Author: Sukardi

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)

We start with the given expression:
$$(a-b)^5 + (b-a)^5.$$Since $ (b-a) = -(a-b) $, we substitute:
$$(b-a)^5 = (-(a-b))^5 = – (a-b)^5.$$Thus, the expression simplifies as follows:
$$(a-b)^5 + (b-a)^5 = (a-b)^5 -(a-b)^5 = 0.$$Therefore, the simplest form of the given expression is $\boxed{0}.$
(Answer A)

Given the equation $2^{2x} = 4^{x+1}.$ Since $4 = 2^2,$ we obtain
$$\begin{aligned} 2^{2x} & = (2^2)^{x+1} \\ 2^{2x} & = 2^{2x+2} \\ & = 2x & = 2x + 2 \\ 0 & = 2. \end{aligned}$$Simplifying this equation implies a false statement, namely $0 = 2.$ Therefore, it can be concluded that the equation has no solution.
(Answer A) 

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)

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)

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)

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)

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)

You should note that every square is also a rectangle. Let:

1. $a$ be the number of blue squares,
2. $b$ be the number of blue rectangles that are not squares,
3. $𝑐$ be the number of red squares,
4. $d$ be the number of red rectangles that are not squares.

From the problem’s conditions, we establish the following equations:

1. The total number of squares is given as $7$, so we set the first equation:
   $$a+c=7$$
2. The number of red rectangles exceeds the number of blue squares by $3$, so we set the second equation:
   $$c+d=a+3$$
3. The number of red squares exceeds the number of blue rectangles by $2$, so we set the third equation:
   $$c = a + b+2$$

Using the first and third equation to eliminate $c,$ we get
$$2a+b = 5.$$Since $c = 7-a$ (from first equation) and $a+3 = c + d \ge c$ (from second equation), we have
$$a + 3 \ge 7-a.$$Consequently, we have $a \ge 2.$For $a$ and $b$ to be nonnegative integers, in $2a+b=5,$ we have $a \le 2.$ Thus, $a$ must be $2$ and then $b = 1.$ Additionally, we obtain $c = 5$ and $d = 0.$
Thus, the number of blue rectangles (both squares and non-square rectangles) is $\boxed{a+b = 2+1=3}.$
(Answer B)