Answer: C

Solution:

$$\begin{align*}&\frac{2+4+6}{1+3+5} - \frac{1+3+5}{2+4+6}\ &= \dfrac {12}9 - \dfrac 9{12} \&= \dfrac 43 - \dfrac 34 \&= \dfrac 7{12}\end{align*}$$

Thus, the correct answer is C .

Answer: A

Solution:

The dimensions of the garden are $2\cdot 15 = 30$ feet by $2\cdot 20=40$ feet. Thus, the square footage is $30\cdot 40 = 1200.$

Therefore, there are $$1200\cdot \dfrac 12 =600$$ pounds of potatoes.

Thus, the correct answer is A .

Answer: C

Solution:

Let the low temperature be represented by $l.$ Then, the high temperature is $l+16.$

This makes the average $$\dfrac{l+l+16}2 = l+8 = 3.$$ Therefore, $l = -5.$

Thus, the correct answer is C .

Answer: D

Solution:

When counting backward, $201$ is the first number, $200$ is the second, and in general $x$ is the $202-x$th number.

Thus $53$ is the $202-53=149$th number counted.

Thus, the correct answer is D .

Answer: B

Solution:

The expression is $2a-ab=a(2-b)$.

To make it as small as possible, choose $b$ as large as possible so that $2-b$ is most negative, and choose $a$ as large as possible. Since $a,b<6$, take $a=b=5$.

The minimum value is $5(2-5)=-15$.

Thus, the correct answer is B .

Answer: C

Solution:

The sum of the ages of all the fifth-graders and their parents is: $$33\cdot 11+55\cdot 33 = 66\cdot 33.$$

Then, as the average is the sum divided by the number of people, the average age must be: $$\dfrac{66\cdot 33}{88} = \dfrac 34 \cdot 33 = 24.75 .$$

Thus, the correct answer is C .

Answer: B

Solution:

Six equally spaced points on the circle form a regular hexagon. A triangle using three of them is not equilateral or isosceles only when its angles are $30^\circ,60^\circ,90^\circ$.

The hypotenuse is a diameter of the unit circle, so it has length $2$. The legs of the $30$-$60$-$90$ triangle are $1$ and $\sqrt3$.

The area is $\frac12\cdot1\cdot\sqrt3=\frac{\sqrt3}{2}$.

Thus, the correct answer is B .

Answer: B

Solution:

Let each car drive $40$ miles. Together they drive $80$ miles.

Ray uses $1$ gallon, while Tom uses $4$ gallons, so together they use $5$ gallons.

The combined rate is $80\div5=16$ miles per gallon.

Thus, the correct answer is B .

Answer: D

Solution:

Only one of them is a multiple of $2,$ only one of them is a multiple of $3,$ and only one of them is a multiple of $5.$

Since each of the positive integers is greater than $1,$ each of them must be a multiple of one of the given primes.

Therefore, since $$27000=2^3\cdot 3^3\cdot 5^3,$$ the numbers must be $2^3,3^3,5^3,$ making their sum $160.$

Thus, the correct answer is D .

Answer: C

Solution:

Let the number of three-point attempts be $t.$

Then, the number of made three-point shots is $0.4 t,$ implying that the number of points made off of three-point shots is $1.2t.$

Similarly, the number of two-point shots is $1.5t,$ so the number of made two-point shots is $0.75t,$ suggesting that the number of points made off of 2 point shots is $1.5t.$

This makes the total number of points scored equal to: $$2.7t=54,$$ and so, $t=20.$

Thus, the correct answer is C .

Answer: B

Solution:

This can be rewritten as $$x^2-10x+y^2+6y+34=0.$$

From this, completing the square yields $$x^2-10x+25+y^2+6y+9$$$$=(x-5)^2+(y+3)^2$$$$=0.$$ Since both of these squared terms must be greater than or equal to $0,$ and their sum equals $0,$ both values must both be $0$ yielding $$(x-5)^2 = (y+3)^2=0.$$

As such, $x=5,y=-3.$ This makes the sum $x+y=2.$

Thus, the correct answer is B .

Answer: B

Solution:

A regular pentagon has $5$ sides of one length and $5$ diagonals of another length.

After the first segment is chosen, there are $9$ segments left, and exactly $4$ of them have the same length as the first chosen segment.

Therefore the probability is $\frac49$.

Thus, the correct answer is B .

Answer: E

Solution:

The sequence $1,2 \cdots n$ is said first after the $n^{th}$ triangular number $T_n,$ which is: $$T_n = \dfrac{n(n+1)}2.$$

(The triangular number $T_n$ is defined as being the sum of the numbers $1$ to $n.$ As for why it has this formula, there's a bunch of ways you can prove it using induction or arithmetic sequences. We'll leave that to you, though!)

