On the number wall shown, the number on each tile is equal to the sum of the numbers on the two tiles directly below it. Which number is on the tile marked with “?”
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Answer: B — 16
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Hint 1 of 2
The top tile equals the sum of everything fed up from the bottom; write each tile from the bottom two unknowns.
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Hint 2 of 2
Use the known tiles 2020 and 2017 to pin down the bottom values first, then read off the marked tile.
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Approach: set up the wall from the bottom two unknown tiles and use the known tiles
Call the two leftmost bottom tiles a (the marked one) and b; the bottom row is a, b, 2017.
The middle-right tile is b + 2017 = 2020, so b = 3.
The top tile is a + 2b + 2017 = 2039, so a + 2b = 22, giving a = 22 - 6 = 16.
Many model railways use the H0-scale 1:87. For his railway, Benjamin owns a 2 cm high model of his brother in H0-scale. How tall is his brother in reality?
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Answer: A — 1.74 m
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Hint 1 of 2
The scale 1:87 means every real centimetre is shrunk 87 times in the model.
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Hint 2 of 2
Multiply the model height by 87 to undo the shrinking, then convert to metres.
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Approach: scale up by the ratio
The model is 2 cm tall and the scale is 1:87, so the real height is 2 x 87 = 174 cm.
In the diagram we see 10 islands that are connected by 15 bridges. What is the minimum number of bridges that need to be closed off so that there is no longer any connection from A to B?
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Answer: C — 3
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Hint 1 of 2
You want the fewest bridges whose removal leaves no path at all from A to B.
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Hint 2 of 2
Look for a bottleneck: the smallest set of bridges that every A-to-B route must use.
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Approach: find a minimum cut (smallest bottleneck of bridges) separating A from B
Every route from A to B has to cross a narrow set of bridges.
Trace the routes and find the smallest group of bridges that all of them share.
Removing that bottleneck of 3 bridges disconnects A from B, and no smaller set works.
A \(4 \times 1 \times 1\) cuboid is made up of 2 white and 2 grey cubes as shown. Which of the following cuboids can be built entirely out of such \(4 \times 1 \times 1\) cuboids?
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Answer: A
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Hint 1 of 2
Each building block is a 4x1x1 bar with a fixed white-white-grey-grey colour pattern.
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Hint 2 of 2
A target box can be built only if its grey and white cubes split into such fixed bars; check the colour layout, not just the shape.
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Approach: check which target colouring can be partitioned into the fixed WWGG bars
Every available bar is 4 cubes long with colours white, white, grey, grey in that order.
Try to cover each candidate box with these bars so that every bar's colour pattern matches.
Only box A has a colour layout that can be cut entirely into such WWGG bars.
Which quadrant contains no points of the graph of the linear function \(f(x) = -3.5x + 7\)? (Quadrants are numbered I, II, III, IV anticlockwise, starting from the upper right.)
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Answer: C — III
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Hint 1 of 2
A line with negative slope and positive y-intercept rises to the upper left and falls to the lower right.
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Hint 2 of 2
Sketch where the line goes: which of the four quadrants does it simply never enter?
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Approach: trace the line by slope and intercept across the quadrants
f(x) = -3.5x + 7 has y-intercept (0,7) and x-intercept (2,0).
For x < 2 the line is above the x-axis (quadrants II then I); for x > 2 it drops below (quadrant IV).
It never reaches the lower-left region, so it has no points in quadrant III.
In each of the five boxes (A) to (E) there are red and blue balls. Benedict wants to take exactly one ball, without looking, out of one of these boxes, and hopes to get a blue ball. In which box is the probability of that happening greatest?
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Answer: B — 6 blue, 4 red
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Hint 1 of 2
For each box the chance of a blue ball is blues divided by total balls.
Three circles with centres A, B, C touch each other in pairs from the outside (see diagram). Their radii are 3, 2 and 1. How big is the area of the triangle ABC?
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Answer: A — 6
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Hint 1 of 2
When two circles touch externally, the distance between their centres is the sum of the radii.
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Hint 2 of 2
Work out the three side lengths of triangle ABC; they may form a familiar right triangle.
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Approach: get the side lengths from touching circles, recognise a 3-4-5 triangle
Touching externally, AB = 3 + 2 = 5, BC = 2 + 1 = 3, CA = 1 + 3 = 4.
Sides 3, 4, 5 form a right triangle (3^2 + 4^2 = 5^2).
