20th International Mathematics and Science Olympiad (IMSO) – Mathematics Short Answer Problems for Primary School

Precisely, Andrea and Charlotte cannot be said to have an exact age difference of 4 years, unless they were born on the same date and month. Differences in birth dates and months could make their age difference less than 4 years, exactly 4 years, or more than 4 years. However, one thing is certain: their age difference is definitely not less than 3 years.
Thus, the age difference of the two sisters is therefore not less than 3 years in any case.
(Answer E)

Given that the average of $20$ numbers is $18.$ This means that the sum of these $20$ numbers can be calculated as $20 \times 18 = 360.$ Now, each number in the sequence is increased according to a pattern: the first number is increased by $2,$ the second by $4,$ the third by $6,$ and so on, until the twentieth number is increased by $40.$ The total increase in the sum can be found by summing the sequence
$$2 + 4 + 6 + \cdots + 40.$$This sequence is an arithmetic series with the first term $a = 2,$ the common difference $d = 2,$ and the number of terms $n = 20.$ The sum of an arithmetic series is given by
$$S_n = \dfrac{n}{2} \times (a + \text{U}_n)$$where $\text{U}_n$ is the last term. Substituting the given values:
$$S_{20} = \dfrac{20}{2} \times (2 + 40) = 10 \times 42 = 420.$$Thus, the new total sum of the numbers after the increases is $360 + 420 = 780.$ The new average is then $\dfrac{780}{20} = 39.$
Thus, the new average of the $20$ numbers is $39.$

We are given twelve consecutive positive integers written on a board. Let these integers be
$$n, n+1, n+2, \cdots, n+11.$$The sum of these twelve consecutive integers can be computed using the formula for the sum of an arithmetic sequence as follows.
$$S = \dfrac{12}{2}(n+(n+11) = 12n+66.$$Maria erases one of the numbers, say $n+y,$ where $y$ is an integer between $0$ and $11$ (inclusively). The sum of the remaining numbers is given as $2023.$ This leads to the equation
$$12n+66-(n+y) = 2023$$which simplifies to
$$11n-y = 1957.$$We need to find an integer $n$ such that the equation holds. Noting that $11 \times 178 = 1958,$ we set
$$11n = 1958 \Rightarrow n = 178.$$Consequently, we have $y = 1.$ Since the erased number is $n + y,$ we find $178 + 1 = 179$ as the erased number.
Verify the solution:
The original sequence is $178, 179, \cdots, 189.$ If we remove $179,$ the sum of the remaining numbers should be $2202-179 = 2023,$ which matches the given condition.
Thus, the number that Maria erased is $\boxed{179}.$

The circular chart provided in option A is the diagram that best corresponds to the given bar chart. We can determine this by matching the sector areas of the four colors in the circular chart with the heights of the bars in the bar chart.
(Answer A)

By applying distributive property of integers and the concept of sum of an arithmetics series, we have
$$\begin{aligned} \dfrac{2001+2002+\cdots+2031}{31} & = \dfrac{(2000+1)+(2000+2)+\cdots+(2000+31)}{31} \\ & = \dfrac{31 \times 2000 + (1+2+\cdots+31)}{31} \\ & = \dfrac{31 \times 2000 + \frac{31}{2}(1+31)}{31} \\ & = 2000 + \dfrac12(32) \\ & = 2016. \end{aligned}$$Thus, the result we get is $\boxed{2016}.$
(Answer D)

We can draw figures A, C, and D by tracing a continuous line without lifting the pen, following the numbered sequence as marked on each figure in order.

However, figure B cannot be drawn using a single continuous line without retracing a segment. Therefore, there are 3 figures that can be drawn with one continuous line without retracing any segment.
(Answer D)

The paper is folded along the dashed lines one after another, in any order or direction. This means that the paper undergoes multiple folds, but all folds originate from a single sheet. As a result, multiple layers may be stacked at the location of the cut. When a single cut is made at the folded corner, it creates only one hole at the intersection of the dashed lines.
(Answer B)

The shape needed to cover the outside of a truncated cone (excluding the base) is a sector of a larger circle with a smaller circular section removed. The reasons are as follow.

1. When unwrapped, the lateral surface of a truncated cone forms a frustum (3D shape that is formed when a cone or pyramid is cut by a plane parallel to its base. This results in a shape with two circular (or polygonal) bases—one larger and one smaller—connected by a slanted surface), which can be developed into a sector of a circle. 
2. The larger arc corresponds to the larger base circumference of the truncated cone.
3. The smaller arc corresponds to the smaller base circumference.
4. The radii of the sector represent the slant height of the frustum.

