In a triangle one side has length 5 and another side has length 2. The length of the third side is an odd whole number. What is the length of the third side?
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Answer: C — 5
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Hint 1 of 2
The third side must be longer than the difference of the other two and shorter than their sum.
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Hint 2 of 2
It has to be more than 5 − 2 = 3 and less than 5 + 2 = 7.
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Approach: triangle inequality
The third side must be greater than 5 − 2 = 3 and less than 5 + 2 = 7.
The distance from the top of the cat sitting on the table to the top of the cat sleeping on the floor is 150 cm. The distance from the top of the cat sleeping on the table to the top of the cat sitting on the floor is 110 cm. How high is the table?
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Answer: C — 130 cm
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Hint 1 of 2
Add the two given distances and see what each cat contributes.
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Hint 2 of 2
Each measurement is one sitting-cat height plus the table minus one sleeping-cat height; adding them cancels the cats.
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Approach: add the two measurements so the cat heights cancel
Let the table height be t, a sitting cat add s above its base and a sleeping cat add p.
First distance: t + s − p = 150. Second distance: t + p − s = 110.
Adding the two equations: 2t = 260, so t = 130 cm.
Maria wants to share 42 apples, 60 peaches and 90 cherries fairly. She puts them into baskets that each hold the same number of apples, the same number of peaches and the same number of cherries, and gives one basket to each friend. At most, how many baskets can she fill?
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Answer: B — 6
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Hint 1 of 2
The number of baskets must divide each of 42, 60 and 90 exactly.
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Hint 2 of 2
Find the greatest common divisor of 42, 60 and 90.
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Approach: greatest common divisor
Each fruit count must split evenly among the baskets, so the number of baskets divides 42, 60 and 90.
The entrances of two student halls are 250 m apart on a straight street. The first hall has 100 students and the second has 150 students. Where should a bus stop be built so that the total of all the students' walking distances is as small as possible?
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Answer: D — directly in front of the second hall
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Hint 1 of 2
Putting the stop nearer one hall trades short walks for the bigger group against longer walks for the smaller group.
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Hint 2 of 2
Place it where the larger group walks zero distance.
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Approach: put the stop where the larger group walks nothing
For any stop between the halls, moving it a metre toward the second hall saves 150 students a metre but costs only 100 students a metre.
So the total distance keeps shrinking as the stop moves toward the second hall.
The best place is directly in front of the second hall (the larger group).
105 numbers are written in a row: 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, … where each number n is written exactly n times. How many of those numbers are divisible by 3?
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Answer: D — 30
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Hint 1 of 2
1 + 2 + 3 + … + 14 = 105, so the row runs from 1 up to 14.
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Hint 2 of 2
Only the multiples of 3 (namely 3, 6, 9, 12) count, and each appears as many times as its value.
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Approach: sum the counts of the multiples of 3 up to 14
Since 1 + 2 + … + 14 = 105, the numbers used run from 1 to 14.
The multiples of 3 in that range are 3, 6, 9 and 12, each written that many times.
Eight congruent semicircles are drawn inside a square with side length 4. How big is the area of the white part?
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Answer: B — 8
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Hint 1 of 2
Find the radius of the semicircles from how they sit along the sides.
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Hint 2 of 2
Pair up the grey and white pieces so the curved parts cancel.
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Approach: match curved pieces so the area is exactly half the square
Two semicircles fit along each side of length 4, so each has diameter 2 and radius 1.
By symmetry the eight semicircles split the square into equal grey and white regions: each curved bite removed from the grey is matched by an equal curved bite added to the white.
So the white area is exactly half of the square: \(\tfrac12\times 16 = \) 8.
On one day there are 40 train trips, each from one of the towns M, N, O, P, Q to exactly one other of those towns. There are 10 trips either from or to M, 10 either from or to N, 10 either from or to O, and 10 either from or to P. How many trips are there either from or to Q?
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Answer: E — 40
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Hint 1 of 2
Each trip touches exactly two towns, so summing “trips touching each town” counts every trip twice.
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Hint 2 of 2
The five town-counts must add up to 2 × 40 = 80.
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Approach: each trip is counted at both of its towns
Every trip has two endpoints, so adding the counts for all five towns gives \(2\times 40 = 80\).
