In three differently sized baskets there are 48 balls in total. Together the smallest and the biggest basket hold twice as many balls as the middle one. The smallest basket holds half as many balls as the middle one. How many balls are there in the biggest basket?
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Answer: C — 24
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Hint 1 of 2
Let the middle basket be your unit and write the others in terms of it.
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Hint 2 of 2
Three quantities add to 48 — turn the words into one equation in the middle amount.
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Approach: express all baskets through the middle one
Let the middle basket hold m balls. The smallest holds m/2.
The diagram shows a solid made up of 6 triangles. Each vertex is assigned a number, two of which are indicated. The total of the three numbers on each triangular face is the same. What is the total of all five numbers?
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Answer: C — 17
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Hint 1 of 2
Adding the sums of all six faces counts every vertex the same number of times.
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Hint 2 of 2
Equal face sums force the three ‘waist’ vertices to be equal and the two tips to be equal.
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Approach: use equal face sums to collapse the unknowns
Each triangular face joins a tip vertex to two neighbouring waist vertices.
Equal sums make all three waist vertices equal and both tip vertices equal.
The figure marks a tip as 1 and a waist vertex as 5, so the five numbers are 1, 1, 5, 5, 5, totalling 17.
The graphs of the functions \(y = x^3 + 3x^2 + ax + 2a + 4\) all pass through a common point independent of the choice of a. How big is the sum of the co-ordinates of this common point?
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Answer: E — another number
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Hint 1 of 2
Group the terms that contain a together.
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Hint 2 of 2
The point must work for every a, so the coefficient of a has to vanish.
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Approach: make the coefficient of the parameter zero
Write y = (x3+3x2+4) + a(x+2); for this to be independent of a, set x+2 = 0, so x = −2.
Then y = (−8 + 12 + 4) = 8, giving the fixed point (−2, 8).
Sum of coordinates = −2 + 8 = 6, which is not among the first four options, so the answer is another number.
a, b and c are real numbers not equal to zero. It is known that the numbers \(-2a^4b^3c^2\) and \(3a^3b^5c^{-4}\) have the same sign. Which of the following statements is definitely correct?
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Answer: E — \(a<0\)
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Hint 1 of 2
Track only the signs: even powers are positive, so focus on the odd-power factors and the leading constants.
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Hint 2 of 2
Set the two signs equal and solve for the sign of a.
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Approach: sign analysis
Sign of −2a^4 b^3 c^2 is −sign(b) (the a^4 and c^2 are positive).
Sign of 3a^3 b^5 c^(−4) is sign(a)·sign(b) (c^(−4) is positive).
Equal signs: −sign(b) = sign(a)·sign(b), so sign(a) = −1.
The arithmetic mean of five numbers is 24. The mean of the three smallest numbers is 19 and that of the three biggest is 28. What is the median of the five numbers?
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Answer: B — 21
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Hint 1 of 2
Write the three totals: all five, the three smallest, the three largest.
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Hint 2 of 2
The median is counted in both the bottom-three and top-three sums.
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Approach: overlap counts the median twice
Sum of all five = 120; smallest three sum to 57; largest three sum to 84.
57 + 84 counts every number once except the median, which is counted twice: 57 + 84 = 120 + median.
If \(A = {]0,1[} \cup {]2,3[}\) and \(B = {]1,2[} \cup {]3,4[}\), what is the set of all numbers of the form \(a+b\) with \(a \in A\) and \(b \in B\)? (Here \(]m,n[\) denotes the open interval from m to n.)
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Answer: D — \(]1,3[ \cup ]3,5[ \cup ]5,7[\)
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Hint 1 of 2
Add each piece of A to each piece of B; adding two open intervals gives another open interval.
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Hint 2 of 2
Take the union of all the resulting intervals.
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Approach: add the intervals pairwise and union the results
A = (0,1)∪(2,3), B = (1,2)∪(3,4). Adding endpoints: (0,1)+(1,2)=(1,3); (0,1)+(3,4)=(3,5); (2,3)+(1,2)=(3,5); (2,3)+(3,4)=(5,7).
The union is (1,3) ∪ (3,5) ∪ (5,7) — note 3 and 5 are never reached.
An infinite list of numbers has the property that, for each positive integer n, the average of the first n terms is n. How many terms are there less than 2021?
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Answer: C — 1010
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Hint 1 of 2
If the average of the first n terms is n, what is their sum?
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Hint 2 of 2
Get a single term by subtracting consecutive sums.
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Approach: turn the average condition into the n-th term
Average of first n is n means the sum of the first n terms is n².
The n-th term is n² − (n−1)² = 2n − 1 (the odd numbers 1, 3, 5, …).