Two cylinders A and B have the same volume. The radius of the base of B is 10% bigger than that of A. By how much is the height of A greater than that of B?
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Answer: E — 21%
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Hint 1 of 2
Equal volumes means (radius squared) x height is the same for both cylinders.
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Hint 2 of 2
If B's radius is 1.1 times A's, compare the heights through the square of that factor.
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Approach: use equal volume to relate the heights through the radius ratio
Volume = pi r^2 h is equal, so r_A^2 h_A = r_B^2 h_B with r_B = 1.1 r_A.
Each face of the polyhedron shown is either a triangle or a square. Each square borders 4 triangles, and each triangle borders 3 squares. The polyhedron has 6 squares. How many triangles does it have?
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Answer: D — 8
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Hint 1 of 2
Count the square-touches-triangle contacts in two different ways.
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Hint 2 of 2
Each square contributes 4 such contacts; each triangle uses up 3 of them.
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Approach: double-count the square-triangle adjacencies
There are 6 squares, each bordering 4 triangles, giving 6 x 4 = 24 square-triangle borders.
Each triangle borders 3 squares, so it accounts for 3 of those borders.
The four faces of a regular tetrahedron are labelled with the four digits 2, 0, 1 and 7 (one digit on each face). For a game, four such tetrahedrons are used as fair dice. All four dice are thrown simultaneously. Three of the four faces of each die can then be seen from above. What is the probability that we can form the number 2017 using exactly one of the three visible digits of each die?
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Answer: B — 6364
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Hint 1 of 2
On each die the three visible faces are simply all faces except the one hidden underneath.
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Hint 2 of 2
It is easier to count the chance you CANNOT form 2017, then subtract from 1.
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Approach: complementary counting over which digit each die hides
Each die hides exactly one of its four digits; the other three are visible. There are 4^4 = 256 equally likely hidden-digit combinations.
You can match the four needed digits (2,0,1,7) to the four dice unless some required digit is hidden on every die.
That fails only when all four dice hide the same one digit: 4 ways out of 256.
So the probability of success is 1 - 4/256 = 63/64.
The polynomial \(5x^3 + ax^2 + bx + 24\) has whole-number coefficients a and b. Which of the following numbers is definitely not a solution to the equation \(5x^3 + ax^2 + bx + 24 = 0\)?
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Answer: D — 5
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Hint 1 of 2
Any integer root of a polynomial with integer coefficients must divide the constant term.
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Hint 2 of 2
The constant term is 24; which listed candidate is not a divisor of 24?
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Approach: rational (integer) root test on the constant term
An integer solution r of 5x^3 + ax^2 + bx + 24 = 0 must divide the constant term 24.
Among 1, -1, 3, 5, 6, all divide 24 except 5.
So 5 can never be a solution, whatever a and b are.
Julia has 2017 round discs available: 1009 black ones and 1008 white ones. Using them, she wants to lay the biggest square pattern possible (as shown) and starts by using a black disc in the upper-left corner. She then lays the discs so that the colours alternate in each row and column. How many discs are left over when she has laid the biggest possible square?
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Answer: E — 40 white and 41 black
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Hint 1 of 2
A checkerboard square with a black corner has roughly half discs of each colour; find the biggest size that fits the supply.
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Hint 2 of 2
Try square sizes n x n and find the largest where black <= 1009 and white <= 1008, then count what is left.
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Approach: find the largest fitting checkerboard, then subtract
For an n x n board starting black, with n even there are n^2/2 of each colour.
n = 44 needs 968 black and 968 white, which fits; n = 46 needs 1058 each, too many.
Leftover: 1009 - 968 = 41 black and 1008 - 968 = 40 white.
Two consecutive positive whole numbers are written on a board. The sum of the digits of each number is divisible by 7. What is the minimum number of digits the smaller of the two numbers has to have?
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Answer: C — 5
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Hint 1 of 2
Normally adding 1 raises the digit sum by 1, so both sums can be multiples of 7 only when carrying happens.
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Hint 2 of 2
Trailing 9s cause big drops in the digit sum; figure out how many 9s are needed for both sums to stay divisible by 7.
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Approach: use trailing nines to control how the digit sum changes
If the smaller number ends in k nines, adding 1 turns them to zeros and bumps the next digit, so the digit sum changes by 1 - 9k.
For both digit sums divisible by 7 we need 9k = 1 (mod 7), i.e. 2k = 1 (mod 7), giving k = 4 as the smallest.
