Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Consider a sequence $x_1,x_2,x_3,\dotsc$ defined by: \begin{align*}&x_1 = \sqrt[3]{3}, \\ &x_2 = \left(\sqrt[3]{3}\right)^{\sqrt[3]{3}},\end{align*} and in general \[x_n = \left(x_{n-1}\right)^{\sqrt[3]{3}} \text{ for } n > 1.\]
What is the smallest value of $n$ for which $x_n$ is an integer?
$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \  } 4 \qquad \mathrm{(D) \  } 9 \qquad \mathrm{(E) \  }27$

Solution: Firstly, we will show by induction that \[x_n = \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\right)}.\] For the base case, we indeed have \begin{align*}x_1 &= \sqrt[3]{3} \\ &= \left(\sqrt[3]{3}\right)^1 \\ &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^0\right)},\end{align*} and for the inductive step, if our claim is true for $x_n$ , then \begin{align*}x_{n+1} &= \left(x_n\right)^{\sqrt[3]{3}} \\ &= \left(\left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\right)}\right)^{\sqrt[3]{3}} \\ &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\cdot\sqrt[3]{3}\right)} \\ &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^n\right)},\end{align*} which completes the proof.
We now rewrite our formula for $x_n$ as follows: \begin{align*}x_n &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\right)} \\ &= \left(\sqrt[3]{3}\right)^{\left(3^{\frac{n-1}{3}}\right)} \\ &=3^{\left(\frac{1}{3} \cdot 3^{\frac{n-1}{3}}\right)} \\ &= 3^{\left(3^{\left(\frac{n-1}{3}-1\right)}\right)} \\ &= 3^{\left(3^{\left(\frac{n-4}{3}\right)}\right)},\end{align*} and as $3$ is not a perfect power, we deduce that $x_n$ is an integer if and only if the exponent, $3^{\left(\frac{n-4}{3}\right)}$ , is itself an integer. By precisely the same argument, this reduces to $\frac{n-4}{3}$ being an integer, so the smallest possible (positive) value of $n$ is $\boxed{4}$

Answer: C 4
Problem: In $\triangle ABC$ , we have $\angle C = 3\angle A$ $a = 27$ and $c = 48$ . What is $b$
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(14,0), C=(10,6); draw(A--B--C--cycle); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$a$", B--C, dir(B--C)*dir(-90)); label("$b$", A--C, dir(C--A)*dir(-90)); label("$c$", A--B, dir(A--B)*dir(-90));[/asy]
$\mathrm{(A)\ } 33 \qquad \mathrm{(B) \ }35 \qquad \mathrm{(C) \  } 37 \qquad \mathrm{(D) \  } 39 \qquad \mathrm{(E) \  }\text{not uniquely determined}$

Solution: Let $\angle A = x^{\circ}$ , so $\angle C = 3x^{\circ}$ , and thus $\angle B = \left(180-4x\right)^{\circ}$ . Now let $D$ be a point on side $AB$ such that $\angle ACD = x^{\circ}$ , so $\angle BCD = 3x^{\circ}-x^{\circ} = 2x^{\circ}$ , which gives \[\angle CDB = 180^{\circ}-2x^{\circ}-\left(180-4x\right)^{\circ} = 2x^{\circ},\] meaning that $\triangle CDB$ and $\triangle CDA$ are both isosceles, with $BC = BD$ and $AD = CD$ . In particular, $BD = BC = 27$ and $CD = AD = AB-BD = 48-27 = 21$ . Hence by Stewart's theorem on triangle $ABC$ \begin{align*}&BD \cdot AB \cdot AD + CD^2 \cdot AB = AC^2 \cdot BD + BC^2 \cdot AD \\ &\iff 27 \cdot 48 \cdot 21 + 21^2 \cdot 48 = AC^2 \cdot 27 + 27^2 \cdot 21 \\ &\iff AC^2 = \frac{27(21)(48-27) + 21^2 \cdot 48}{27} \\ &\iff AC^2 = \frac{21^2(27+48)}{27} \\ &\iff AC^2 = \frac{21^2 \cdot 25}{9} \\ &\iff AC = \frac{21 \cdot 5}{3} \qquad \text{(as } AC > 0\text{)} \\ &\iff AC = \boxed{35}

