twice and adds her points. What can her total not be?
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Answer: D — 90
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Hint 1 of 3
Each single throw can only score one of the numbers on the target (or 0 for a miss).
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Hint 2 of 3
Try adding two of those numbers together in every way you can.
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Hint 3 of 3
Check each answer choice to find the one total you can never build.
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Approach: add the two throws in every way and find the missing total
One throw scores 0, 30, 50 or 70, and the total is two throws added.
Gerhard has the same number of white, grey and black counters. He has thrown some of these round pieces together onto a pile. All the pieces he used can be seen in the picture. He has, however, got 5 counters left that will not stay on the pile. How many black counters did he have to begin with?
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Answer: B — 6
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Hint 1 of 3
He began with the same number of white, grey and black, so think of them in equal groups.
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Hint 2 of 3
Count the counters on the pile, colour by colour, from the picture.
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Hint 3 of 3
The 5 left over are the extras that did not fit, so add them back to find each starting group.
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Approach: count the pile by colour, then add back the leftovers to make equal groups
Count how many white, grey and black counters are actually on the pile in the picture.
He started with the same number of each colour, and 5 counters were left over that did not stay on.
Sharing everything back into three equal colour groups, each group had 6 counters.
Grandmother baked 20 ginger biscuits for her grandchildren. She decorated them with raisins and nuts. First she decorated 15 with raisins, and then 15 with nuts. No biscuit was left plain. How many biscuits were decorated with both raisins and nuts?
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Answer: E — 10
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Hint 1 of 2
Adding the raisin biscuits and the nut biscuits double-counts the ones with both.
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Hint 2 of 2
The overlap is (15 + 15) - 20.
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Approach: overlap of two groups
15 biscuits got raisins and 15 got nuts, a total of 15 + 15 = 30 'decorations'.
But there are only 20 biscuits and none was left plain, so the extra 30 - 20 = 10 decorations come from biscuits counted twice.
Joana has several sheets of paper, each with a drawing of a parrot. She wants to paint only the head, tail and wing of the parrot, using red, blue or green. The head and tail may be the same colour, but the wing must not be the same colour as the head or the tail. How many sheets can she paint so that no two parrots are painted the same way?
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Answer: D — 12
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Hint 1 of 2
Pick colours for head, tail and wing in turn, remembering the wing's restriction.
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Hint 2 of 2
Split into 'head and tail same colour' versus 'head and tail different'.
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Approach: count by cases on the head/tail colours
If head and tail share a colour (3 ways), the wing has 2 allowed colours: 3×2 = 6.
If head and tail differ (3×2 = 6 ways), the wing must avoid both, leaving 1 choice: 6.
The Kangaroo Hotel has 30 floors, numbered 1 to 30, and each floor has 20 rooms, numbered 1 to 20. The code to enter a room is formed by writing the floor number followed by the room number, in that order. But a code can be confusing: for example, the code 111 could mean floor 11 room 1 or floor 1 room 11. Note that the code 101 is not confusing, since it can only mean floor 10 room 1 (floor 1 room 1 has the code 11, not 101). How many codes are confusing, including the one in the example?
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Answer: E — 18
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Hint 1 of 3
A code is confusing when the same digits can be cut into a floor and a room in two different correct ways.
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Hint 2 of 3
For three digits, you can cut after the first digit or after the second digit, so look for codes where both cuts give a real floor and a real room.
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Hint 3 of 3
Think about which three-digit codes start with a small floor but could also start with a teens floor.
Show solution
Approach: find three-digit codes that can be cut two ways, both giving a real floor and room
A three-digit code can be cut after the 1st digit (1-digit floor, 2-digit room) or after the 2nd digit (2-digit floor, 1-digit room); it is confusing when BOTH cuts give a real floor (1 to 30) and room (1 to 20).
For both cuts to work, the middle digit must be 1, so the codes look like floor-1-room, and they are exactly 11c and 21c where c is 1 to 9.
11c reads as floor 1 room 1c or floor 11 room c, and 21c reads as floor 2 room 1c or floor 21 room c, and c can be 1 to 9.
That is 9 codes of the form 11c and 9 codes of the form 21c, so 9 + 9 = 18 confusing codes, choice E.
In a vase there is one red, one blue, one yellow and one white flower. Maja the bee visits each flower exactly once. She begins with the red flower and she never flies directly from the yellow to the white flower. In how many different ways can she visit each flower?
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Answer: D — 4
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Hint 1 of 2
She always starts at the red flower, so list the orders of the other three.
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Hint 2 of 2
Then cross out any order where yellow comes immediately before white.
Show solution
Approach: list the routes and remove the forbidden ones
Starting at red, the other three flowers (blue, yellow, white) can be ordered in 6 ways.
Remove every order in which yellow is immediately followed by white.
Two of the six orders are forbidden, leaving 4 allowed routes.
