Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: The sum of two natural numbers is $17402$ . One of the two numbers is divisible by $10$ . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(E)} ~15426$

Solution: Since the unit's place of a multiple of $10$ is $0$ , the other integer must end with a $2$ , for both integers sum up to a number ending in a $2$ . Thus, the unit's place of the difference must be $10-2=8$ , and the only answer choice that ends with an $8$ is $\boxed{14238}$

Answer: D 14238
Problem: The sum of two natural numbers is $17402$ . One of the two numbers is divisible by $10$ . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(E)} ~15426$

Solution: Let the larger number be $\underline{ABCD0}.$ It follows that the smaller number is $\underline{ABCD}.$ Adding vertically, we have \[\begin{array}{cccccc}   & A & B & C & D & 0 \\ +\quad &   & A & B & C & D \\ \hline   &   &   &   &   &   \\ [-2.5ex]   & 1 & 7 & 4 & 0 & 2 \\ \end{array}\] Working from right to left, we get \[D=2\implies C=8 \implies B=5 \implies A=1.\] The larger number is $15820$ and the smaller number is $1582.$ Their difference is $15820-1582=\boxed{14238}.$

Answer: D 14238
Problem: The sum of two natural numbers is $17402$ . One of the two numbers is divisible by $10$ . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(E)} ~15426$

Solution: We know that the larger number has a units digit of $0$ since it is divisible by 10. If $D$ is the ten's digit of the larger number, then $D$ is the units digit of the smaller number. Since the sum of the natural numbers has a unit's digit of $2$ $D=2$
The units digit of the larger number is $0$ and the units digit of the smaller number is $2$ , so the positive difference between the numbers is 8. There is only one answer choice that has this units digit, and that is $\boxed{14238}.$

Answer: D 14238
Problem: A cart rolls down a hill, travelling $5$ inches the first second and accelerating so that during each successive $1$ -second time interval, it travels $7$ inches more than during the previous $1$ -second interval. The cart takes $30$ seconds to reach the bottom of the hill. How far, in inches, does it travel?
$\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242$

Solution: Since \[\mathrm{Distance}=\mathrm{Speed}\cdot\mathrm{Time},\] we seek the sum \[5\cdot1+12\cdot1+19\cdot1+26\cdot1+\cdots=5+12+19+26+\cdots,\] in which there are $30$ terms.
The last term is $5+7\cdot(30-1)=208.$ Therefore, the requested sum is \[5+12+19+26+\cdots+208=\frac{5+208}{2}\cdot30=\boxed{3195}.\] ~MRENTHUSIASM

Answer: D 3195
Problem: A cart rolls down a hill, travelling $5$ inches the first second and accelerating so that during each successive $1$ -second time interval, it travels $7$ inches more than during the previous $1$ -second interval. The cart takes $30$ seconds to reach the bottom of the hill. How far, in inches, does it travel?
$\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242$

Solution: The distance (in inches) traveled within each $1$ -second interval is: \[5,5+1(7),5+2(7), \dots , 5+29(7).\] This is an arithmetic sequence so the total distance travelled, found by summing them up is: \[\text{number of terms} \cdot \text{average of terms} = \text{number of terms} \cdot \dfrac{\text{first term}+\text{last term}}{2}.\] Or, \[30 \cdot \dfrac{5+5+29(7)}{2} = 15 \cdot 213 = \boxed{3195}.\] ~BakedPotato66

Answer: D 3195
Problem: A cart rolls down a hill, travelling $5$ inches the first second and accelerating so that during each successive $1$ -second time interval, it travels $7$ inches more than during the previous $1$ -second interval. The cart takes $30$ seconds to reach the bottom of the hill. How far, in inches, does it travel?
$\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242$

Solution: From the $30$ -term sum \[5+12+19+26+\cdots\] in Solution 1, taking modulo $10$ gives \[5+12+19+26+\cdots \equiv 3\cdot(5+2+9+6+3+0+7+4+1+8) = 3\cdot45\equiv5 \pmod{10}.\] The only answer choices congruent to $5$ modulo $10$ are $\textbf{(A)}$ and $\textbf{(D)}.$ By a quick estimation, $\textbf{(A)}$ is too small, leaving us with $\boxed{3195}.$