We need 2n − 1 < 2021, i.e. n ≤ 1010, so 1010 terms.
In the \(5 \times 5\) square shown the sum of the numbers in each row and in each column is the same. There is a number in every cell, but some of the numbers are not shown. What is the number in the cell marked with a question mark?
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Answer: B — 10
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Hint 1 of 2
The grand total of a 1–25 square fixes the common row/column sum.
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Hint 2 of 2
Use row and column equations through the marked cell to pin its value.
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Approach: use the magic sum, then solve the right cells
The numbers 1–25 total 325, so each row and column sums to 65.
Solving the equations for row 2 and the involved columns forces the entries 7, 15, 13, 17 around column 4.
Column 4 then gives 22 + e + 1 + j + ? = 65 with e=15, j=17, so ? = 10.
Let a, b, c be nonzero real numbers such that \((a - a^{-1})^2 + (b - b^{-1})^2 + (c - c^{-1})^2 = 0\). Which of the following can NOT be the value of \(a + b + c\)?
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Answer: C — 0
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Hint 1 of 2
A sum of three squares equals zero only if each square is zero.
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Hint 2 of 2
Solve x − 1/x = 0; what can x be?
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Approach: zero sum of squares forces each term zero
Since each squared term is ≥ 0 and they sum to 0, each must be 0: a = 1/a, and likewise for b and c.
So a, b, c each equal +1 or −1.
Then a+b+c is one of −3, −1, 1, 3 — never 0, so the impossible value is 0.
The sequence \(f_n\) is given by \(f_1 = 1\), \(f_2 = 2\) and \(f_n = f_{n-1} \cdot f_{n+1}\) for \(n \ge 2\). How many of the first 2020 terms of this sequence are even numbers?
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Answer: B — 674
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Hint 1 of 2
Rearrange the rule to f(n+1) = f(n)/f(n−1) and list a few terms.
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Hint 2 of 2
The sequence repeats with a short period — find it.
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Approach: detect the period and count evens within it
From the rule we get f(n+1) = f(n)/f(n−1), giving 1, 2, 2, 1, 1/2, 1/2, then repeating with period 6.
Each period of 6 has exactly two even terms (the two 2's).
2020 = 336×6 + 4; the 336 periods give 672 evens and the leading 1,2,2,1 add 2 more, totalling 674.
Matias wrote 15 numbers around the wheel shown. Only one of them is visible, the 10 at the top. The sum of the numbers in any seven consecutive positions (such as the gray positions in the figure) is always the same. When seven numbers in consecutive positions are added, which of the following results is possible?
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Answer: B — 70
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Hint 1 of 2
Equal 7-sums force entry n to equal entry n−7 around the 15-position wheel.
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Hint 2 of 2
Since gcd(7,15)=1, that links all the numbers.
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Approach: constant window sum forces all entries equal
If every 7 consecutive entries have the same sum, then sliding by one shows entry n = entry n−7 (indices mod 15).
Because gcd(7, 15) = 1, this makes all 15 numbers equal; the visible one is 10, so all are 10.
Michael invents a new operation \(\diamond\) for real numbers, defined by \(x \diamond y = y - x\). Which of the following statements is definitely true if numbers \(a\), \(b\) and \(c\) satisfy \((a \diamond b) \diamond c = a \diamond (b \diamond c)\)?
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Answer: D — \(a = 0\)
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Hint 1 of 2
Carefully apply \(x \diamond y = y - x\) to each side of the equation.
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Hint 2 of 2
Both sides become a simple expression in \(a\), \(b\), \(c\); the difference shows which variable is forced.
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Approach: expand both sides of the condition
With \(x \diamond y = y - x\), the left side is \((a \diamond b) \diamond c = c - (b - a) = c - b + a\).
The right side is \(a \diamond (b \diamond c) = (c - b) - a = c - b - a\).
Setting them equal gives \(c - b + a = c - b - a\), so \(2a = 0\), i.e. \(a = 0\).
a, b, c, d are positive whole numbers for which \(a + 2 = b - 2 = c \times 2 = d \div 2\) holds true. Which of the four numbers a, b, c and d is biggest?
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Answer: D — d
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Hint 1 of 2
Set the common value equal to k and write each of a, b, c, d in terms of k.
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Hint 2 of 2
Compare a = k-2, b = k+2, c = k/2, d = 2k.
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Approach: express all four through the common value
Let a+2 = b-2 = 2c = d/2 = k.
Then a = k-2, b = k+2, c = k/2, d = 2k.
For positive whole numbers 2k exceeds the others, so d is biggest.
We know the following about five positive whole numbers a, b, c, d, e. All the numbers are different, b = c : e, d = a + b and a = e − d. Which of the numbers a, b, c, d, e is the largest?