(Answer E)

Suppose the diameters of the semicircles with areas $X$ cm², $Y$ cm², dan $Z$ cm² are $a, b,$ dan $c,$ respectively. Since the triangle in the figure is a right triangle, the Pythagorean theorem applies, namely $c^2 = a^2 + b^2.$
Next, note that
$$\begin{aligned} X + Y & = \dfrac12\left(\dfrac14\pi a^2\right) + \dfrac12\left(\dfrac14\pi b^2\right) \\ & = \dfrac12 \cdot \dfrac14 \pi(a^2 + b^2) \\ & = \dfrac12 \cdot \dfrac14 \pi c^2 \\ & = Z. \end{aligned}$$Thus,, the necessarily true statement is $\boxed{X+Y=Z}.$
(Answer C)

A convex quadrilateral is a four-sided polygon in which all interior angles are less than $180^\circ,$ and its diagonals lie entirely within the quadrilateral. This means that for any two points inside the quadrilateral, the line segment connecting them is also completely inside the quadrilateral.
Suppose the four interior angles of a convex quadrilateral are $a, b, c,$ and $d$ such that $a+b+c+d=360^\circ.$
Case 1: $0$ acute angles
This occurs when $a = b = c = d = 90^\circ,$ meaning the quadrilateral is actually a rectangle.
Case 2: $1$ acute angle
This is possible when $a + b= 180^\circ$ and $c = d = 90^\circ$ where one angle $(a)$ is less than $90^\circ,$ and the other $(b)$ is greater than $90^\circ.$
Case 3: $2$ acute angles
This is possible when $a + b= 180^\circ$ and $c + d = 180^\circ,$ where two angles ($a$ and $c$) are less than $90^\circ$, while the other two ($b$ and $d$) are greater than $90^\circ.$
Case 4: $3$ acute angles
This is possible when three angles $(a, b, c)$ are less than $90^\circ,$ while the fourth angle $(d)$ is greater than $90^\circ.$
Case 5: $4$ acute angles
This is impossible. If all four angles were acute $(a, b, c, d < 90^\circ)$, then their sum would be less than $360^\circ,$ which contradicts the required sum of a quadrilateral’s interior angles.
Thus, the complete list of the number of acute angles a convex quadrilateral can have is $\boxed{0,1,2,3}.$
(Answer B)

By applying some basic algebra and the identity $(a+1)^2 = a^2 + 2a + 1,$ we obtain:
$$\begin{aligned} & \sqrt{(2015+2015)+(2015-2015)+(2015\cdot2015)+(2015:2015)} \\ & = \sqrt{2 \cdot 2015 + 0 + 2015^2 + 1} \\ & = \sqrt{2015^2 + 2 \cdot 2015 + 1} \\ & = \sqrt{(2015 + 1)^2} \\ & = 2016. \end{aligned}$$Thus, the value of the expression is $\boxed{2016}.$
(Answer C)

The Cartesian plane is divided by the $X$-axis, the graphs of the functions defined by $f(x)=2x-2$ (a straight line) and $g(x) = x^2-1$ (a parabola). To determine the number of regions formed, we analyze their intersections and how they split the plane. First, the horizontal axis itself divides the plane into two main halves. Then, the parabola $g(x) = x^2-1$ intersects the $X$-axis at $x = \pm 1,$ creating a curved boundary. Lastly, the line $f(x) = 2x-2$ intersects the parabola at two points and the $X$-axis at $x = 1.$
By examining the intersections and considering how each function divides the plane, the total number of distinct regions formed is 7, as seen in the given image.
(Answer A)

Ella wants to assign a number to each circle in the diagram such that each number is the sum of its two neighboring numbers. Suppose the numbers in the circles, in clockwise order, are:
$$3, a, 5, b, c, d, e, f.$$The goal is to determine the value of $d,$ which is marked with a question mark in the figure. Since each number must be the sum of its two neighbors, we set up the following equations:
$$\begin{aligned} a & = 3 + 5 = 8 \\ b & = 5-8 = -3 \\ c & = -3-5 = -8 \\ d & = -8-(-3) = -5 \\ e & = -5-(-8) = 3 \\ f & = 3-(-5) = 8, \end{aligned}$$ respectively. However, this results in a contradiction, as $3 \neq = 8 + 8 = 16.$
Therefore, it is impossible for Ella to assign numbers to the circles while satisfying the given condition.
(Answer E)

We are given five different positive integers: $a, b, c, d,$ and $e.$ The equation $a+b=d$ implies that $d > a$ and $d > b.$ Similarly, $e-d=a$ (or $e = a+d$) implies $e > a$ and $e > d.$ Additionally, this confirms that $e \neq 1$ since $e$ must be greater than both $a$ and $d.$ The equation $c : e = b$ (interpreted as $c = b \cdot e$) implies that $c > e$ and $c > b.$
From the inequalities established:

1. $d$ is larger than $a$ and $b.$
2. $e$ is larger than $a$ and $d,$ meaning $e$ is larger than all numbers defined before it.
3. $c$ is larger than $e$ and $b,$ meaning $c$ is larger than all numbers mentioned.