Answer: D 3195
Problem: A cart rolls down a hill, travelling $5$ inches the first second and accelerating so that during each successive $1$ -second time interval, it travels $7$ inches more than during the previous $1$ -second interval. The cart takes $30$ seconds to reach the bottom of the hill. How far, in inches, does it travel?
$\textbf{(A)} ~215 \qquad\textbf{(B)} ~360\qquad\textbf{(C)} ~2992\qquad\textbf{(D)} ~3195\qquad\textbf{(E)} ~3242$

Solution: This problem can be solved by physics method. This method is perhaps the quickest too and shows the beauty of the problem. The average speed increases $7 \ \text{in/s}$ per second. So, the acceleration $a=7 \ \text{in/s\textsuperscript{2}}.$ The average speed of the first second is $5 \ \text{in/s}.$ We can know the initial velocity $v_0=5-0.5\cdot7=1.5.$ The displacement at $t=30$ is \[s=\frac{1}{2}at^2+v_0t=\frac{1}{2}\cdot7\cdot30^2+1.5\cdot30= \boxed{3195}.\] ~Bran_Qin

Answer: D 3195
Problem: When a student multiplied the number $66$ by the repeating decimal, \[\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},\] where $a$ and $b$ are digits, he did not notice the notation and just multiplied $66$ times $\underline{1}.\underline{a} \ \underline{b}.$ Later he found that his answer is $0.5$ less than the correct answer. What is the $2$ -digit number $\underline{a} \ \underline{b}?$
$\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75$

Solution: We are given that $66\Bigl(\underline{1}.\overline{\underline{a} \ \underline{b}}\Bigr)-0.5=66\Bigl(\underline{1}.\underline{a} \ \underline{b}\Bigr),$ from which \begin{align*} 66\Bigl(\underline{1}.\overline{\underline{a} \ \underline{b}}\Bigr)-66\Bigl(\underline{1}.\underline{a} \ \underline{b}\Bigr)&=0.5 \\ 66\Bigl(\underline{1}.\overline{\underline{a} \ \underline{b}} - \underline{1}.\underline{a} \ \underline{b}\Bigr)&=0.5 \\ 66\Bigl(\underline{0}.\underline{0} \ \underline{0} \ \overline{\underline{a} \ \underline{b}}\Bigr)&=0.5 \\ 66\left(\frac{1}{100}\cdot\underline{0}.\overline{\underline{a} \ \underline{b}}\right)&=\frac12 \\ \underline{0}.\overline{\underline{a} \ \underline{b}}&=\frac{25}{33} \\ \underline{0}.\overline{\underline{a} \ \underline{b}}&=0.\overline{75} \\ \underline{a} \ \underline{b}&=\boxed{75} ~MRENTHUSIASM

Answer: E 75
Problem: When a student multiplied the number $66$ by the repeating decimal, \[\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},\] where $a$ and $b$ are digits, he did not notice the notation and just multiplied $66$ times $\underline{1}.\underline{a} \ \underline{b}.$ Later he found that his answer is $0.5$ less than the correct answer. What is the $2$ -digit number $\underline{a} \ \underline{b}?$
$\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75$

Solution: It is known that $\underline{0}.\overline{\underline{a} \ \underline{b}}=\frac{\underline{a} \ \underline{b}}{99}$ and $\underline{0}.\underline{a} \ \underline{b}=\frac{\underline{a} \ \underline{b}}{100}.$
Let $x=\underline{a} \ \underline{b}.$ We have \[66\biggl(1+\frac{x}{99}\biggr)-66\biggl(1+\frac{x}{100}\biggr)=0.5.\] Expanding and simplifying give $\frac{x}{150}=0.5,$ so $x=\boxed{75}.$

Answer: E 75
Problem: When a student multiplied the number $66$ by the repeating decimal, \[\underline{1}.\underline{a} \ \underline{b} \ \underline{a} \ \underline{b}\ldots=\underline{1}.\overline{\underline{a} \ \underline{b}},\] where $a$ and $b$ are digits, he did not notice the notation and just multiplied $66$ times $\underline{1}.\underline{a} \ \underline{b}.$ Later he found that his answer is $0.5$ less than the correct answer. What is the $2$ -digit number $\underline{a} \ \underline{b}?$
$\textbf{(A) }15 \qquad \textbf{(B) }30 \qquad \textbf{(C) }45 \qquad \textbf{(D) }60 \qquad \textbf{(E) }75$