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Answer: C — c
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Hint 1 of 2
Express everything in terms of the smaller quantities, then see which is built up the most.
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Hint 2 of 2
c is a product, while the others are sums or differences.
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Approach: rewrite each in terms of a and b
From a = e−d and d = a+b, we get e = 2a+b, so e and d are modest sums.
b = c÷e means c = b·e, a product of two positive whole numbers larger than 1.
A product of such factors beats the sums, so c is the largest (C).
The geometric mean of n numbers is defined as the nth root of the product of all n numbers, that is n√x1 · x2 · … · xn. We have six numbers. The geometric mean of three of them is 3, the geometric mean of the other three is 12. How big is the geometric mean of all six numbers?
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Answer: B — 6
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Hint 1 of 2
A geometric mean of 3 numbers tells you their product.
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Hint 2 of 2
Multiply the two products, then take the sixth root.
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Approach: combine the two products under one sixth root
Geometric mean 3 of three numbers means their product is 3³ = 27; geometric mean 12 means the other product is 12³ = 1728.
All six multiply to 27×1728 = 46656 = 6⁶.
The geometric mean of all six is the sixth root, 6 (B).
Let \(a,b,c\) be different real numbers, none equal to zero, and let \(n\) be a positive whole number. It is known that the numbers \((-2)^{2n+3} imes a^{2n+2} imes b^{2n-1} imes c^{3n+2}\) and \((-3)^{2n+2} imes a^{4n+1} imes b^{2n+5} imes c^{3n-4}\) have the same sign. Which of the following statements is definitely true?
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Answer: D — \(a<0\)
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Hint 1 of 2
Even powers are always positive, so only the odd-powered factors carry a sign.
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Hint 2 of 2
Compare the parities of the exponents in the two products to see what must be true.
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Approach: track signs through even/odd exponents
In a product, only factors with odd exponents affect the sign; even powers are positive.
Matching the two products' signs forces the contribution of a to flip consistently, and the only sign that is pinned down in every case is that of a.
Working it through shows a must be negative, so (D): a < 0 is definitely true.
Peter drew the graph of a function \(f : \mathbb{R} \to \mathbb{R}\) consisting of two rays and a line segment, as shown. How many solutions does the equation \(f(f(f(x))) = 0\) have?
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Answer: A — 4
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Hint 1 of 2
First find every input that f sends to 0, then work outward one layer at a time.
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Hint 2 of 2
Track preimages: solve f = 0, then f = those values, then again.
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Approach: count preimages layer by layer
f(x)=0 at x = 0 and x = −4.
Pulling back once more, f(x) ∈ {0,−4} adds x = −8; pulling back again adds x = −12.
The solution set is {0,−4,−8,−12}: 4 solutions, A.
Michael wants to write whole numbers into the empty cells of the 3×3 table on the right so that the numbers in every 2×2 square add up to 10. Four numbers are already filled in. Which of the following could be the sum of the remaining five numbers?
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Answer: E — None of these numbers is possible.
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Hint 1 of 3
Call the centre cell c and write each 2×2 sum-equals-10 condition in terms of c.
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Hint 2 of 3
Add the four border cells and the centre — the total comes out as 20 − 3c.
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Hint 3 of 3
Check whether any listed value has the form 20 − 3c for a whole number c.
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Approach: express the five unknowns through the centre value
With the centre = c, the four 2×2 conditions give the corners as 7−c, 5−c, 5−c, 3−c.
Their sum with c is (7−c)+(5−c)+c+(5−c)+(3−c) = 20 − 3c.
That is always 2 more than a multiple of 3, but 9, 10, 12, 13 are not.
How many of the functions \(y=x^{2}\), \(y=-x^{2}\), \(y=+\sqrt{x}\), \(y=-\sqrt{x}\), \(y=+\sqrt{-x}\), \(y=-\sqrt{-x}\), \(y=+\sqrt{|x|}\), \(y=-\sqrt{|x|}\) have graphs that appear in the sketch on the right?
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Answer: D — 6
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Hint 1 of 2
Sort the eight functions into square-type parabolas and square-root-type curves.
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Hint 2 of 2
The sketch shows the sideways square-root branches, not the upward/downward parabolas.
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Approach: match the sketch to the square-root family of curves
The drawing shows curves that flatten near the axis like square roots, meeting at the origin.
Six of the listed functions (the √-type ones) reproduce exactly those branches; the two parabolas do not.
In the (x, y)-plane the coordinate axes are drawn as usual. The point A(1, −10), which lies on the parabola \(y=ax^{2}+bx+c\), was marked. Then the coordinate axes and most of the parabola were erased, leaving the sketch on the right. Which of the following statements could be false?