Answer: B 35
Problem: In $\triangle ABC$ , we have $\angle C = 3\angle A$ $a = 27$ and $c = 48$ . What is $b$
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(14,0), C=(10,6); draw(A--B--C--cycle); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$a$", B--C, dir(B--C)*dir(-90)); label("$b$", A--C, dir(C--A)*dir(-90)); label("$c$", A--B, dir(A--B)*dir(-90));[/asy]
$\mathrm{(A)\ } 33 \qquad \mathrm{(B) \ }35 \qquad \mathrm{(C) \  } 37 \qquad \mathrm{(D) \  } 39 \qquad \mathrm{(E) \  }\text{not uniquely determined}$

Solution: We apply the law of sines in the form \[\frac{\sin(A)}{a} = \frac{\sin(C)}{c},\] yielding \[\frac{\sin(A)}{27} = \frac{\sin(3A)}{48} \iff 9\sin(3A) = 16\sin(A).\]
Now, the angle sum and double angle identities give \begin{align*}\sin(3A) &= \sin(2A+A) \\ &= \sin(2A)\cos(A)+\cos(2A)\sin(A) \\ &= \left(2\sin(A)\cos(A)\right)\cos(A)+\left(\cos^2(A)-\sin^2(A)\right)\sin(A) \\ &= 2\sin(A)\cos^2(A)+\sin(A)\cos^2(A)-\sin^3(A) \\ &= 3\sin(A)\left(1-\sin^2(A)\right)-\sin^3(A) \\ &\text{(using the further identity } \cos^2(\theta)+\sin^2(\theta) = 1\text{)} \\ &= 3\sin(A)-4\sin^3(A).\end{align*}
Thus our equation becomes \begin{align*}9\left(3\sin(A)-4\sin^3(A)\right) = 16\sin(A) &\iff 27\sin(A)-36\sin^3(A) = 16\sin(A) \\ &\iff 36\sin^3(A) = 11\sin(A) \\ &\iff \sin(A) = 0 \text{ or } \pm\frac{\sqrt{11}}{6}.\end{align*} Notice, however, that we must have $0^{\circ} < A < 45^{\circ}$ , the latter because otherwise $A+3A \geq 180^{\circ}$ , which would contradict the fact that $A$ and $3A$ are angles in a (non-degenerate) triangle. This means $\sin(A) > 0$ , so the only valid solution is \[\sin(A) = \frac{\sqrt{11}}{6},\] and the fact that $A$ is acute also means $\cos(A) > 0$ , so we deduce \begin{align*}\cos(A) &= \sqrt{1-\left(\frac{\sqrt{11}}{6}\right)^2} \\ &= \sqrt{1-\frac{11}{36}} \\ &=\frac{5}{6}.\end{align*} Accordingly, using the double angle identities again, \begin{align*}\sin(4A) &= \sin(2 \cdot 2A) \\ &= 2\sin(2A)\cos(2A) \\ &= 2\left(2\sin(A)\cos(A)\right)\left(\cos^2(A)-\sin^2(A)\right) \\ & =2\left(2 \cdot \frac{\sqrt{11}}{6} \cdot\frac{5}{6}\right)\left(\left(\frac{5}{6}\right)^2-\left(\frac{\sqrt{11}}{6}\right)^2\right) \\ &= \frac{5\sqrt{11}}{9} \cdot \frac{25-11}{36} \\ &= \frac{5\sqrt{11}}{9} \cdot \frac{7}{18} \\ &= \frac{35\sqrt{11}}{162}.\end{align*} Finally, the law of sines now gives \begin{align*}\frac{\sin(A)}{27} &= \frac{\sin(B)}{b} \\ &= \frac{\sin(180^{\circ}-3A-A)}{b} \\ &= \frac{\sin(4A)}{b} \qquad \text{(using the identity } \sin\left(180^{\circ}-\theta\right) = \sin(\theta)\text{)},\end{align*} so, substituting the above results, \[\frac{\left(\frac{\sqrt{11}}{6}\right)}{27} = \frac{\left(\frac{35\sqrt{11}}{162}\right)}{b} \iff b = \frac{6 \cdot 27 \cdot 35}{162} = \boxed{35}.\]