Thus, among $a, b, c, d,$ and $e,$ the number $c$ is the largest.
(Answer C)

Suppose the first set consists of the numbers $a, b,$ and $c$ and the second set consists of $d, e,$ and $f.$ We are given $\sqrt[3]{abc} = 3$ and $\sqrt[3]{def} = 12.$ Now, the geometric mean of all six numbers is
$$\begin{aligned} \sqrt[6]{abcdef} & = \sqrt{ \sqrt[3]{abcdef}} \\ & = \sqrt{\sqrt[3]{abc} \cdot \sqrt[3]{def}} \\ & = \sqrt{3 \cdot 12} \\ & = \sqrt{36} \\ & = 6. \end{aligned}$$Thus, the geometric mean of the combined set of six numbers is $\boxed{6}.$
(Answer B)

Based on the given figure, there are three concentric circles, namely small, medium, and large circles, which have a radius of $r_1 = 1,$ $r_2,$ and $r_3,$ respectively. The area of the innermost region is one-fourth of the area of the small circle with a radius of $1,$ which is $$A_1 = \dfrac14 \pi (1)^2 = \dfrac14 \pi.$$The area of the middle shaded region can be calculated by subtracting the area of the small circle from the area of the medium circle, then dividing the result by four. Since this area is equal to the innermost area we just found, we get
$$\begin{aligned} \dfrac14\left(\pi r_2^2-\pi(1)^2\right) & = \dfrac14\pi = A_1 \\ r_2^2-1 & = 1 \\ r_2 & = \sqrt2. \end{aligned}$$Similarly, the outermost shaded area can be calculated by subtracting the area of the medium circle from the area of the large circle, then dividing the result by four.
$$\begin{aligned} \dfrac14\left(\pi r_3^2-\pi(\sqrt2)^2\right) & = \dfrac14\pi = A_1 \\ r_3^3-2 & = 1 \\ r_3 & = \sqrt3. \end{aligned}$$Therefore, the radii are $r_1 = 1,$ $r_2 = \sqrt2,$ and $r_3 = \sqrt3.$ Thus, the product of the three radii is $\boxed{\sqrt6}.$
(Answer A)

Let $x$ and $y$ be the amount paid for the first and second car, respectively. Since the dealer sold the first car for $140\%$ of its cost and the second car for $160\%$ of its cost, the total revenue from selling both cars is $140\%x + 160\%y.$ We are also given that the total revenue equals $154\%$ of the total cost, namely $154\%(x + y).$ Then, we set up the equation:
$$\begin{aligned} 140\%x + 160\%y & = 154\%(x + y) \\ 140x + 160y & = 154x + 154y \\ 6y & = 14x \\ \dfrac{x}{y} & = \dfrac{6}{14} = \dfrac37. \end{aligned}$$Thus, ratio of the prices the dealer paid for the first and the second car was $\boxed{3 : 7}.$
(Answer C)

There are five conditions in which Tina wins, corresponding to the following pairs of dice rolls from Bibi and Tina, respectively: $(1, 2),$ $(1, 5),$ $(2, 5),$ $(3, 5),$ and $(4, 5).$ Since Tina’s die contains the numbers $2$ and $5$ each appearing three times, each of these five winning conditions occurs in three different ways. Thus, the total number of favorable outcomes for Tina is $3 \times 5 = 15.$ Since there are $6 \times 6 = 36$ total possible outcomes when rolling both dice, the probability that Tina wins is $\dfrac{15}{36} = \dfrac{5}{12}.$
Thus, the probability that Tina wins is $\boxed{\dfrac{5}{12}}.$
(Answer C)

Each marble is numbered from $1$ to $2015.$ The color of a marble depends on its digit sum, which is the sum of its individual digits. The smallest possible digit sum occurs for the number 1, which has a digit sum of $1.$ To find the maximum digit sum within the given range ($1$ to $2015$), we consider the number where each digit is as large as possible. The largest number in the range is $2015$, but the number with the highest digit sum is $1999,$ namely $1+9+9+9=28.$ Since the digit sum can take any integer value from $1$ to $28$, there are $28$ different colors of marbles.
Thus, there are $\boxed{28}$ different colors of marbles in the cane.
(Answer C)

In the given figure, two identical standard dice are shown. The numbers $6$ and $4$ are visible on the top and front faces of the die, respectively. Since opposite faces add up to $7,$ the numbers on the bottom and back faces must be $1$ and $3,$ as $6+1=7$ and $4+3=7.$ Now, considering the right face (marked with a “?”), the two possible numbers are $2$ or $5,$ since the other die covers the left face. Therefore, the exact number on the right face cannot be determined.
(Answer C)

