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There is symmetry, and so whatever fraction of the top-left quarter is shaded, is the same as the fraction of the entire square that is shaded.

Focus on the top-left quarter. Consider moving a single triangle. The shaded area in the problem is exactly the same as the shaded area in this diagram (and the top-left quarter is outlined in bold):

But then it is obvious that exactly half of the top-left quarter is shaded, and so the answer is $50\%,$ which is choice B.

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We just need to count how many of each type of glyph there are.

There is $1$ glyph worth $10,000.$

There are $4$ glyphs worth $100$ each.

There are $2$ glyphs worth $10$ each.

There are $3$ glyphs worth $1$ each.

So, the answer is $10,423,$ which is choice B.

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At the start, there are $4$ total people (Annika plus $3$ friends), each with $15$ cards, so there are $4 \times 15 = 60$ cards in total. If $2$ more friends join, there will be $6$ people in total, and so each should get $60 \div 6 = 10$ cards, which is choice C.

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There will be $10 - 1 = 9$ gaps between his first number $100$ and his $10$th number. Each gap has size $7.$

So, he will subtract a total of $9 \times 7 = 63$ from $100,$ leaving an answer of $100 - 63 = 37,$ which is choice B.

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The key idea is that if driving from coordinates $(x_1, y_1)$ to $(x_2, y_2),$ then the shortest distance is $|x_2 - x_1| + |y_2 - y_1|.$ This is often called the Manhattan distance. It is also equal to the number of horizontal blocks between the locations, plus the number of vertical blocks between the locations.

The shortest distance from $F$ to $A$ is then $1 + 2 = 3.$

The shortest distance from $A$ to $B$ is $7 + 3 = 10.$

The shortest distance from $B$ to $C$ is $2 + 4 = 6.$

The shortest distance from $C$ to $F$ is $4 + 1 = 5.$

Adding up all of these numbers, we get $3 + 10 + 6 + 5 = 24,$ which is choice C.

One small possible shortcut for the solution is to notice that when going from $B$ to $C$ to $F,$ the visit to $C$ is conveniently along a shortest path from $B$ to $F$ anyway, so we can even remove the requirement to stop at $C$ from the problem.

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The remainders of the five numbers after dividing by $4$ are: $3,$ $0,$ $1,$ $2,$ and $3.$ So, the sum of all five numbers modulo $4$ is the same as $3 + 0 + 1 + 2 + 3 = 9$ which has remainder $1$ modulo $4.$ Therefore, in order to erase a single number and get a sum that is $0$ modulo $4,$ we must erase the number which was $1$ modulo $4,$ which was $17.$ Therefore, the answer is C.

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Notice that the numbers of students in each category keeps increasing, which makes sense because the categories are getting broader.

We want to include all $50$ of the students who earned a score of at least $80\%$ but exclude all $13$ of the students who earned a score of at least $90\%.$ So, the answer is $50 - 13 = 37,$ which is choice D.

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There are $6$ squares, so each has area $18 \div 6 = 3.$ Then the side length of the cube is $\sqrt{3}.$ The volume is $\sqrt{3} \times \sqrt{3} \times \sqrt{3} = 3\sqrt{3},$ which is choice A.

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The answer is the same as the average of all $12$ numbers. To understand why this is the case, it is actually generally true that if a whole bunch of numbers is split up into pairs, and each pair is averaged, and then all those pair-averages are averaged, the answer is the average of all the numbers.

To see why this is true, consider a smaller example of $6$ numbers $a,$ $b,$ $c,$ $d,$ $e,$ and $f.$ The average of the pair-averages is: $$\frac{1}{3} \left( \frac{a+b}{2} + \frac{c+d}{2} + \frac{e+f}{2} \right)$$ and that is equal to $$\frac{a+b+c+d+e+f}{6}.$$ The same type of simplification happens with $12$ numbers.

Since the $12$ numbers are equally spaced (in an arithmetic progression), the answer is the same as the average of the first and last number, which is $$\frac{1+12}{2} = \frac{13}{2} = 6.5,$$ or choice B.

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The easiest way to solve this problem is using the Inclusion-Exclusion formula. That is: add the areas of the two rectangles, and then subtract the overlapping (square) area.

Each rectangle has area $5 \times 3 = 15.$

Their overlap is a square that has side length $2.5,$ and so its area is $2.5^2 = 6.25.$

Therefore, the total area is $15 + 15 - 6.25 = 23.75,$ which is choice D.

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A little trial-and-error suggests placing the S tetromino in a way that does not block too much, like this:

It is then easy to see that the remainder can be partitioned into two L tetrominoes, and so the answer is C.

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The corners of the region which are closest to the center are the $8$ points which lie on this circle:

By the Pythagorean Theorem, each of these points has this distance from the center: $$\sqrt{1^2 + 2^2} = \sqrt{5}.$$ The area of the circle is then $$\pi (\sqrt{5})^2 = 5\pi,$$ which is choice C.