Answer: B 35
Problem: In their base $10$ representations, the integer $a$ consists of a sequence of $1985$ eights and the integer $b$ consists of a sequence of $1985$ fives. What is the sum of the digits of the base $10$ representation of the integer $9ab$
$\mathrm{(A)\ } 15880 \qquad \mathrm{(B) \ }17856 \qquad \mathrm{(C) \  } 17865 \qquad \mathrm{(D) \  } 17874 \qquad \mathrm{(E) \  }19851$

Solution: By the formula for the sum of a geometric series, \begin{align*}a &= 8 \cdot 10^0 + 8 \cdot 10^1 + \dotsb + 8 \cdot 10^{1984} \\ &= \frac{8\left(10^{1985}-1\right)}{10-1} \\ &= \frac{8\left(10^{1985}-1\right)}{9},\end{align*} and similarly \[b = \frac{5\left(10^{1985}-1\right)}{9},\] so \begin{align*}9ab &= 9\cdot\frac{8\left(10^{1985}-1\right)}{9}\cdot\frac{5\left(10^{1985}-1\right)}{9} \\ &= \frac{40\left(10^{1985}-1\right)^2}{9} \\ &= \frac{40\left(10^{3970}-2 \cdot 10^{1985}+1\right)}{9} \\ &= \frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9}.\end{align*}
We now compute the decimal expansion of this expression. Firstly, $10^{3971} = 100 \dotsb 0$ , with $1$ one and $3971$ zeroes, and $2 \cdot 10^{1986} = 200 \dotsb 0$ , with $1$ two and $1986$ zeroes. Subtracting therefore gives \[10^{3971}-2 \cdot 10^{1986} = 99 \dotsb 9800 \dotsb 0,\] where there are $3971-1986-1 = 1984$ nines followed by $1$ eight and then $1986$ zeroes. Adding $10$ transforms this to $99 \dotsb 9800 \dotsb 010$ , now with $1984$ nines followed by $1$ eight, $1984$ zeroes, $1$ one, and a final zero.
Using long division, and noting that $80 = 8 \cdot 9 + 8$ and $81 = 9 \cdot 9$ , it follows that \[\frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9} = 11 \dotsb 1088 \dotsb 890,\] with $1984$ ones, $1$ zero, then $1984$ eights, $1$ nine, and a final zero. Lastly, using long multiplication and noting that $9 \cdot 4 = 36$ $8 \cdot 4 = 32$ , and $8 \cdot 4 + 3 = 35$ , we obtain \[\frac{4\left(10^{3971}-2 \cdot 10^{1986}+10\right)}{9} = 44 \dotsb 4355 \dotsb 560,\] where there are $1984$ fours, $1$ three, $1984$ fives, $1$ six, and a final zero, so the sum of the digits is \begin{align*}1984 \cdot 4 + 3 + 1984 \cdot 5 + 6 + 0 &= 1984 \cdot 9 + 9 \\ &= 1985 \cdot 9 \\ &= \boxed{17865}

Answer: C 17865
Problem: Let $\left\lfloor x\right\rfloor$ be the greatest integer less than or equal to $x$ . Then the number of real solutions to $4x^2-40\left\lfloor x\right\rfloor+51 = 0$ is
$\mathrm{(A)\ } 0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \  } 2 \qquad \mathrm{(D) \  } 3 \qquad \mathrm{(E) \  }4$