Moving on, writing out some triangular numbers, we notice that $T_9 = 45$ is the closest triangular number less than $53.$ This means that the sequence $1,2 \cdots 9$ is first said after $$T_9 = \dfrac{9\cdot 10}2 = 45$$ numbers.

Therefore, the $46^{th}$ number starts this new sequence at $1,$ and counting forwards, we can see that the $53rd$ number is $7$ more than this, with a value of $8.$

Thus, the correct answer is E .

Answer: E

Solution:

The condition is $x\clubsuit y=y\clubsuit x$, so $x^2y-xy^2=y^2x-yx^2$.

Combining like terms gives $2x^2y-2xy^2=2xy(x-y)=0$.

Thus $x=0$, $y=0$, or $x=y$. These are three lines.

Thus, the correct answer is E .

Answer: B

Solution:

Let the side length of the equilateral triangle be $s.$ Then, let its area be $A.$ This would make $a=3s.$

As such, a hexagon with side $s$ would have $6$ equilateral triangles, with side length $s,$ making its area $6A.$

Therefore, the side length of the hexagon with area $A$ is equal to $\dfrac{s}{\sqrt 6} .$ As such, $$b = 6\cdot \dfrac{s}{\sqrt 6} = s\sqrt 6 .$$

This makes $$\dfrac ab = \dfrac{3s}{s\sqrt 6} = \dfrac{3 \sqrt 6}6 = \dfrac{\sqrt 6} 2.$$

Thus, the correct answer is B .

Answer: B

Solution:

Since $PE:PD:DE=1.5:2:2.5=3:4:5$, triangle $DPE$ is right at $P$. Thus medians $AD$ and $CE$ are perpendicular.

The centroid divides each median in a $2:1$ ratio, so $CE=3\cdot PE=4.5$ and $AD=3\cdot PD=6$.

Quadrilateral $AEDC$ has perpendicular diagonals $AD$ and $CE$, so its area is $\frac12(AD)(CE)=\frac12\cdot6\cdot4.5=13.5$.

Thus, the correct answer is B .

Answer: E

Solution:

Suppose Alex makes $m$ exchanges at the red-token booth and $n$ exchanges at the blue-token booth.

He then has $75-2m+n$ red tokens and $75+m-3n$ blue tokens. At the end he must have fewer than $2$ red tokens and fewer than $3$ blue tokens.

Solving these terminal possibilities gives only two candidate final token counts: $(1,2)$, which comes from $(m,n)=(59,44)$, or $(0,0)$, which comes from $(m,n)=(60,45)$.

The final count $(0,0)$ is impossible, because the last exchange would always create either one blue token or one red token. The final count $(1,2)$ is attainable, for example by the exchange sequence described in the official construction.

Therefore Alex ends with $59+44=103$ silver tokens, and the correct answer is E .

Answer: D

Solution:

Given the first three numbers, if their sum is less than or equal to $9,$ it creates one number with the property.

Now, we can case on the $1$st digit.

If it is 1, then the sum of the $2$nd and $3$rd digit must be less than or equal to $8.$ For each possible sum $s,$ there are $s+1$ ways to choose the other numbers as the 2nd number can be anywhere from $0$ to $s.$

Thus, the total is the $9^{th}$ triangular number: $$\dfrac{9\cdot 10}2 =45.$$

If it is 2, then the only way we can get a number that works less than $2013$ is $2002,$ making a total of $46$ cases.

Thus, the correct answer is D .

Answer: D

Solution:

Let the common difference be $d$, so $c=b-d$ and $a=b+d$.

A quadratic with exactly one real root has discriminant $0$, so $b^2-4ac=0$. Substituting gives $b^2=4(b+d)(b-d)=4b^2-4d^2$, hence $4d^2=3b^2$.

Since $b,d\ge0$, $2d=\sqrt3 b$. The double root is $\frac{-b}{2a}=\frac{-b}{2(b+d)}=\frac{-1}{2+\sqrt3}=-2+\sqrt3$.

Thus, the correct answer is D .

Answer: B

Solution:

The prime factorization is $2013=3\cdot11\cdot61$, so the numerator must contain a factor of $61$. Hence $a_1\ge61$.

But $61!$ also contains the prime factor $59$, which is not in $2013$, so the denominator must contain a factor of $59$. Hence $b_1\ge59$.

The lower bound $a_1+b_1\ge120$ is attainable because $2013=\frac{61!\,11!\,3!}{59!\,10!\,5!}$.

Thus $|a_1-b_1|=61-59=2$, and the correct answer is B .

Answer: C

Solution:

A sequence starting with $u,v$ has seventh term $5u+8v$.

For two sequences $(a_1,a_2)$ and $(b_1,b_2)$ with different first terms, assume $a_1\lt b_1$. Then $5a_1+8a_2=5b_1+8b_2$, so $5(b_1-a_1)=8(a_2-b_2)$.