At a humanistic university you can study languages, history or philosophy. Some students study exactly one language (nobody studies several at once). Among those language students, 35% study English. Among all students of the university, 13% study a language other than English. What percentage of all students study a language?
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Answer: B — 20 %
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Hint 1 of 2
If 35% of language students study English, then 65% study a non-English language.
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Hint 2 of 2
That 65% of the language group equals 13% of all students.
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Approach: the non-English language students link the two percentages
Among language students, 35% study English, so 65% study another language.
Those non-English language students are 13% of all students, so 65% of the language group = 13% of everyone.
Thus the language group is \(13\% \div 0.65 = \) 20% of all students.
Peter wants to buy a book but has no money. His father and his two brothers help. His father gives him half as much as his two brothers give jointly. His older brother gives a third of what the other two give him. The youngest brother gives 10 €. How expensive is the book?
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Answer: A — 24 €
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Hint 1 of 2
Write the total as father + older brother + youngest brother and use each clue as a fraction of the total.
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Hint 2 of 2
Father is one third of the whole; the older brother is one quarter of the whole.
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Approach: express each gift as a fraction of the total
Father gives half of what the brothers give, so father is \(\tfrac13\) of the total cost.
The older brother gives a third of what the other two give, so he is \(\tfrac14\) of the total.
The youngest brother's 10 € is the remaining \(1-\tfrac13-\tfrac14 = \tfrac{5}{12}\) of the total.
How many three-digit numbers have the property that the two-digit number obtained by deleting the middle digit is exactly one ninth of the original number?
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Answer: D — 4
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Hint 1 of 2
Write the number as \(100a+10b+c\); deleting the middle digit gives \(10a+c\).
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Hint 2 of 2
The condition \(100a+10b+c = 9(10a+c)\) simplifies to a relation between the digits.
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Approach: set up the place-value equation and count solutions
Let the number be \(100a+10b+c\); deleting the middle digit gives \(10a+c\).
The condition \(100a+10b+c = 9(10a+c)\) becomes \(10a+10b = 8c\), i.e. \(5(a+b)=4c\).
So \(c\) is a multiple of 5, and \(c\neq 0\) forces \(c=5\), giving \(a+b=4\).
In a regular 2018-sided polygon the vertices are numbered 1 to 2018 in order. Two diagonals are drawn: one joins vertices 18 and 1018, the other joins vertices 1018 and 2000. How many vertices do the three resulting polygons have?
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Answer: A — 38, 983, 1001
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Hint 1 of 2
The two diagonals share vertex 1018, cutting the polygon into three pieces.
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Hint 2 of 2
Count the vertices on each piece inclusively (shared endpoints belong to both adjoining pieces).
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Approach: count the vertices on each arc inclusively
Diagonal 18–1018 bounds the arc 18, 19, …, 1018: that is \(1018-18+1 = 1001\) vertices.
Diagonal 1018–2000 bounds the arc 1018, …, 2000: that is \(2000-1018+1 = 983\) vertices.
The third piece runs 2000, …, 2018, 1, …, 18 together with vertex 1018: \(19 + 18 + 1 = 38\) vertices.
So the three polygons have 38, 983, 1001 vertices.
Some whole numbers are written on a board, among them the number 2018. The sum of all of them is 2018, and the product of all of them is also 2018. Which of the following could be how many numbers are on the board?
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Answer: B — 2017
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Hint 1 of 2
The product is already 2018 from that one number, so every other number must be \(\pm 1\).
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Hint 2 of 2
The sum is already 2018, so the extra \(+1\)'s and \(-1\)'s must cancel; that forces an exact count.
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Approach: every other number is +1 or -1, then match the sum and product constraints
The list already contains 2018, whose value equals both the required sum and the required product, so every other number must be a \(+1\) or a \(-1\) (these don't change a product and are the only integers that can).
For the product to stay \(+2018\), the number of \(-1\)'s must be even; for the sum to stay 2018, the \(+1\)'s and \(-1\)'s must cancel, so there are equally many of each, say \(k\) of each.
Then the board holds \(1 + k + k = 2k+1\) numbers, which is always odd, so among the options only 2017 is possible (take \(k = 1008\)).