Solution: We rearrange the equation as $4x^2 = 40\left\lfloor x\right\rfloor-51$ , where the right-hand side is now clearly an integer, meaning that $4x^2 = n$ for some non-negative integer $n$ . Therefore, in the case where $x \geq 0$ , substituting $x = \frac{\sqrt{n}}{2}$ gives \[40\left\lfloor\frac{\sqrt{n}}{2}\right\rfloor-51 = n.\] To proceed, let $a$ be the unique non-negative integer such that $a \leq \frac{\sqrt{n}}{2} < a+1$ , so that \begin{align*}&\left\lfloor \frac{\sqrt{n}}{2}\right\rfloor = a, \text{ and} \\ &4a^2 \leq n < 4a^2+8a+4,\end{align*} and our equation reduces to \[40a-51 = n.\]
The above inequalities therefore become \[4a^2 \leq 40a-51 < 4a^2+8a+4  \iff 4a^2-40a+51 < 0 \text{ and } 4a^2-32a+55 > 0,\] where the first inequality can now be rewritten as $(2a-10)^2 \leq 49$ , i.e. $\left\lvert 2a-10\right\rvert \leq 7$ . Since $(2a-10)$ is even for all integers $a$ , we must in fact have \begin{align*}\left\lvert 2a-10\right\rvert \leq 6 &\iff \left\lvert a-5\right\rvert \leq 3 \\ &\iff 2 \leq a \leq 8.\end{align*} The second inequality similarly simplifies to $(2a-8)^2 > 9$ , i.e. $\left\lvert 2a-8\right\rvert > 3$ . As $(2a-8)$ is even, this is equivalent to \begin{align*}\left\lvert 2a-8 \right\rvert \geq 4 &\iff \left\lvert a-4\right\rvert \geq 2 \\ &\iff a \geq 6 \text{ or } a \leq 2,\end{align*} so the values of $a$ satisfying both inequalities are $2$ $6$ $7$ , and $8$ . Since $n = 40a-51$ , each of these distinct values of $a$ gives a distinct solution for $n$ , and thus for $x = \frac{\sqrt{n}}{2}$ , giving a total of $4$ solutions in the $x \geq 0$ case.
As $4$ is already the largest of the answer choices, this suffices to show that the answer is $\text{(E)}$ , but for completeness, we will show that the $x < 0$ case indeed gives no other solutions. If $x = -\frac{\sqrt{n}}{2}$ (and so $n > 0$ ), we require \[40\left\lfloor -\frac{\sqrt{n}}{2}\right\rfloor-51 = n,\] and recalling that $\left\lfloor -x\right\rfloor = -\left\lceil x\right\rceil$ for all $x$ , this equation can be rewritten as \[-40\left\lceil \frac{\sqrt{n}}{2}\right\rceil-51 = n.\] Since $n$ is positive, the least possible value of $\left\lceil \frac{\sqrt{n}}{2}\right\rceil$ is $1$ , but this means \begin{align*}n &= -40\left\lceil\frac{\sqrt{n}}{2}\right\rceil-51 \\ &\leq -40 \cdot 1 - 51 \\ &= -91,\end{align*} which is a contradiction. Therefore the $x < 0$ case indeed gives no further solutions, confirming that the total number of solutions is precisely $\boxed{4}$