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Answer: E — \(c<0\)
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Hint 1 of 2
Use the point A(1, −10): substituting x = 1 gives a + b + c = −10.
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Hint 2 of 2
That fixes some statements as always true; look for the one that the picture does not force.
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Approach: substitute the known point and test each statement
A(1,−10) gives a+b+c = −10 < 0, so that statement is always true; the upward shape forces a > 0.
But c is the value at x = 0, which the trimmed picture does not pin down — it may be positive.
Which of the following graphs represents the solution set of \((x-|x|)^2 + (y-|y|)^2 = 4\)?
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Answer: A
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Hint 1 of 2
The value of x − |x| depends on whether x is negative.
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Hint 2 of 2
Split the plane into the four sign-quadrants and simplify in each.
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Approach: case-split on the signs of x and y
If a coordinate is non-negative, t−|t| is 0; if it is negative, t−|t| = 2t.
First quadrant gives 0 = 4 (nothing); the second and fourth quadrants give the rays x = −1 (for y ≥ 0, pointing up) and y = −1 (for x ≥ 0, pointing right).
The third quadrant gives 4x² + 4y² = 4, a quarter circle x² + y² = 1 joining (−1, 0) to (0, −1).
The quarter arc together with the upward ray at x = −1 and the rightward ray at y = −1 matches graph A.
The sum of the numbers in each row, column and diagonal in the “magic square” on the right is always constant. Only two numbers are visible. Which number is missing in field a?
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Answer: D — 55
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Hint 1 of 2
Use that every line, column and both diagonals share the same total S.
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Hint 2 of 2
Combine a few of those equal-sum lines so the unknown cell pops out on its own.
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Approach: add and subtract equal-sum lines to isolate the corner
Writing each row, column and diagonal as equal to S gives a linear system in the cells.
Combining the equations forces the corner cell a to a single value regardless of the others.
The parabola in the figure has an equation of the form \(y = ax^{2} + bx + c\) for some distinct real numbers a, b and c. Which of the following equations could be an equation of the line in the figure?
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Answer: D — \(y = ax + c\)
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Hint 1 of 2
The parabola opens upward, so what is the sign of a?
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Hint 2 of 2
Match the line's slope and y-intercept to combinations of a, b and c.
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Approach: read the sign of a and the intercept from the picture
The parabola opens upward, so a > 0, giving a positive slope; its y-intercept is c.
The line in the figure has positive slope and the same y-intercept c.
An equation with slope a and intercept c is y = ax + c.
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.
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.
The sum of the ages of Tom and Johann is 23. The sum of the ages of Johann and Alex is 24, and the sum of the ages of Alex and Tom is 25. How old is the oldest of them?
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Answer: D — 13
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Hint 1 of 3
Add all three given pair-sums together.
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Hint 2 of 3
The grand total counts every person twice, so half of it is the sum of all three ages.
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Hint 3 of 3
Subtract a known pair from that whole-group total to isolate one person's age.
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Approach: add the equations to get the total, then back out each age
Adding the three pair-sums: 23+24+25 = 72.
Each person is counted twice, so Tom+Johann+Alex = 36.
Alex = 36 - (Tom+Johann) = 36 - 23 = 13; similarly Tom = 12, Johann = 11.
Ella wants to write a number into each circle in the diagram on the right, in such a way that each number is equal to the sum of its two direct neighbours. Which number does Ella need to write into the circle marked with “?”?
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Answer: E — This question has no solution.
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Hint 1 of 2
‘Each number equals the sum of its two neighbours’ rearranges to ‘next = this − previous’, which repeats with period 6.
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Hint 2 of 2
Follow that pattern around the 8-circle ring and see whether the two given numbers, 3 and 5, can both fit.
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Approach: chase the neighbour-sum rule around the ring
Writing each circle as the sum of its neighbours rearranges to ‘next neighbour = this − previous’, a rule that repeats every 6 steps.
On a ring of 8 circles this period-6 repetition forces two of the circles (here the ones holding 3 and 5) to carry equal values.
Since 3 ≠ 5, no consistent filling exists, so the answer is ‘no solution’ (E).
In a list of five numbers the first number is 2 and the last one is 12. The product of the first three numbers is 30, of the middle three 90 and of the last three 360. What is the middle number in that list?
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Answer: C — 5
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Hint 1 of 2
The first three multiply to 30 and the first number is 2, so you know the product of numbers 2 and 3.
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Hint 2 of 2
Use the last-three product and the known last number to pin down number 4, then chain back.
Show solution
Approach: peel products from both ends
Label the list \(a,b,c,d,e\) with \(a=2\) and \(e=12\). Since \(abc=30\), we get \(bc = 15\).