Solution: We have \[66 \cdot \left(1 + \frac{10a+b}{100}\right) + \frac{1}{2} = 66 \cdot \left(1+ \frac{10a+b}{99}\right).\] Expanding both sides, we have \[66 + \frac{33(10a+b)}{50} + \frac{1}{2} = 66 + \frac{2(10a+b)}{3}.\] Subtracting $66$ from both sides, we have \[\frac{33(10a+b)}{50} + \frac{1}{2} = \frac{2(10a+b)}{3}.\] Multiplying both sides by $50 \cdot 3 = 150,$ we have \[99(10a+b) + 75 = 100(10a+b).\] Thus, the answer is $10a+b = \boxed{75}.$

Answer: E 75
Problem: What is the least possible value of $(xy-1)^2+(x+y)^2$ for real numbers $x$ and $y$
$\textbf{(A)} ~0\qquad\textbf{(B)} ~\frac{1}{4}\qquad\textbf{(C)} ~\frac{1}{2} \qquad\textbf{(D)} ~1 \qquad\textbf{(E)} ~2$

Solution: Expanding, we get that the expression is $x^2+2xy+y^2+x^2y^2-2xy+1$ or $x^2+y^2+x^2y^2+1$ . By the Trivial Inequality (all squares are nonnegative) the minimum value for this is $\boxed{1}$ , which can be achieved at $x=y=0$

Answer: D 1
Problem: What is the least possible value of $(xy-1)^2+(x+y)^2$ for real numbers $x$ and $y$
$\textbf{(A)} ~0\qquad\textbf{(B)} ~\frac{1}{4}\qquad\textbf{(C)} ~\frac{1}{2} \qquad\textbf{(D)} ~1 \qquad\textbf{(E)} ~2$

Solution: We expand the original expression, then factor the result by grouping: \begin{align*} (xy-1)^2+(x+y)^2&=\left(x^2y^2-2xy+1\right)+\left(x^2+2xy+y^2\right) \\ &=x^2y^2+x^2+y^2+1 \\ &=x^2\left(y^2+1\right)+\left(y^2+1\right) \\ &=\left(x^2+1\right)\left(y^2+1\right). \end{align*} Clearly, both factors are positive. By the Trivial Inequality, we have \[\left(x^2+1\right)\left(y^2+1\right)\geq\left(0+1\right)\left(0+1\right)=\boxed{1}.\] Note that the least possible value of $(xy-1)^2+(x+y)^2$ occurs at $x=y=0.$

Answer: D 1
Problem: What is the least possible value of $(xy-1)^2+(x+y)^2$ for real numbers $x$ and $y$
$\textbf{(A)} ~0\qquad\textbf{(B)} ~\frac{1}{4}\qquad\textbf{(C)} ~\frac{1}{2} \qquad\textbf{(D)} ~1 \qquad\textbf{(E)} ~2$

Solution: Like solution 1, expand and simplify the original equation to $x^2+y^2+x^2y^2+1$ and let $f(x, y) = x^2+y^2+x^2y^2+1$ . To find local extrema, find where $\nabla f(x, y) = \boldsymbol{0}$ . First, find the first partial derivative with respect to x and y and find where they are $0$ \[\frac{\partial f}{\partial x} = 2x + 2xy^{2} = 2x(1 + y^{2}) = 0 \implies x = 0\] \[\frac{\partial f}{\partial y} = 2y + 2yx^{2} = 2y(1 + x^{2}) = 0 \implies y = 0\] Thus, there is a local extremum at $(0, 0)$ . Because this is the only extremum, we can assume that this is a minimum because the problem asks for the minimum (though this can also be proven using the partial second derivative test) and the global minimum since it's the only minimum, meaning $f(0, 0)$ is the minimum of $f(x, y)$ . Plugging $(0, 0)$ into $f(x, y)$ , we find 1 $\implies \boxed{1}$

Answer: D 1
Problem: For which of the following integers $b$ is the base- $b$ number $2021_b - 221_b$ not divisible by $3$
$\textbf{(A)} ~3 \qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~6\qquad\textbf{(D)} ~7\qquad\textbf{(E)} ~8$

Solution: We have \begin{align*} 2021_b - 221_b &= (2021_b - 21_b) - (221_b - 21_b) \\ &= 2000_b - 200_b \\ &= 2b^3 - 2b^2 \\ &= 2b^2(b-1), \end{align*} which is divisible by $3$ unless $b\equiv2\pmod{3}.$ The only choice congruent to $2$ modulo $3$ is $\boxed{8}.$