Answer: E 4
Problem: A rectangle intersects a circle as shown: $AB=4$ $BC=5$ , and $DE=3$ . Then $EF$ equals:
[asy]defaultpen(linewidth(0.7)+fontsize(10)); pair D=origin, E=(3,0), F=(10,0), G=(12,0), H=(12,1), A=(0,1), B=(4,1), C=(9,1), O=circumcenter(B,C,F); draw(D--G--H--A--cycle); draw(Circle(O, abs(O-C))); label("$A$", A, NW); label("$B$", B, NW); label("$C$", C, NE); label("$D$", D, SW); label("$E$", E, SE); label("$F$", F, SW);  label("4", (2,0.85), N); label("3", D--E, S); label("5", (6.5,0.85), N); [/asy] $\mathbf{(A)}\; 6\qquad \mathbf{(B)}\; 7\qquad \mathbf{(C)}\; \frac{20}3\qquad \mathbf{(D)}\; 8\qquad \mathbf{(E)}\; 9$

Solution: [asy] defaultpen(linewidth(0.7)+fontsize(10)); pair D=origin, E=(3,0), F=(10,0), X=(12,0), Y=(12,1), A=(0,1), B=(4,1), C=(9,1), O=circumcenter(B,C,F), G=foot(E,A,C), H=foot(B,D,F), I=foot(C,D,F); draw(D--X--Y--A--cycle); draw(Circle(O, abs(O-C))); label("$A$", A, NW); label("$B$", B, N); label("$C$", C, NE); label("$D$", D, SW); label("$E$", E, S); label("$F$", F, S); label("$G$", G, N); label("$H$", H, S); label("$I$", I, S);  label("4", (2,0.85), N); label("3", D--E, S); label("5", (6.5,0.85), N); draw(E--G^^H--B^^I--C, linetype("4 4")); [/asy]
Draw $BE$ and $CF$ , forming a trapezoid . Since it's cyclic, this trapezoid must be isosceles . Also, drop altitudes from $E$ to $AC$ $B$ to $DF$ , and $C$ to $DF$ , and let the feet of these altitudes be $G$ $H$ , and $I$ respectively. $AGED$ is a rectangle since it has $4$ right angles . Therefore, $AG=DE=3$ , and $GB=4-3=1$ . By the same logic, $GBHE$ is also a rectangle, and $EH=GB=1$ $BH=CI$ since they're both altitudes to a trapezoid, and $BE=CF$ since the trapezoid is isosceles. Therefore, $\triangle BHE \cong \triangle CIF$ by HL congruence , so $IF=EH=1$ . Also, $BCIH$ is a rectangle from $4$ right angles, and $HI=BC=5$ . Therefore, $EF=EH+HI+IF=1+5+1=\boxed{7}$

Answer:  7
Problem: The largest integer $n$ for which $n^{200}<5^{300}$ is
$\mathrm{(A) \ }8 \qquad \mathrm{(B) \ }9 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ }11 \qquad \mathrm{(E) \ } 12$

Solution: Since both sides are positive, we can take the $100th$ root of both sides to find the largest integer $n$ such that $n^2<5^3$ . Fortunately, this is simple to evaluate: $5^3=125$ , and the largest square less than $125$ is $11^2=121$ , so the largest $n$ is $11, \boxed{11}$

Answer: D 11
Problem: The number of digits in $4^{16}5^{25}$ (when written in the usual base $10$ form) is
$\mathrm{(A) \ }31 \qquad \mathrm{(B) \ }30 \qquad \mathrm{(C) \ } 29 \qquad \mathrm{(D) \ }28 \qquad \mathrm{(E) \ } 27$

Solution: We can rewrite this as $2^{32}5^{25}$ . We can also combine some of the factors to introduce factors of $10$ , whose digit count is simple to evaluate because it simply adds $0$ s. Thus, we have $2^{32}5^{25}=2^72^{25}5^{25}=2^710^{25}$ . We can see that this final number is $2^7$ with $25$ $0$ s annexed onto it. $2^7=128$ , which has $3$ digits, so the entire number has $25+3=28$ digits, $\boxed{28}$