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The remainders go in order $2,$ $4,$ $6,$ $1,$ $3,$ $5,$ $0,$ and then repeat.

All of the answer choices have $3$ bars of height $3$ and $4$ bars of height $4.$

So, we should pick the answer choice where the taller bars are on the first remainders to appear in our order, which are $2,$ $4,$ $6,$ and $1.$ That is option A.

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After inserting $N$ into the list, there will be $6$ total numbers. That is even, so the median will be the average of the middle two numbers.

All of the answer choices are at least $7,$ so when they are inserted into the list, the middle two numbers will be $7$ and $7.$ It is convenient that the median will always be $7,$ no matter which answer choice is picked.

The mean becomes twice the median, which is $2 \times 7 = 14.$

To have a total of $6$ numbers with mean $14,$ their sum must become $6 \times 14 = 84.$

The sum of the original $5$ numbers is $$2 + 6 + 7 + 7 + 28 = 50,$$ so the answer is $84 - 50 = 34,$ which is choice E.

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The number of gold squares is $$6 \times 6 - 13 = 36 - 13 = 23.$$ The $36$ total squares overlap as $18$ pairs.

To minimize the number of pairs with two gold squares, the gold squares should first be spread out across all pairs. That uses up $18$ of them. The remaining $23 - 18 = 5$ gold squares double-up and create a total of $m = 5$ gold-on-gold pairs.

To maximize the number of pairs with two gold squares, the $23$ gold squares should first be paired up as much as possible. That can be done to create $M = 11$ pairs, with $1$ gold square left over, because $23 \div 2$ is $11$ with a remainder of $1.$

The answer is $m + M = 5 + 11 = 16,$ which is choice C.

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Call the integers from $1$ to $10$ inclusive the lower range, and call the integers from $11$ to $20$ inclusive the higher range.

Each of the $5$ distinct numbers chosen from the lower range blocks out the number in the higher range that is exactly $10$ more than itself. There are only $10$ numbers in the higher range, so there are only $10 - 5 = 5$ numbers not yet blocked.

We need to choose $5$ distinct numbers from the higher range, so the numbers chosen from the higher range are precisely those which are not yet blocked. They are each exactly $10$ more than a not-chosen number in the lower range.

So, the sum of the $5$ distinct numbers chosen from the higher range is exactly $5 \times 10 = 50$ more than the sum of the $5$ not-chosen numbers in the lower range.

The sum of all $10$ chosen numbers is therefore equal to $50$ plus the sum of all chosen and not-chosen numbers in the lower range $1 + 2 + \dots + 10.$

The sum of the numbers from $1$ to $10$ is $\frac{10 (10+1)}{2} = 55,$ so the answer is $50 + 55 = 105,$ or choice C.

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The number of people who live in $A$ and work in $A$ is $$100 - 100 \times \frac{1}{4} - 100 \times \frac{1}{5}$$ which is $$100 - 25 - 20 = 55.$$ The number of people who live in $B$ and work in $A$ is $$120 \times \frac{1}{3} = 40$$ The number of people who live in $C$ and work in $A$ is $$160 \times \frac{1}{8} = 20$$ So, the answer is $$55 + 40 + 20 = 115,$$ which is choice D.

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The diagram on the right is similar to the diagram on the left, but the corresponding areas in the diagram on the right are $4$ times the areas on the left. So, each length on the right is $sqrt{4} = 2$ times the corresponding length on the left. This gives $R = 2,$ which is choice B.

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Think of the road as having three sections: left, middle, and right. Each section is $5$ miles long.

The car from $A$ reaches the middle section in $$\frac{5}{25} = \frac{1}{5}$$ hours.

The car from $B$ reaches the middle section in $$\frac{5}{20} = \frac{1}{4}$$ hours. By that time, the car from $A$ has already driven in the middle section for $$\frac{1}{4} - \frac{1}{5} = \frac{1}{20}$$ hours. During this time, that car from $A$ has traveled $$40 \times \frac{1}{20} = 2$$ miles in the middle section.

That leaves $5 - 2 = 3$ miles between the two cars in the middle section.

At that moment, the car from $A$ is $5 + 2 = 7$ miles from $A.$

Since the cars drive at the same speed of $40$ mph in the middle section, they meet after each driving $1.5$ more miles. This takes the car from $A$ a total distance of $7 + 1.5 = 8.5$ miles, which is choice D.

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Sarika gets $\frac{1}{2}$ of the cheese in the first step.

Then Dev gets $\frac{1}{2^2}$ of the cheese.

Then Rajiv gets $\frac{1}{2^3}$ of the cheese.

Sarika then gets another $\frac{1}{2^4}$ of the cheese.