Answer: E 8
Problem: For which of the following integers $b$ is the base- $b$ number $2021_b - 221_b$ not divisible by $3$
$\textbf{(A)} ~3 \qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~6\qquad\textbf{(D)} ~7\qquad\textbf{(E)} ~8$

Solution: Vertically subtracting $2021_b - 221_b,$ we see that the ones place becomes $0,$ and so does the $b^1$ place. Then, we perform a carry (make sure the carry is in base $b$ ). Let $b-2 = A.$ Then, we have our final number as \[1A00_b.\] Now, when expanding, we see that this number is simply $b^3 - (b - 2)^2.$
Now, notice that the final number will only be congruent to \[b^3-(b-2)^2\equiv0\pmod{3}.\] If either $b\equiv0\pmod{3},$ or if $b\equiv1\pmod{3}$ (because note that $(b - 2)^2$ would become $\equiv1\pmod{3},$ and $b^3$ would become $\equiv1\pmod{3}$ as well, and therefore the final expression would become $1-1\equiv0\pmod{3}.$ Therefore, $b$ must be $\equiv2\pmod{3}.$ Among the answers, only $8$ is $\equiv2\pmod{3},$ and therefore our answer is $\boxed{8}.$

Answer: E 8
Problem: For which of the following integers $b$ is the base- $b$ number $2021_b - 221_b$ not divisible by $3$
$\textbf{(A)} ~3 \qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~6\qquad\textbf{(D)} ~7\qquad\textbf{(E)} ~8$

Solution: By the definition of bases, we have \[2021_b - 221_b = \left(2b^3+2b+1\right) - \left(2b^2+2b+1\right).\] For values $b_1$ and $b_2$ such that $b_1\equiv b_2\pmod{3},$ we get \[\left(2b_1^3+2b_1+1\right) - \left(2b_1^2+2b_1+1\right) \equiv \left(2b_2^3+2b_2+1\right) - \left(2b_2^2+2b_2+1\right) \pmod{3}.\] Note that answer choices $\textbf{(A)},\textbf{(B)},\textbf{(C)},\textbf{(D)},\textbf{(E)}$ are congruent to $0,1,0,1,2$ modulo $3,$ respectively. So, $\textbf{(A)}$ and $\textbf{(C)}$ are either both correct or both incorrect. Since there is only one correct answer, $\textbf{(A)}$ and $\textbf{(C)}$ are both incorrect. Similarly, $\textbf{(B)}$ and $\textbf{(D)}$ are both incorrect. This leaves us with $\boxed{8},$ the answer choice with a unique residue modulo $3.$

Answer: E 8
Problem: Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are $3$ cm and $6$ cm. Into each cone is dropped a spherical marble of radius $1$ cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?
[asy] size(350); defaultpen(linewidth(0.8)); real h1 = 10, r = 3.1, s=0.75; pair P = (r,h1), Q = (-r,h1), Pp = s * P, Qp = s * Q; path e = ellipse((0,h1),r,0.9), ep = ellipse((0,h1*s),r*s,0.9); draw(ellipse(origin,r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill(origin--Pp--Qp--cycle,gray(0.8)); draw((-r,h1)--(0,0)--(r,h1)^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(0,Qp.y),Arrows(size=8)); draw(origin--(0,12),linetype("4 4")); draw(origin--(r*(s-0.1),0)); label("$3$",(-0.9,h1*s),N,fontsize(10));  real h2 = 7.5, r = 6, s=0.6, d = 14; pair P = (d+r-0.05,h2-0.15), Q = (d-r+0.05,h2-0.15), Pp = s * P + (1-s)*(d,0), Qp = s * Q + (1-s)*(d,0); path e = ellipse((d,h2),r,1), ep = ellipse((d,h2*s+0.09),r*s,1); draw(ellipse((d,0),r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill((d,0)--Pp--Qp--cycle,gray(0.8)); draw(P--(d,0)--Q^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(d,Qp.y),Arrows(size=8)); draw((d,0)--(d,10),linetype("4 4")); draw((d,0)--(d+r*(s-0.1),0)); label("$6$",(d-r/4,h2*s-0.06),N,fontsize(10)); [/asy]
$\textbf{(A) }1:1 \qquad \textbf{(B) }47:43 \qquad \textbf{(C) }2:1 \qquad \textbf{(D) }40:13 \qquad \textbf{(E) }4:1$

Solution: The heights of the cones are not given, so suppose the heights are very large (i.e. tending towards infinity) in order to approximate the cones as cylinders with base radii $3$ and $6$ and infinitely large height. Then the base area of the wide cylinder is $4$ times that of the narrow cylinder. Since we are dropping a ball of the same volume into each cylinder, the water level in the narrow cone/cylinder should rise $\boxed{4}$ times as much.

