[i]20 problems for 25 minutes.[/i]


[b]p1.[/b] Compute the value of $2 + 20 + 201 + 2016$.


[b]p2.[/b] Gleb is making a doll, whose prototype is a cube with side length $5$ centimeters. If the density of the toy is $4$ grams per cubic centimeter, compute its mass in grams.


[b]p3.[/b] Find the sum of $20\%$ of $16$ and $16\%$ of $20$.


[b]p4.[/b] How many times does Akmal need to roll a standard six-sided die in order to guarantee that two of the rolled values sum to an even number?


[b]p5.[/b] During a period of one month, there are ten days without rain and twenty days without snow. What is the positive difference between the number of rainy days and the number of snowy days?


[b]p6.[/b] Joanna has a fully charged phone. After using it for $30$ minutes, she notices that $20$ percent of the battery has been consumed. Assuming a constant battery consumption rate, for how many additional minutes can she use the phone until $20$ percent of the battery remains?


[b]p7.[/b] In a square $ABCD$, points $P$, $Q$, $R$, and $S$ are chosen on sides $AB$, $BC$, $CD$, and $DA$ respectively, such that $AP = 2PB$, $BQ = 2QC$, $CR = 2RD$, and $DS = 2SA$. What fraction of square $ABCD$ is contained within square $PQRS$?


[b]p8.[/b] The sum of the reciprocals of two not necessarily distinct positive integers is $1$. Compute the sum of these two positive integers.


[b]p9.[/b] In a room of government officials, two-thirds of the men are standing and $8$ women are standing. There are twice as many standing men as standing women and twice as many women in total as men in total. Find the total number of government ocials in the room.


[b]p10.[/b] A string of lowercase English letters is called pseudo-Japanese if it begins with a consonant and alternates between consonants and vowels. (Here the letter "y" is considered neither a consonant nor vowel.) How many $4$-letter pseudo-Japanese strings are there?


[b]p11.[/b] In a wooden box, there are $2$ identical black balls, $2$ identical grey balls, and $1$ white ball. Yuka randomly draws two balls in succession without replacement. What is the probability that the first ball is strictly darker than the second one?


[b]p12.[/b] Compute the real number $x$ for which $(x + 1)^2 + (x + 2)^2 + (x + 3)^2 = (x + 4)^2 + (x + 5)^2 + (x + 6)^2$.


[b]p13.[/b] Let $ABC$ be an isosceles right triangle with $\angle C = 90^o$ and $AB = 2$. Let $D$, $E$, and $F$ be points outside $ABC$ in the same plane such that the triangles $DBC$, $AEC$, and $ABF$ are isosceles right triangles with hypotenuses $BC$, $AC$, and $AB$, respectively. Find the area of triangle $DEF$.


[b]p14.[/b] Salma is thinking of a six-digit positive integer $n$ divisible by $90$. If the sum of the digits of n is divisible by $5$, find $n$.


[b]p15.[/b] Kiady ate a total of $100$ bananas over five days. On the ($i + 1$)-th day ($1 \le i \le 4$), he ate i more bananas than he did on the $i$-th day. How many bananas did he eat on the fifth day?


[b]p16.[/b] In a unit equilateral triangle $ABC$; points $D$,$E$, and $F$ are chosen on sides $BC$, $CA$, and $AB$, respectively. If lines $DE$, $EF$, and $FD$ are perpendicular to $CA$, $AB$ and $BC$, respectively, compute the area of triangle $DEF$.


[b]p17.[/b] Carlos rolls three standard six-sided dice. What is the probability that the product of the three numbers on the top faces has units digit 5?


[b]p18.[/b] Find the positive integer $n$ for which $n^{n^n}= 3^{3^{82}}$.


[b]p19.[/b] John folds a rope in half five times then cuts the folded rope with four knife cuts, leaving five stacks of rope segments. How many pieces of rope does he now have?


[b]p20.[/b] An integer $n > 1$ is conglomerate if all positive integers less than n and relatively prime to $n$ are not composite. For example, $3$ is conglomerate since $1$ and $2$ are not composite. Find the sum of all conglomerate integers less than or equal to $200$.


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To solve the problem, we need to find the area of triangle \( DEF \) given the specific geometric conditions. Let's break down the problem step by step.

1. **Understanding the Geometry:**
   - We are given an isosceles right triangle \( \triangle ABC \) with \( \angle C = 90^\circ \) and \( AB = 2 \).
   - Points \( D, E, \) and \( F \) are such that \( \triangle DBC, \triangle AEC, \) and \( \triangle ABF \) are isosceles right triangles with hypotenuses \( BC, AC, \) and \( AB \) respectively.

2. **Determine the Side Lengths:**
   - Since \( \triangle ABC \) is an isosceles right triangle with \( AB = 2 \), the legs \( AC \) and \( BC \) are equal. Using the Pythagorean theorem:
     \[
     AC = BC = \frac{AB}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}
     \]

3. **Constructing the Isosceles Right Triangles:**
   - \( \triangle DBC \) is an isosceles right triangle with hypotenuse \( BC = \sqrt{2} \). Therefore, the legs \( BD \) and \( DC \) are:
     \[
     BD = DC = \frac{BC}{\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2}} = 1
     \]
   - Similarly, \( \triangle AEC \) and \( \triangle ABF \) are isosceles right triangles with hypotenuses \( AC \) and \( AB \) respectively. Thus:
     \[
     AE = EC = 1 \quad \text{and} \quad AF = BF = \frac{AB}{\sqrt{2}} = \sqrt{2}
     \]

4. **Forming the Rectangle:**
   - Since \( \triangle ABC \) is isosceles and right, and the triangles \( \triangle DBC, \triangle AEC, \) and \( \triangle ABF \) are isosceles right triangles, the points \( D, E, \) and \( F \) form a rectangle \( AFBDE \) with:
     \[
     AB = 2 \quad \text{and} \quad BD = 1
     \]

5. **Calculating the Area of \( \triangle DEF \):**
   - The triangle \( DEF \) is formed within the rectangle \( AFBDE \). The base \( EF \) of \( \triangle DEF \) is equal to \( AB = 2 \), and the height from \( D \) to \( EF \) is also \( 2 \) (since \( D \) is directly above the midpoint of \( EF \)).
   - The area of \( \triangle DEF \) is given by:
     \[
     \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 2 = 2
     \]

\(\blacksquare\)

The final answer is \( \boxed{2} \).