Answer: D 28
Problem: If the sequence $\{a_n\}$ is defined by
$a_1=2$
$a_{n+1}=a_n+2n$
where $n\geq1$
Then $a_{100}$ equals
$\mathrm{(A) \ }9900 \qquad \mathrm{(B) \ }9902 \qquad \mathrm{(C) \ } 9904 \qquad \mathrm{(D) \ }10100 \qquad \mathrm{(E) \ } 10102$

Solution: We begin to evaluate the first couple of terms of the sequence, hoping to find a pattern: $2, 4, 8, 14, 22, ....$ . We notice that the difference between succesive terms of the sequence are $2, 4, 6, 8, ....$ , a clear pattern. We can see that this pattern continues infinitely because of the recursive definition: each term is the previous term plus the next even number. Therefore, since the differences of consecutive terms form an arithmetic sequence, then the terms satisfy a quadratic , specifically, the one that contains the points $(1, 2), (2, 4),$ , and $(3, 8)$ . Let the quadratic be $f(x)=ax^2+bx+c$ , so:
$a+b+c=2$ (1)
$4a+2b+c=4$ (2)
$9a+3b+c=8$ (3)
Subtracting (1) from (2) and (2) from (3) yields the two-variable system of equations
$3a+b=2$ (4)
$5a+b=4$ (5)
We can subtract (4) from (5) to find that $2a=2$ , so $a=1$ . Substituting this back in yields $b=-1$ , and substituting these back into one of the original equations yields $c=2$ , so the closed form for the terms is
$f(x)=x^2-x+2$ , or
$a_n=n^2-n+2$
Substituting in $n=100$ yields $a_{100}=100^2-100+2=9902, \boxed{9902}$

Answer: B 9902
Problem: If $\sin{2x}\sin{3x}=\cos{2x}\cos{3x}$ , then one value for $x$ is
$\mathrm{(A) \ }18^\circ \qquad \mathrm{(B) \ }30^\circ \qquad \mathrm{(C) \ } 36^\circ \qquad \mathrm{(D) \ }45^\circ \qquad \mathrm{(E) \ } 60^\circ$

Solution: We divide both sides of the equation by $\cos{2x}\times\cos{3x}$ to get $\frac{\sin{2x}\times\sin{3x}}{\cos{2x}\times\cos{3x}}=1$ , or $\tan{2x}\times\tan{3x}=1$
This looks a lot like the formula relating the slopes of two perpendicular lines , which is $m_1\times m_2=-1$ , where $m_1$ and $m_2$ are the slopes . It's made even more relatable by the fact that the tangent of an angle can be defined by the slope of the line that makes that angle with the x-axis
We can make this look even more like the slope formula by multiplying both sides by $-1$
$\tan{2x}\times-\tan{3x}=-1$ , and using the trigonometric identity $-\tan{x}=\tan{-x}$ , we have $\tan{2x}\times\tan{-3x}=-1$
Now it's time for a diagram:
[asy] unitsize(2.54cm); draw(unitcircle); draw((0,-1.25)--(0,1.25)); draw((-1.25,0)--(1.25,0)); draw((0,0)--(cos(2pi/10),sin(2pi/10))); draw((0,0)--(cos(-3pi/10),sin(-3pi/10))); label("$\tan{-3x}$",(cos(-3pi/10),sin(-3pi/10)),SE); label("$\tan{2x}$",(cos(2pi/10),sin(2pi/10)),NE); label("$2x$",(.125,.03),ENE); label("$3x$",(.125,-.06),ESE); [/asy] Since the product of the two slopes, $\tan{2x}$ and $\tan{-3x}$ , is $-1$ , the lines are perpendicular, and the angle between them is $\frac{\pi}{2}$ . The angle between them is also $2x+3x=5x$ , so $5x=\frac{\pi}{2}$ and $x=\frac{\pi}{10}$ , or $18^\circ, \boxed{18}$