From \(bcd=90\): \(d = \frac{90}{bc} = \frac{90}{15} = 6\); and \(cde=360\) gives \(cd = \frac{360}{12} = 30\).
Then \(c = \frac{cd}{d} = \frac{30}{6} = 5\), so the middle number is \(5\), choice C.
For a ski race consecutive starting numbers are handed out. One number was accidentally given out twice. The sum of all the numbers handed out is 857. Which number was given out twice?
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Answer: D — 37
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Hint 1 of 2
First add up 1+2+…+n and see which n lands just below 857.
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Hint 2 of 2
The leftover above that triangular number is the repeated starting number.
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Approach: match a triangular number, then read the extra
If the numbers run \(1\) to \(n\), their sum is \(\frac{n(n+1)}{2}\) and the repeated number adds a little extra.
\(1+\cdots+40 = 820\), and \(857 - 820 = 37\), which is a valid number between 1 and 40 (while \(1+\cdots+41 = 861\) is already too big).
In one class a test did not yield a very successful result because the average mark was exactly 4. The boys have done slightly better with an average mark of 3.6, while the girls have received an average mark of 4.2. Which of the following statements is correct?
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Answer: C — There are twice as many girls as boys.
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Hint 1 of 2
Write the total of all marks two ways: as 4×(everyone) and as boys' total plus girls' total.
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Hint 2 of 2
Set them equal and the ratio of boys to girls drops out.
Show solution
Approach: balance the weighted average
With \(b\) boys and \(g\) girls, the total of all marks is \(3.6b + 4.2g\) and also \(4(b+g)\).
Fritz fills out a table with two columns and 51 rows. In the first row, he writes 5 on the left and 3 on the right. In each subsequent row he writes the sum of the two numbers from the row above on the left and the positive difference of these two numbers on the right. Which two numbers does he write in the bottom row?
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Answer: D — \(5\cdot 2^{25}\) and \(3\cdot 2^{25}\)
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Hint 1 of 2
Compute a few rows and watch the left and right entries separately.
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Hint 2 of 2
Each pair (L, R) becomes (L+R, L−R); the left entry doubles every two rows starting from 5.
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Approach: track the recurrence two rows at a time
Rows give left entries 5, 8, 10, 16, 20, 32, 40… and right entries 3, 2, 6, 4, 12, 8, 24…
I have a four-digit number \(N=\overline{pqrs}\). If I place a decimal point between the digits q and r, I obtain the number \(\overline{pq}.\overline{rs}\). This is exactly the average of the two numbers \(\overline{pq}\) and \(\overline{rs}\). What is the sum of the digits of N?
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Answer: B — 18
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Hint 1 of 2
Turning pq.rs into an average gives an equation linking the two-digit blocks pq and rs.
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Hint 2 of 2
From pq + rs/100 = (pq + rs)/2 you get 50·pq = 49·rs, which pins down pq and rs.
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Approach: translate the average condition into an equation
The number with the decimal is pq + rs/100, and it equals (pq + rs)/2.
A function \(f:\mathbb{R}\to\mathbb{R}\) fulfils the condition \(f(20-x)=f(22+x)\) for all real numbers x. It is known that f has exactly two real zeros. What is the sum of the two zeros?
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Answer: E — another number
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Hint 1 of 2
The condition says f takes equal values at points that are mirror images of each other.
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Hint 2 of 2
Find the axis of symmetry by averaging 20 − x and 22 + x; the two zeros are symmetric about it.
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Approach: locate the axis of symmetry
f(20 − x) = f(22 + x) means f is symmetric about the midpoint of 20−x and 22+x.
That midpoint is (20−x + 22+x)/2 = 21, so the graph is symmetric about x = 21.
Two zeros symmetric about 21 add to 2×21 = 42, which is another number.
A part of a polynomial of degree five is illegible due to an ink stain (see diagram). It is known that all zeros of the polynomial are integers. What is the highest power of \(x - 1\) that divides this polynomial?
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Answer: D — \((x-1)^4\)
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Hint 1 of 2
Vieta's formulas link the visible coefficients to the sum and product of the roots.
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Hint 2 of 2
All roots are integers, the product is 7 and the sum is 11 — that pins them down.
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Approach: recover the integer roots with Vieta's formulas
For x5 − 11x4 + ... − 7, the integer roots have product 7 and sum 11.
The only integer multiset is 7, 1, 1, 1, 1 (product 7, sum 11).
So (x−1) appears four times, and the highest power dividing it is (x−1)4.