Observe that $$kx_1 + kx_2 + \cdots + kx_n = k(x_1 + x_2 + \cdots + x_n).$$ We will use this factorization to simplify the calculation.
First, note that $x = 1+2+\cdots+10=55.$
Now, consider the sum:
$$(1+2+\cdots+10) + (2+4+\cdots +20) + \cdots + (10+20+\cdots+100).$$Rewrite it using factorization,
$$x + 2x +\cdots + 10x = x(1+2+\cdots+10).$$Since $1+2+\cdots+10=x,$ we get
$$x \cdot x = 55 \times 55 = 3025.$$Thus, the sum of all $100$ products in the complete table is $\boxed{3025}.$
(Answer D)

The given equation of the curve is $$(x^2+y^2-2x)^2=2(x^2+y^2).$$To determine which of the lines represents the $Y$-axis, we analyze the symmetry of the equation. Since the equation contains only even powers of $y$ (i.e. $y^2$), replacing $y$ with $-y$ results in the same equation. This means the curve is symmetric with respect to the $X$-axis. Additionally, we rewrite $x^2+y^2-2x$ as $(x-1)^2 + y^2-1.$ This suggests that the equation is centered around $(1, 0)$ rather than the origin, but still has vertical symmetry because the terms remain unchanged when reflecting across the $Y$-axis.
From the figure, the vertical symmetry line corresponds to line a, as it divides the curve into two symmetric halves along the vertical direction. Since the $Y$-axis is the natural vertical symmetry axis in the Cartesian plane, we conclude that the correct answer is line a.
(Answer A)

When reading the statements from option A to option E, we determine the first true statement as follows:

1. If option A is true, then option C must also be true, which in turn implies that option E is false. However, this contradicts the fact that $1+1=2$ is true. Therefore, option A must be false.
2. If option B is true, then option A must also be true, leading to the same contradiction as before. Thus, option B must be false.
3. If option C is true, then option E must be false, again resulting in the same contradiction. Therefore, option C must be false.
4. If option D is true, then option B must be false, which implies that option A is false and option C is also false. As a result, option E must be true, and no contradiction arises, since $1+1=2$ is indeed true.

Thus, the first true statement when reading the options from A to E is the statement in option D.
(Answer D)

Notice that a positive integer $$n = p_1^{k_1} \cdot p_2^{k_2} \cdot p_m^{k_m}$$ has $(k_1+1)(k_2+1)\cdots(k_m+1)$ positive factors, where $p_i$ is a prime number and $k_i$ is a natural number for $i = 1, 2, \cdots, m.$ We will use this fact to solve the problem.
Since an $n$-gon (a regular polygon with $n$ sides) has a total sum of interior angles given by
$$(n-2) \cdot 180^\circ,$$each interior angle measures
$$\dfrac{(n-2) \cdot 180^\circ}{n} = 180^\circ-\dfrac{360^\circ}{n} = k.$$For $k$ to be a positive integer, $n$ must be a divisor of $360$ with the condition that $n > 2.$ Since
$$360 = 2^3 \cdot 3^2 \cdot 5,$$the number of positive factors of $360$ is
$$(3+1)(2+1)(1+1)=4\cdot 3 \cdot 2 = 24.$$Excluding $1$ and $2,$ there are $22$ possible values of $n.$ Thus, there are $22$ regular polygons where each interior angle (in degrees) is an integer.
(Answer C)

To determine how many three-digit positive integers can be represented as the sum of exactly nine different powers of $2,$ we analyze the properties of binary representations. Since we are summing nine distinct powers of $2,$ the integer must have exactly nine 1’s in its binary form. This means it must be of the form:
$$2^{a_1} + 2^{a_2} + \cdots + 2^{a_9}$$where $$a_1 > a_2 > \cdots > a_9 \ge 0.$$The smallest possible three-digit number with exactly nine 1’s in binary occurs when we take the nine smallest consecutive powers of 2 within the three-digit range. This gives
$$2^8 + 2^7 + \cdots + 2^1 + 2^0 = 511.$$The smallest possible three-digit number with exactly nine 1’s in binary occurs when we take the smallest eight consecutive powers of $2,$ plus $2^9$ as the starting number (since upper exponents would form a four-digit number). This gives
$$2^9 + 2^7 + 2^6 + \cdots + 2^1 + 2^0 = 767.$$Thus, there are only two three-digit numbers that can be expressed as the sum of exactly nine different powers of $2,$ which is $511$ and $767.$ Hence, the answer is $\boxed{2}.$
(Answer B)