Four positive numbers are given. Take three of them, find their mean, then add the fourth number. This can be done in four ways, giving the results 17, 21, 23 and 29. Which is the biggest of the four numbers?
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Answer: C — 21
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Hint 1 of 2
Each result is “mean of three” plus the one left out; adding the four results relates them to the total.
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Hint 2 of 2
The largest result corresponds to leaving out (adding back) the largest number.
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Approach: add all four results to find the total, then back out each number
Let the total of the four numbers be T. Each result is \(\frac{T-x}{3}+x = \frac{T+2x}{3}\), where x is the added number.
Summing the four results: \(\frac{4T + 2T}{3} = 2T\), and \(17+21+23+29 = 90\), so \(2T = 90\), giving \(T = 45\).
The biggest result 29 comes from the biggest x: \(\frac{45+2x}{3}=29\Rightarrow x = 21\).
The points \(A_0, A_1, A_2, \ldots\) all lie on a straight line. It is given that \(A_0A_1 = 1\) and that \(A_n\) is the midpoint of segment \(A_{n+1}A_{n+2}\) for every non-negative index n. How long is the segment \(A_0A_{11}\)?
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Answer: E — 683
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Hint 1 of 2
The midpoint rule rearranges to \(A_{n+2} = 2A_n - A_{n+1}\), so the signed step doubles and flips sign each time.
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Hint 2 of 2
Track the signed steps \(d_n = A_{n+1}-A_n\); they follow \(d_{n+1} = -2d_n\).
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Approach: find the recurrence for the directed steps, then sum
Since \(A_n\) is the midpoint of \(A_{n+1}A_{n+2}\), the signed steps satisfy \(d_{n+1} = -2d_n\) with \(d_0 = 1\), so \(d_n = (-2)^n\).
Two concentric circles with radii 1 and 9 form an annulus. n non-overlapping circles are drawn inside this annulus, each touching both circles of the annulus. (The diagram shows an example for n = 1.) What is the biggest possible value of n?
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Answer: C — 3
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Hint 1 of 2
Each inscribed circle has radius 4, with its centre 5 from the common centre.
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Hint 2 of 2
Compare the angle each circle takes up at the centre with the full 360°.
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Approach: fit equal circles around the ring by their central angles
A circle touching both the radius-1 and radius-9 circles has radius \(\tfrac{9-1}{2}=4\), and its centre lies on the circle of radius \(1+4 = 5\).
Two such neighbouring circles just touch when the half-angle \(\theta\) at the centre satisfies \(\sin\theta = \tfrac{4}{5}\), so each circle takes up about \(106^\circ\).
Three circles use about \(318^\circ < 360^\circ\) (they fit), but four would need about \(424^\circ\) (too much).
A number is written at every vertex of the 18-sided shape so that each number equals the sum of the numbers at its two neighbouring vertices. Two of the numbers are given (see picture). Which number is written at vertex A?
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Answer: D — 38
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Hint 1 of 2
The rule “each = sum of its two neighbours” makes the sequence of vertex values repeat with period 6.
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Hint 2 of 2
Write a few terms: the pattern is \(x,\ y,\ y-x,\ -x,\ -y,\ x-y\) and then it repeats.
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Approach: use the period-6 pattern forced by the neighbour rule
If a vertex equals the sum of its neighbours, going around gives the repeating block \(x,\,y,\,y-x,\,-x,\,-y,\,x-y\) (period 6).
Over the 18 vertices this block repeats exactly three times.
Placing the two given values (20 and 18) at their vertices and following the period-6 pattern forces vertex A to 38.
Diana draws a rectangle made of squares on grid paper and colours some squares black. In every white square she writes the number of black squares next to it (sharing an edge). The diagram shows an example. She now does the same with a rectangle made of 2018 squares. What is the biggest possible sum of all the numbers in the white squares?
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Answer: D — 3025
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Hint 1 of 2
The total written down is just the number of shared edges that have a black square on one side and a white square on the other.
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Hint 2 of 2
Since \(2018 = 2\times 1009\), the rectangle is either \(1\times 2018\) or \(2\times 1009\); compare which shape allows more black–white edges.
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Approach: the sum equals the count of black–white shared edges; maximise that
Each number in a white square counts its black neighbours, so adding them all gives exactly the number of edges that separate a black square from a white square.