The numbers 1 to 10 were written into the ten circles in the pattern shown in the picture. The sum of the four numbers in the left and the right column is 24 each and the sum of the three numbers in the bottom row is 25. Which number is in the circle with the question mark?
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Answer: E — another number
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Hint 1 of 2
All ten numbers add to 55; the two columns already use 24 + 24 = 48.
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Hint 2 of 2
That leaves 7 for the two circles outside the columns; the bottom-row total of 25 then pins each one.
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Approach: use total 55 and the column/row sums
1+2+···+10 = 55. The left and right columns (four each) take 24 + 24 = 48.
The remaining two circles sum to 55 − 48 = 7; with the bottom row equal to 25 they are forced to be 6 and 1.
The question-mark circle is the one equal to 1, which is none of 2, 4, 5, 6, so the answer is another number.
The function f is such that \(f(x+y) = f(x) \cdot f(y)\) and \(f(1) = 2\). What is the value of \(\dfrac{f(2)}{f(1)} + \dfrac{f(3)}{f(2)} + \cdots + \dfrac{f(2021)}{f(2020)}\)?
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Answer: E — none of the previous
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Hint 1 of 2
The rule f(x+y)=f(x)f(y) with f(1)=2 forces a familiar formula for f.
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Hint 2 of 2
Simplify a typical term f(n+1)/f(n) before adding.
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Approach: identify the exponential and collapse each term
From f(x+y)=f(x)f(y) and f(1)=2 we get f(n)=2ⁿ.
Each term f(n+1)/f(n) = 2ⁿ⁺¹/2ⁿ = 2, and there are 2020 terms (from f(2)/f(1) to f(2021)/f(2020)).
The sum is 2020 × 2 = 4040, which is not among A–D, so none of the previous.
Let \(M(k)\) be the maximum value of \(\left|\,4x^{2} - 4x + k\,\right|\) for x in the interval \([-1,1]\), where k can be any real number. What is the minimum possible value of \(M(k)\)?
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Answer: B — \(\tfrac{9}{2}\)
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Hint 1 of 2
Find the range of g(x) = 4x² − 4x on [−1,1]; then 4x² − 4x + k just shifts that range by k.
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Hint 2 of 2
The maximum of the absolute value is the larger of the two endpoint distances; choose k to balance them.
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Approach: find the range, then minimise the larger absolute endpoint
On [−1,1], g(x)=4x²−4x ranges from −1 (at x=½) to 8 (at x=−1), so 4x²−4x+k lies in [k−1, k+8].
Thus M(k) = max(|k−1|, |k+8|); this is smallest when k−1 = −(k+8), i.e. k = −7/2.
Four different straight lines pass through the origin of the coordinate system. They intersect the parabola \(y = x^{2} - 2\) at eight points. What could the product of the \(x\)-coordinates of these eight points be?
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Answer: A — only 16
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Hint 1 of 2
A line \(y = mx\) through the origin meets \(y = x^{2} - 2\) where \(x^{2} - mx - 2 = 0\); by Vieta the two \(x\)-roots multiply to \(-2\).
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Hint 2 of 2
Each of the four lines contributes a root-product of \(-2\), so just multiply across the four lines.
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Approach: use Vieta on each line–parabola intersection
A line \(y = mx\) meets \(y = x^{2} - 2\) where \(x^{2} - mx - 2 = 0\), whose two roots multiply to \(-2\) (the constant term).
The four lines give four such pairs, each with root-product \(-2\).
So the product of all eight \(x\)-coordinates is \((-2)^{4} = 16\), no matter which four lines are chosen.
The sequence \(a_1, a_2, a_3, \ldots\) starts with \(a_1 = 49\). To find \(a_{n+1}\) for \(n \ge 1\), you add 1 to the digit sum of \(a_n\) and square the result. For example, \(a_2 = (4 + 9 + 1)^{2} = 196\). Find \(a_{2019}\).
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Answer: C — 64
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Hint 1 of 2
Compute terms in order — \(a_1 = 49\), \(a_2 = 196\), \(a_3 = 289\), \(a_4 = 400\), \(a_5 = 25\), … — and watch for a repeating cycle.
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Hint 2 of 2
Once the sequence cycles, locate term 2019 by its position within the cycle.
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Approach: iterate until the sequence settles into a cycle
The faces of the prism shown are made up of two triangles and three squares. The six vertices are labelled using the numbers 1 to 6. The sum of the four numbers around each square is always the same. The numbers 1 and 5 are given in the diagram. Which number is written at vertex X?
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Answer: A — 2
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Hint 1 of 3
Add up all six labels, and use that each vertex sits on exactly two of the three square faces.
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Hint 2 of 3
That forces each pair of vertices joined by a vertical edge to add to the same total.