This pattern continues. Ultimately, Sarika gets $$\frac{1}{2} + \frac{1}{2^4} + \frac{1}{2^7} + \dots$$ which is the sum of an infinite geometric series with first term $a = \frac{1}{2}$ and common ratio $r = \frac{1}{2^3}.$ That sum is $$\frac{1}{1-r} = \frac{\frac{1}{2}}{1 - \frac{1}{8}} = \frac{\frac{1}{2}}{\frac{7}{8}} = \frac{4}{7}$$ which is choice A.

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Among pods $A,$ $B,$ $C,$ and $F,$ each pair is directly connected by a walkway. So, they have to be at least $2$ grades apart. The only way to select $4$ numbers from $1$ through $7$ inclusive, which are at least $2$ apart, is to select $\{1, 3, 5, 7\}$ in some order.

Try dealing with $G$ next, because it looks less complex. What if grade $2$ is in pod $G?$

Then $A$ and $F$ cannot be $1$ or $3.$ That forces $1$ and $3$ to be among $B$ and $C,$ and $5$ and $7$ to be among $A$ and $F$.

The remaining undetermined pods $D$ and $E$ only are directly connected to $C$ and $F,$ so this suggests putting grade $1$ in pod $C$ and grade $7$ in pod $F,$ as the extreme grades only have one conflict each.

Then grade $6$ can go in pod $D$ and grade $4$ can go in pod $E.$

In order to satisfy the constraints we have made so far, grade $3$ goes in pod $B$, and grade $5$ goes in pod $A.$ This actually works, and so the answer is $1 + 4 + 7 = 12,$ which is choice A.

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Imagine adding an extra coat hook with a coat on it after the last coat. Now, there will be $36$ coat hooks, and a repeating pattern, where each block of the pattern has a bunch of empty hooks followed by a coat. Let $b$ be the number of items in each block. Let $d$ be the number of blocks. Note that $d$ is exactly one more than the number of coats, because we added an extra coat at the end.

We then have $bd = 36.$

The constraint on $b$ is that $b \geq 2$ because each block has at least one empty hook, and ends with a coat.

The constraint on $d$ is that $d \geq 2$ because there was at least one coat before, and we added one extra coat at the end.

So, we just need to find out how many ways there are to factorize $36$ into the product of two integers that are at least $2.$

The number of ways to factorize $36$ into the product of two positive integers is exactly equal to the number of factors of $36.$ There is a formula for that: since $36 = 2^2 \times 3^2$, the number of factors of $36$ is $$(2+1)(2+1) = 9.$$

Out of these factorizations, exactly two are disqualified: $1 \times 36$ and $36 \times 1.$ So, the answer is $9 - 2 = 7,$ which is choice D.

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The number has the form $XX99,$ and so the perfect square that is $1$ more ends in $00,$ and so it is the square of a number ending in $0.$ Suppose that square is $a^2.$

In order for $a^2 - 1$ to be a $4$-digit number, the only possibilities for $a$ are $\{4, 5, \dots, 10\}.$

Since $a^2 - 1 = (a-1)(a+1),$ we also need both $a-1$ and $a+1$ to be prime. We are then looking for pairs of prime numbers that are right around $\{40, 50, \dots, 100\}.$

Going through all the possibilities, the only ones that work are $59$ and $61$, and so there is exactly $1$ way to do this. The answer is B.

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Draw a line through $A$ parallel to line $CD,$ and observe that some lengths are automatically equal to each other:

That is because $\angle AEB = \angle DCB = 60^\circ,$ and so triangle $ABE$ is equilateral. All lengths labeled $x$ are always equal.

Also, $ADCE$ is a parallelogram (not necessarily a rhombus), so the remaining lengths labeled $y$ are always equal to each other, but not necessarily equal to $x.$

The perimeter is $3x + 2y,$ but it is also supposed to be $30.$

The problem the amounts to counting how many positive integer solutions there are to the equation $3x + 2y = 30.$

As we consider positive integers for $x,$ observe that it is precisely the even integers which give an integer solution for $y,$ because $30 - 3x$ is the even number $2y.$

And, once we get to $x = 10,$ the $y$ that satisfies the equation is $y = \frac{30 - 3 \times 10}{2} = 0,$ which is not a positive integer.

So, the values that work for $x$ are the positive even integers less than $10,$ or $\{2, 4, 6, 8\}.$ There are $4$ options, which gives the answer of E.

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Let $X$ be the answer.

By symmetry, if the question asked for the sum of areas between each path and the left side of the grid, then the answer would be exactly the same $X.$

But if that answer is added to the original answer, that is exactly the same as the sum over all paths, of the sum of areas to the left and to the right.

For each path, that sum of areas is exactly $25.$

The number of paths is equal to the number of ways to rearrange $LLLLLRRRRR,$ where $L$ stands for Left and $R$ stands for Right, as the path walks up. The number of rearrangements is $10$ choose $5,$ denoted $\binom{10}{5}.$

So, $2X = 25 \times \binom{10}{5}.$ Dividing by $2,$ we get that $X$ equals $$\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \times \frac{25}{2},$$ which is 3150, or choice B.