Because \(2018 = 2\times 1009\), the rectangle is \(1\times 2018\) or \(2\times 1009\); colouring whole columns black/white alternately so that no two same-colour columns touch makes nearly every internal edge a black–white edge.
For the \(2\times 1009\) strip this alternating pattern gives \(3\times 1009 - 2 = 3025\) such edges (the \(1\times 2018\) strip gives only \(2017\)), so the largest total is 3025.
Seven little dice were removed from a 3 × 3 × 3 die, as shown in the diagram. The remaining (completely symmetrical) figure is cut along a plane through the centre and perpendicular to one of the four space diagonals. What does the cross-section look like?
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Answer: A
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Hint 1 of 2
Looking straight down a space diagonal of a cube, the outline you see is a regular hexagon.
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Hint 2 of 2
Track which little cubes were removed and where their gaps fall on that hexagonal cross-section.
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Approach: take the cross-section perpendicular to a space diagonal
Viewed along a space diagonal, the cube's cross-section through the centre is a regular hexagon.
Removing the seven little cubes (the centre and the six face-centres) leaves gaps that, on this cross-section, form a six-pointed star.
So the cross-section looks like option A (the star inside the hexagon).
Each number of the set {1, 2, 3, 4, 5, 6} is written into exactly one cell of a 2 × 3 table. In how many ways can this be done so that the sum of the numbers in every column and every row is divisible by 3?
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Answer: D — 48
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Hint 1 of 2
Group the numbers by remainder mod 3: {3,6} give 0, {1,4} give 1, {2,5} give 2.
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Hint 2 of 2
Decide the remainder pattern of the table first, then count how to place the actual numbers.
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Approach: work modulo 3 on the remainder pattern, then fill in
By remainder mod 3 the numbers are two 0's (3,6), two 1's (1,4) and two 2's (2,5).
Each row of three and each column of two must sum to a multiple of 3; find the valid remainder patterns.
For each valid pattern, the two numbers in each remainder class can be swapped, and counting all arrangements gives 48.
Ed builds a big cube from several identical small white dice and colours some of the big cube's faces red. His sister Nicole drops it and it breaks back into the small dice. 45 of them have no red face. How many faces of the big cube did Ed colour red?
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Answer: C — 4
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Hint 1 of 2
A small die stays all-white only if it touches none of the coloured big-cube faces.
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Hint 2 of 2
Find a cube size where exactly 45 small dice can avoid the coloured faces.
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Approach: count the small dice that avoid every coloured face
Take the big cube as \(5\times5\times5\) (125 small dice) and colour the 4 side faces, leaving the top and bottom uncoloured.
The dice touching no red face are the inner \(3\times3\) of each of the 5 layers: \(3\times 3\times 5 = 45\), exactly as stated.
Two chords AB and AC are drawn in a circle with diameter AD. \(\angle BAC = 60^\circ\), \(AB = 24\) cm, point E lies on AC with \(EC = 3\) cm, and BE is perpendicular to AC. How long is the chord BD?
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Answer: D — \(2\sqrt{3}\) cm
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Hint 1 of 2
Drop into right triangle ABE to get the two legs \(AE\) and \(BE\) from the \(60^\circ\) angle.
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Hint 2 of 2
AD is a diameter, so \(\angle ABD = 90^\circ\); also the inscribed angle \(\angle ADB\) equals \(\angle ACB\) because both stand on arc \(AB\).
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Approach: find angle ACB from the right triangle, then use the right angle at B in triangle ABD
In right triangle BEC (right angle at E): \(BE = 24\sin 60^\circ = 12\sqrt3\) and \(EC = 3\), so \(\tan(\angle ACB) = \dfrac{BE}{EC} = \dfrac{12\sqrt3}{3} = 4\sqrt3\).
Because AD is a diameter, \(\angle ABD = 90^\circ\); and \(\angle ADB = \angle ACB\) since both inscribed angles stand on the same arc \(AB\).
In right triangle ABD then \(BD = \dfrac{AB}{\tan(\angle ADB)} = \dfrac{24}{4\sqrt3} = \dfrac{6}{\sqrt3} = 2\sqrt3\).