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Hint 3 of 3
Find that common edge-sum, then look at the edge through vertex 5.
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Approach: double counting forces each vertical edge of the prism to have the same vertex-sum
The labels 1–6 total 21. Each vertex lies on exactly two of the three square faces, so the three equal square sums total 2·21 = 42, giving each square sum 14.
Each square face is built from two vertical edges, and comparing the three faces shows the two endpoints of every vertical edge must add to the same value; with total 21 over three edges that value is 21/3 = 7.
So the vertical edges pair the numbers as (1,6), (2,5), (3,4); in the picture X sits directly above 5 on one vertical edge, so X + 5 = 7.
A function \(f\) fulfils the property \(f(x + y) = f(x) \cdot f(y)\) for all whole numbers \(x\) and \(y\). Furthermore \(f(1) = \tfrac{1}{2}\). Determine the value of the expression \(f(0) + f(1) + f(2) + f(3)\).
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Answer: D — \(\tfrac{15}{8}\)
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Hint 1 of 2
Plug in x = y = 0 to pin down f(0).
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Hint 2 of 2
Build f(2) and f(3) by repeatedly using f(a+b) = f(a)f(b).
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Approach: exploit the multiplicative functional equation
Setting x = y = 0 gives f(0) = f(0)², so f(0) = 1.
A quadratic function of the form \(f(x) = x^2 + px + q\) intersects the x-axis and the y-axis in three different points. The circle through these three points intersects the graph of \(f\) in a fourth point. What are the coordinates of this fourth point of intersection?
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Answer: C — \((-p \mid q)\)
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Hint 1 of 2
The four intersection x-values of circle and parabola are the roots of one quartic.
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Hint 2 of 2
Use the sum of those roots; three of them you already know.
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Approach: sum of roots of the circle-parabola quartic
Substituting y = x²+px+q into the circle equation gives a quartic in x whose four roots sum to −2p.
Three known intersection x-values are the two parabola roots (summing to −p) and 0.
So the fourth x-value is −2p −(−p) = −p, and y = f(−p) = q.
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 \(9m = 504\).
The equations \(x^2 + ax + b = 0\) and \(x^2 + bx + a = 0\) both have real solutions. It is known that the sum of the squares of the solutions of the first equation is equal to the sum of the squares of the solutions of the second equation, and that \(a \ne b\). Then \(a + b\) equals
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Answer: B — -2
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Hint 1 of 3
By Vieta, the sum of the squares of the roots is \((\text{sum})^2 - 2(\text{product})\).
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Hint 2 of 3
Set the two sums-of-squares equal, then factor the resulting symmetric equation.
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Hint 3 of 3
The condition \(a \ne b\) lets you cancel one factor.
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Approach: Vieta plus factoring
For \(x^2+ax+b\) the roots have sum \(-a\) and product \(b\), so their squares sum to \(a^2 - 2b\); for \(x^2+bx+a\) it is \(b^2 - 2a\).
The mapping \(f:\mathbb{Z} o\mathbb{Z}\) fulfils the conditions \(f(4)=6\) and \(xf(x)=(x-3)f(x+1)\). What is the value of the expression \(f(4) imes f(7) imes f(10) imes\ldots imes f(2011) imes f(2014)\)?
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Answer: D — \(2013!\)
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Hint 1 of 2
The relation x·f(x) = (x−3)·f(x+1) lets you step f from one integer to the next.
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Hint 2 of 2
When you multiply the wanted terms, look for a massive telescoping cancellation.
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Approach: use the recurrence and telescope the product
The condition gives f(x+1) = x·f(x)/(x−3), so with f(4)=6 every value of f is determined.
Forming f(4)·f(7)·f(10)·…·f(2014) and simplifying, the fractions telescope.
After an especially intense lesson the graph of the function y = x² was still on the board as well as 2012 straight lines parallel to the straight line with the equation y = x, which each intersected the parabola in two points. How big is the sum of all x-coordinates of the intersections of the straight lines with the parabola?
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Answer: D — 2012
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Hint 1 of 2
Each line parallel to \(y = x\) has the form \(y = x + c\); intersect it with \(y = x^2\).
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Hint 2 of 2
The two \(x\)-values on one line are the roots of \(x^2 - x - c = 0\) — use the sum of roots, which is the same for every line.
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Approach: sum of roots per line, added over all lines
A line \(y = x + c\) meets \(y = x^2\) where \(x^2 - x - c = 0\), whose two roots sum to \(1\) (independent of \(c\)).
So each of the 2012 lines contributes \(1\) to the running total of \(x\)-coordinates.
The overall sum is \(2012 \cdot 1 = 2012\), choice D.