Four trailing nines plus at least one leading digit (e.g. 69999) is needed, which has 5 digits.
In a convex quadrilateral ABCD the diagonals are perpendicular to each other. The lengths of the edges are AB = 2017, BC = 2018 and CD = 2019 (diagram not to scale). How long is side AD?
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Answer: D — \(\sqrt{2018^2 + 2}\)
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Hint 1 of 2
With perpendicular diagonals, the four sides satisfy a neat relation between opposite pairs.
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Hint 2 of 2
Use AB^2 + CD^2 = BC^2 + AD^2 to solve for AD.
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Approach: apply the perpendicular-diagonal identity for the sides
For perpendicular diagonals, AB^2 + CD^2 = BC^2 + AD^2.
So AD^2 = 2017^2 + 2019^2 - 2018^2.
Since 2017^2 + 2019^2 = 2*2018^2 + 2, this gives AD^2 = 2018^2 + 2, so AD = sqrt(2018^2 + 2).
Lilli tries to be a well-behaved kangaroo, but she is having just too much fun not to lie every now and then. So every third statement of hers is a lie and the rest are true; sometimes she starts with a lie and sometimes with one or two true statements. Lilli thinks of a two-digit number and says to her friend: 1: “One digit of the number is a 2.” 2: “The number is greater than 50.” 3: “It is an even number.” 4: “The number is less than 30.” 5: “The number is divisible by 3.” 6: “One digit of the number is a 7.” What is the sum of the digits of the number Lilli is thinking of?
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Answer: D — 15
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Hint 1 of 2
Exactly one statement in each block of three is a lie, so the lies sit at positions {1,4}, {2,5}, or {3,6}.
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Hint 2 of 2
Test each lie-pattern: only one makes all the 'true' statements consistent for a real two-digit number.
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Approach: casework on which statements are the lies, then find the number
The two lies are at positions {1,4}, {2,5}, or {3,6}. The first two cases give contradictions (e.g. '>50' and '<30' both true).
With lies at {1,4}: the number is even, greater than 50, divisible by 3, contains a 7, contains no 2, and is at least 30.
That number is 78 (even, >50, 7+8=15 divisible by 3, has a 7, no 2).
How many positive whole numbers have the property that, if you delete the last digit, you obtain a new number that is exactly equal to 114 of the original number?
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Answer: C — 2
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Hint 1 of 2
Deleting the last digit of N leaves the number formed by the other digits; call that part a and the last digit d.
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Hint 2 of 2
Write N = 10a + d and set a = N/14, then see which digits d are possible.
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Approach: set up the place-value equation and solve for valid digits
Let the number be 10a + d, where a is what remains after deleting the last digit d.
The condition a = (10a + d)/14 gives 14a = 10a + d, so 4a = d.
Since d is a single digit, a = 1 (d = 4, number 14) or a = 2 (d = 8, number 28).
The diagram shows a regular hexagon with side length 1. The grey flower is outlined by circular arcs of radius 1 whose centres lie at the vertices of the hexagon. How big is the area of the grey flower?
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Answer: E — \(2\pi - 3\sqrt{3}\)
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Hint 1 of 2
Each petal is built from two circular arcs of radius 1; relate it to a 60-degree sector of a unit circle.
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Hint 2 of 2
Subtract the straight triangular pieces from the arc sectors to isolate the petal area, then multiply by the number of petals.
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Approach: decompose the flower into arc-sectors minus triangles
Each petal is the overlap of two unit circles centred at adjacent vertices; that lens is two \(60^\circ\) sectors minus the equilateral triangle counted twice, i.e. \(2\cdot\frac{\pi}{6} - 2\cdot\frac{\sqrt{3}}{4} = \frac{\pi}{3} - \frac{\sqrt{3}}{2}\).
The flower is made of six such petals, so its area is \(6\left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right)\).
That simplifies to \(2\pi - 3\sqrt{3}\), which is answer E.
We look at a regular tetrahedron with volume 1. Its four vertices are cut off by planes that go through the midpoints of the respective edges (see diagram). How big is the volume of the remaining solid?
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Answer: D — 12
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Hint 1 of 2
Each cut through edge midpoints slices off a small tetrahedron similar to the whole, at half scale.
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Hint 2 of 2
A half-scale tetrahedron has 1/8 the volume; account for all four corners.
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Approach: subtract four half-scale corner tetrahedra
Cutting through the midpoints removes a corner tetrahedron with edges half as long, so each has volume (1/2)^3 = 1/8.