Since $5$ and $8$ are relatively prime, $b_1-a_1$ is at least $8$, and then nondecreasing order gives $b_2\ge b_1\ge a_1+8$.

The smallest construction is $a_1=0$, $b_1=b_2=8$, and $a_2=13$. This gives $N=5\cdot0+8\cdot13=104$.

Thus, the correct answer is C .

Answer: C

Solution:

Let $S$ be defined as: $$\begin{align*}S &= A+J+E\&=B+J+F\&=C+J+G\&=D+J+H\end{align*}$$ $$\begin{align*}4S &= A+B+C+D+E\&+F+G+H+4J \end{align*}$$ $$4S = 45+3J$$ $$45+3J \equiv 0 \mod 4$$ $$3J \equiv 3 \mod 4$$ $$J \equiv 1 \mod 4$$

This means that $J=1,5,9.$ From here, let's assume $J=1.$ We will see that the other cases are similar enough to omit.

If $J=1,$ then we know that the pairs of numbers that satisfy the equality above are: $$2+9 = 3+8 = 4+7 = 5+6$$ There are $4!$ ways to distribute the pairs over the four groups, and then $2^4$ ways for these groups to swap elements (i.e. $2+9\iff 9+2$).

Now, if we look at the $J=5$ and $J=9$ cases, we see a similar pattern in the number of groupings and swaps. As such, we have: $$3\cdot 4! \cdot 2^4 = 1152$$ possibilities.

Thus, the correct answer is C .

Answer: B

Solution:

First, we can deduce that $BD =5, AD = 12, CD = 9$ by inspecting Pythagorean triples.

This yields the following diagram:

Then, we get $\angle ADE = \angle ACD$ by the similarity of \triangleriangle ADC and \triangleriangle AED . Since \triangleriangle ABF and \triangleriangle ADF are both right triangles, they both have circumcircles with diameter $AB,$ making $ABDF$ cyclic. Thus, $$\angle ABF = \angle ADF = \angle ACD,$$ making $$\cos (\angle ABF) = \dfrac 35,$$$$\sin (\angle ABF) = \dfrac 45 .$$ As such, $$AF = \dfrac 35 \cdot 13,$$$$BF = \dfrac 45 \cdot 13.$$

By Ptolemy's Theorem, we get $$AB \cdot DF + DB \cdot AF = BF\cdot AD .$$ Therefore, $$13\cdot DF + 5\cdot 13\dfrac 45 =12 \cdot 13\dfrac 35 .$$

This makes $$DF + 5\cdot \dfrac 45 = 12 \cdot \dfrac 35 ,$$ so $DF + 4 = \dfrac {36}5.$ As such, $DF=\dfrac {16}5,$ making $m+n=21.$

Thus, the correct answer is B .

Answer: A

Solution:

An integer with exactly four positive divisors is either $p^3$ or $pq$, where $p$ and $q$ are distinct primes.

If $m=p^3$, the divisor sum is $1+p+p^2+p^3$. The values for $p=11$ and $p=13$ fall below and above the interval $2010$ to $2019$, so this case gives none.

In the $m=pq$ case, the divisor sum is $1+p+q+pq=(p+1)(q+1)$. If one prime is $2$, the sum is divisible by $3$; only $2010$ and $2016$ qualify, but $2010/3-1=669$ and $2016/3-1=671$ are not prime.

If both primes are odd, then the sum is divisible by $4$, leaving $2012$ and $2016$. The factorization $2012=4\cdot503$ would give primes $3$ and $502$, impossible, while $2016=4\cdot504=(3+1)(503+1)$ works.

Thus exactly one number is nice, and the correct answer is A .

Answer: E

Solution:

It is enough to work modulo $900=\operatorname{lcm}(25,36,100)$, because the last two base-$5$, base-$6$, and decimal digits repeat with that period.

Let the last two base-$5$ digits be $a_1a_0$ and the last two base-$6$ digits be $b_1b_0$. The units digit condition gives $a_0+b_0\equiv2N\equiv2a_0\pmod{10}$, so $a_0=b_0$.

Writing $N=150N_3+30a_1+a_0$, the base-$6$ tens digit condition gives $N_3\equiv a_1+b_1\pmod6$, so $N=900N_4+180a_1+150b_1+a_0$.

The tens digit condition for $S$ and $2N$ reduces to $5a_1\equiv b_1\pmod{10}$. With $0\le a_1\le4$ and $0\le b_1\le5$, the valid pairs are $(0,0),(2,0),(4,0),(1,5),(3,5)$.

There are $5$ choices for $a_0=b_0$, so there are $5\cdot5=25$ choices of $N$.

Thus, the correct answer is E .