In the sequence 1, 1, 0, 1, −1, … the first two terms a1 and a2 are each 1. The third term is the difference of the previous two and a3 = a1 − a2 holds true. The fourth one is the sum of the previous two with a4 = a2 + a3, the fifth is the difference a5 = a3 − a4, a6 = a4 + a5, and so on, as well as the alternating difference and the sum. How big is the sum of the first 100 terms of this sequence?
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Answer: B — 3
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Hint 1 of 2
Just generate terms, alternately subtracting then adding the previous two, until the pattern repeats.
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Hint 2 of 2
Find the period and the sum over one full period, then handle the leftover terms.
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Approach: find the period, then sum 100 terms
Listing terms gives \(1,1,0,1,-1,0,-1,-1,0,-1,1,0\), after which \(a_{13}=1, a_{14}=1\) repeat the start, so the period is 12.
One full period sums to \(0\), so the first \(96 = 8\times12\) terms contribute \(0\).
The remaining four terms \(a_{97},\dots,a_{100}\) match \(a_1,\dots,a_4 = 1,1,0,1\), summing to \(3\), choice B.
Consider the two arithmetic sequences 5, 20, 35, … and 35, 61, 87, …. How many different arithmetic sequences of positive whole numbers have both of these as subsequences?
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Answer: C — 5
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Hint 1 of 2
For an arithmetic sequence to contain another as a subsequence, its common difference must divide the other's difference.
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Hint 2 of 2
Here the difference must divide both 15 and 26, whose gcd is 1.
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Approach: the super-sequence's step must divide both 15 and 26
To contain the step-15 sequence its difference d divides 15; to contain the step-26 sequence d divides 26.
Since gcd(15, 26) = 1, d = 1, so the super-sequence runs through consecutive integers.
Starting at any of 1, 2, 3, 4, 5 (it must reach 5) gives 5 such sequences.
(Note: the official key also accepts 'infinite' under a looser reading of the question.)
The sequence of functions \(f_{1}(x),\,f_{2}(x),\,\ldots\) satisfies \(f_{1}(x)=x\) and \(f_{n+1}(x)=\dfrac{1}{1-f_{n}(x)}\). Determine the value of \(f_{2011}(2011)\).
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Answer: A — 2011
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Hint 1 of 2
Compute f₂, f₃, f₄ and watch for a repeat.
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Hint 2 of 2
The map cycles with period 3, so reduce 2011 modulo 3.
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Approach: detect the period-3 cycle
f₁(x)=x, f₂=1/(1−x), f₃=(x−1)/x, and f₄=x again — period 3.
An airline does not charge for luggage below a certain weight; for each additional kg there is a charge. Mr. and Mrs. Raiss had 60 kg of luggage and paid 3 €. Mr. Wander also had 60 kg of luggage but had to pay 10.50 €. How many kg of luggage per passenger were carried free?
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Answer: D — 25
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Hint 1 of 2
Mr. and Mrs. Raiss are two passengers; Mr. Wander is one — they share the same free allowance and rate.
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Hint 2 of 2
Set up the two charge equations and divide them to remove the rate.
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Approach: two charge equations, eliminate the per-kg rate by dividing
Let f be the free kg per passenger and c the rate. Raiss: c(60 − 2f) = 3; Wander: c(60 − f) = 10.5.
Dividing: (60 − f)/(60 − 2f) = 3.5, so 60 − f = 210 − 7f, giving 6f = 150.
The numbers from 1 to 10 are written 10 times each on a board. Now the children play the following game: one child deletes two numbers off the board and writes instead the sum of the two numbers minus 1. Then a second child does the same, and so forth until there is only one number left on the board. The last number is
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Answer: B — 451.
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Hint 1 of 2
Each move replaces two numbers with one, so track how the count and the total change.
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Hint 2 of 2
The total drops by exactly 1 every move, regardless of which numbers are chosen.
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Approach: track the invariant: total minus number of moves
The starting numbers sum to 10×(1+...+10) = 550, and there are 100 numbers.
Each move removes one number and lowers the total by 1; reaching one number takes 99 moves.
A function maps all positive real numbers to real numbers. For all \(x\in\mathbb{R}^{+}\) the following holds true: \(2f(x)+3f\!\left(\dfrac{2010}{x}\right)=5x\). Determine the value of f(6).
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Answer: A — 993
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Hint 1 of 2
Replace x with 2010/x to get a second equation in the same two unknowns.
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Hint 2 of 2
Solve the pair for f(x), then plug in x = 6.
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Approach: substitute x -> 2010/x and solve the system
The given equation is 2f(x) + 3f(2010/x) = 5x.
Swapping x and 2010/x gives 2f(2010/x) + 3f(x) = 5·2010/x.