The four corner pieces do not overlap, removing 4 x 1/8 = 1/2 of the volume.
The sum of the three side lengths of a right-angled triangle equals 18. The sum of the squares of these three side lengths equals 128. How big is the area of the triangle?
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Answer: E — 9
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Hint 1 of 2
For a right triangle the sum of the two leg-squares equals the hypotenuse-square, so the squared-sum simplifies.
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Hint 2 of 2
Use the perimeter and the squared identity to find the legs' product, which gives the area.
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Approach: use the Pythagorean relation to find the legs' product
With legs a, b and hypotenuse c: a^2 + b^2 = c^2, so a^2 + b^2 + c^2 = 2c^2 = 128, giving c = 8.
Then a + b = 18 - 8 = 10, and (a + b)^2 = a^2 + b^2 + 2ab = 64 + 2ab = 100, so ab = 18.
Anna has five boxes, as well as five black balls and five white balls. She is allowed to decide how she shares out the balls between the boxes, as long as she puts at least one ball into each box. Beate randomly chooses one box and takes one ball without looking. Beate wins if she draws a white ball; otherwise Anna wins. How should Anna distribute the balls to get the highest probability of winning?
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Answer: D — Anna puts all of the white balls into one box and then puts one black ball into each box.
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Hint 1 of 2
Anna wins when Beate draws black, so Anna wants to minimise the chance of a white draw.
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Hint 2 of 2
Beate picks a box uniformly first; compute the white-draw probability for each option and pick the smallest.
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Approach: compute Beate's white-draw probability for each distribution
Beate first picks one of the 5 boxes with equal chance \(\frac{1}{5}\), then a ball from it; Anna wants the white-draw probability as small as possible.
Option D buries all 5 white balls in one box that also gets a black ball (6 balls, \(\frac{5}{6}\) white) while the other four boxes hold only black, giving white chance \(\frac{1}{5}\cdot\frac{5}{6} = \frac{1}{6}\).
Every other option leaves more boxes containing white balls, so its white chance exceeds \(\frac{1}{6}\) (e.g. option C gives \(\frac{1}{5}\)).
The smallest white chance, hence Anna's best play, is option D.
Nine whole numbers were written into the cells of a 3 × 3 table. The sum of these nine numbers is 500. We know that the numbers in two adjacent cells (sharing a common side) differ by exactly 1. Which number is in the middle cell?
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Answer: D — 56
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Hint 1 of 2
Adjacent cells differ by 1, so the grid splits into two parity classes like a checkerboard around the centre.
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Hint 2 of 2
Express all nine entries in terms of the centre value and set the total equal to 500.
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Approach: write all cells relative to the centre, then use the sum
Colour the grid like a checkerboard; neighbours differ by 1, so the centre and four corners share one parity while the four edge cells share the other.
A valid tight filling is centre \(m\), each edge cell \(m-1\), and each corner \(m\) (every adjacent pair then differs by exactly 1).
The total is \(m + 4(m-1) + 4m = 9m - 4\); setting \(9m - 4 = 500\) gives \(9m = 504\).
2017 people live on an island. Each person is either a liar (who always lies) or a nobleman (who always tells the truth). Over a thousand of them attend a banquet where they all sit together around one big round table. Everyone says, “Of my two neighbours, one is a liar and one is a nobleman.” What is the maximum number of noblemen on the island?
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Answer: A — 1683
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Hint 1 of 2
The banquet statement constrains only the people seated at the round table; the others on the island are free.
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Hint 2 of 2
Work out the densest valid liar/nobleman pattern around the table, then add the unseated islanders.
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Approach: maximise noblemen given the round-table constraint plus free islanders
A seated nobleman truly has one liar neighbour, so two noblemen can sit together but never three in a row; a seated liar lies, so its two neighbours match (both noblemen or both liars).
The densest legal seating repeats the block nobleman-nobleman-liar, making at most \(\frac{2}{3}\) of the seated people noblemen, so a table of \(n\) (a multiple of 3) seats up to \(\frac{2n}{3}\) noblemen.
Everyone not at the banquet can be a nobleman, so the island total is \((2017-n) + \frac{2n}{3} = 2017 - \frac{n}{3}\), maximised by the smallest legal \(n\) over a thousand, namely \(n = 1002\).
That gives \(2017 - 334 = 1683\) noblemen, answer A.