Answer: E 4
Problem: What is the volume of tetrahedron $ABCD$ with edge lengths $AB = 2$ $AC = 3$ $AD = 4$ $BC = \sqrt{13}$ $BD = 2\sqrt{5}$ , and $CD = 5$
$\textbf{(A)} ~3 \qquad\textbf{(B)} ~2\sqrt{3} \qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~3\sqrt{3}\qquad\textbf{(E)} ~6$

Solution: Drawing the tetrahedron out and testing side lengths, we realize that the $\triangle ACD, \triangle ABC,$ and $\triangle ABD$ are right triangles by the Converse of the Pythagorean Theorem. It is now easy to calculate the volume of the tetrahedron using the formula for the volume of a pyramid. If we take $\triangle ADC$ as the base, then $\overline{AB}$ must be the altitude. The volume of tetrahedron $ABCD$ is $\dfrac{1}{3} \cdot \dfrac{3 \cdot 4}{2} \cdot 2=\boxed{4}.$

Answer: C 4
Problem: What is the volume of tetrahedron $ABCD$ with edge lengths $AB = 2$ $AC = 3$ $AD = 4$ $BC = \sqrt{13}$ $BD = 2\sqrt{5}$ , and $CD = 5$
$\textbf{(A)} ~3 \qquad\textbf{(B)} ~2\sqrt{3} \qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~3\sqrt{3}\qquad\textbf{(E)} ~6$

Solution: We will place tetrahedron $ABCD$ in the $xyz$ -plane. By the Converse of the Pythagorean Theorem, we know that $\triangle ACD$ is a right triangle. Without the loss of generality, let $A=(0,0,0), C=(3,0,0), D=(0,4,0),$ and $B=(x,y,z).$
We apply the Distance Formula to $\overline{BA},\overline{BC},$ and $\overline{BD},$ respectively: \begin{align*} x^2+y^2+z^2&=2^2, &(1) \\ (x-3)^2+y^2+z^2&=\sqrt{13}^2, &(2) \\ x^2+(y-4)^2+z^2&=\left(2\sqrt5\right)^2. &\hspace{1mm} (3) \end{align*} Subtracting $(1)$ from $(2)$ gives $-6x+9=9,$ from which $x=0.$
Subtracting $(1)$ from $(3)$ gives $-8y+16=16,$ from which $y=0.$
Substituting $(x,y)=(0,0)$ into $(1)$ produces $z^2=4,$ or $|z|=2.$
Let the brackets denote areas. Finally, we find the volume of tetrahedron $ABCD$ using $\triangle ACD$ as the base: \begin{align*} V_{ABCD}&=\frac13\cdot[ACD]\cdot h_B \\ &=\frac13\cdot\left(\frac12\cdot AC \cdot AD\right)\cdot |z| \\ &=\boxed{4} ~MRENTHUSIASM

Answer: C 4
Problem: All the roots of the polynomial $z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16$ are positive integers, possibly repeated. What is the value of $B$
$\textbf{(A) }{-}88 \qquad \textbf{(B) }{-}80 \qquad \textbf{(C) }{-}64 \qquad \textbf{(D) }{-}41\qquad \textbf{(E) }{-}40$

Solution: By Vieta's formulas, the sum of the six roots is $10$ and the product of the six roots is $16$ . By inspection, we see the roots are $1, 1, 2, 2, 2,$ and $2$ , so the function is $(z-1)^2(z-2)^4=(z^2-2z+1)(z^4-8z^3+24z^2-32z+16)$ . Therefore, calculating just the $z^3$ terms, we get $B = -32 - 48 - 8 = \boxed{88}$

Answer: A 88
Problem: All the roots of the polynomial $z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16$ are positive integers, possibly repeated. What is the value of $B$
$\textbf{(A) }{-}88 \qquad \textbf{(B) }{-}80 \qquad \textbf{(C) }{-}64 \qquad \textbf{(D) }{-}41\qquad \textbf{(E) }{-}40$

Solution: Using the same method as Solution 1, we find that the roots are $2, 2, 2, 2, 1,$ and $1$ . Note that $B$ is the negation of the 3rd symmetric sum of the roots. Using casework on the number of 1's in each of the $\binom {6}{3} = 20$ products $r_a \cdot r_b \cdot r_c,$ we obtain \[B= - \left(\binom {4}{3} \binom {2}{0} \cdot 2^{3} + \binom {4}{2} \binom{2}{1} \cdot 2^{2} \cdot 1 + \binom {4}{1} \binom {2}{2} \cdot 2 \right) = -\left(32+48+8 \right) = \boxed{88}.\] ~ike.chen

Answer: A 88
Problem: Values for $A,B,C,$ and $D$ are to be selected from $\{1, 2, 3, 4, 5, 6\}$ without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves $y=Ax^2+B$ and $y=Cx^2+D$ intersect? (The order in which the curves are listed does not matter; for example, the choices $A=3, B=2, C=4, D=1$ is considered the same as the choices $A=4, B=1, C=3, D=2.$
$\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360$

Solution: Visualizing the two curves, we realize they are both parabolas with the same axis of symmetry. Now assume that the first equation is above the second, since order doesn't matter. Then $C>A$ and $B>D$ . Therefore the number of ways to choose the four integers is $\tbinom{6}{2}\tbinom{4}{2}=90$ , and the answer is $\boxed{90}$

Answer: C 90
Problem: Values for $A,B,C,$ and $D$ are to be selected from $\{1, 2, 3, 4, 5, 6\}$ without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves $y=Ax^2+B$ and $y=Cx^2+D$ intersect? (The order in which the curves are listed does not matter; for example, the choices $A=3, B=2, C=4, D=1$ is considered the same as the choices $A=4, B=1, C=3, D=2.$
$\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360$

Solution: Setting $y = Ax^2+B = Cx^2+D$ , we find that $Ax^2-Cx^2 = x^2(A-C) = D-B$ , so $x^2 = \frac {D-B}{A-C} \ge 0$ by the trivial inequality. This implies that $D-B$ and $A-C$ must both be positive or negative. If two distinct values are chosen for $(A, C)$ and $(B, D)$ respectively, there are $2$ ways to order them so that both the numerator and denominator are positive/negative (increasing and decreasing). We must divide by $2$ at the end, however, since the $2$ curves aren't considered distinct. Calculating, we get \[\frac {1}{2} \cdot \binom {6}{2} \binom {4}{2} \cdot 2 = \boxed{90}.\] ~ike.chen

Answer: C 90
Problem: Values for $A,B,C,$ and $D$ are to be selected from $\{1, 2, 3, 4, 5, 6\}$ without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves $y=Ax^2+B$ and $y=Cx^2+D$ intersect? (The order in which the curves are listed does not matter; for example, the choices $A=3, B=2, C=4, D=1$ is considered the same as the choices $A=4, B=1, C=3, D=2.$
$\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360$

Solution: Like in Solution 2 , we find $\frac {D-B}{A-C} \ge 0$ . Notice that, since $D \ne B$ , this expression can never equal $0$ , and since $A \ne C$ , there won't be a divide-by- $0$ . This means that every choice results in either a positive or a negative value.
For every choice of $(A,B,C,D)$ that results in a positive value, we can flip $B$ and $D$ to obtain a corresponding negative value. This is a bijection (we could flip $B$ and $D$ again to obtain the original choice (injectivity) and we could flip $B$ and $D$ from any negative choice to obtain the corresponding positive choice (surjectivity)), so half of the choices are positive (where the curves intersect) and half are negative (where they don't).
This means that of the $\frac{6\cdot5\cdot4\cdot3}{2} = 180$ total choices (dividing by $2$ because the order of the curves does not matter), half of them, or $\frac{180}{2} = \boxed{90}$ , lead to intersecting curves.

Answer: C 90
Problem: In the following list of numbers, the integer $n$ appears $n$ times in the list for $1 \leq n \leq 200$ \[1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots , 200\] What is the median of the numbers in this list?
$\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\textbf{(E)} ~167$

Solution: The $x$ th number of this sequence is $\left\lceil\frac{-1\pm\sqrt{1+8x}}{2}\right\rceil$ via the quadratic formula. We can see that if we halve $x$ we end up getting $\left\lceil\frac{-1\pm\sqrt{1+4x}}{2}\right\rceil$ . This is approximately the number divided by $\sqrt{2}$ $\frac{200}{\sqrt{2}} = 141.4$ and since $142$ looks like the only number close to it, it is answer $\boxed{142}$ ~Lopkiloinm

Answer: C 142