Answer: A 18
Problem: If $\sin{2x}\sin{3x}=\cos{2x}\cos{3x}$ , then one value for $x$ is
$\mathrm{(A) \ }18^\circ \qquad \mathrm{(B) \ }30^\circ \qquad \mathrm{(C) \ } 36^\circ \qquad \mathrm{(D) \ }45^\circ \qquad \mathrm{(E) \ } 60^\circ$

Solution: Start by subtracting $\sin{2x}\sin{3x}$ from both sides to get $\cos{2x}\cos{3x}-\sin{2x}\sin{3x}=0$ . We recognize that this is of the form $\cos{a}\cos{b}-\sin{a}\sin{b}=\cos{(a+b)}$ , so $\cos{2x}\cos{3x}-\sin{2x}\sin{3x}=\cos{(2x+3x)}=0$ $\cos{90^\circ}=0$ , so $x=\boxed{18}$

Answer:  18
Problem: The function $f(x)$ satisfies $f(2+x)=f(2-x)$ for all real numbers $x$ . If the equation $f(x)=0$ has exactly four distinct real roots , then the sum of these roots is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8$

Solution: Let one of the roots be $r_1$ . Also, define $x$ such that $2+x=r_1$ . Thus, we have $f(2+x)=f(r_1)=0$ and $f(2+x)=f(2-x)$ . Therefore, we have $f(2-x)=0$ , and $2-x$ is also a root. Let this root be $r_2$ . The sum $r_1+r_2=2+x+2-x=4$ . Similarly, we can let $r_3$ be a root and define $y$ such that $2+y=r_3$ , and we will find $2-y$ is also a root, say, $r_4$ , so $r_3+r_4=2+y+2-y=4$ . Therefore, $r_1+r_2+r_3+r_4=4+4=8, \boxed{8}$

Answer: E 8
Problem: The number of the distinct solutions to the equation
$|x-|2x+1||=3$ is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ } 4$

Solution: We can create a tree of possibilities, progressively eliminating the absolute value signs by creating different cases.
[asy] unitsize(2cm); draw((0,0)--(2,-2)); draw((0,0)--(-2,-2)); label("$|x-|2x+1||=3$",(0,0),N); label("$x-|2x+1|=3$",(-2,-2),S); label("$x-|2x+1|=-3$",(2,-2),S); label("$|2x+1|=x-3$",(-2,-2.25),S); label("$|2x+1|=x+3$",(2,-2.25),S); draw((2,-2.5)--(3,-3.5)); draw((2,-2.5)--(1,-3.5)); draw((-2,-2.5)--(-3,-3.5)); draw((-2,-2.5)--(-1,-3.5)); label("$2x+1=x-3$",(-3,-3.5),S); label("$x=-2$",(-3,-3.75),S); label("$2x+1=-x+3$",(-1,-3.5),S); label("$x=\frac{2}{3}$",(-1,-3.75),S); label("$2x+1=x+3$",(1,-3.5),S); label("$x=2$",(1,-3.75),S); label("$2x+1=-x-3$",(3,-3.5),S); label("$x=-\frac{4}{3}$",(3,-3.75),S); [/asy]
So we have $4$ possible solutions: $-2, \frac{2}{3}, 2,$ and $-\frac{4}{3}$ . Checking for extraneous solutions, we find that the only ones that work are $2$ and $-\frac{4}{3}$ , so there are $2$ solutions, $\boxed{2}$

Answer: C 2
Problem: The number of triples $(a, b, c)$ of positive integers which satisfy the simultaneous equations
$ab+bc=44$
$ac+bc=23$
is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ } 4$

Solution: We can factor the second equation to get $c(a+b)=23$ , so we see that $c$ must be a factor of $23$ , and since this is prime $c=1$ or $c=23$ . However, if $c=23$ , then $a+b=1$ , which is impossible for the field of positive integers. Therefore, $c=1$ for all possible solutions. Substituting this into the original equations gives
$ab+b=44$
and
$a+b=23$
From the second equation, $a=23-b$ , and substituting this into the first equation yields $b(23-b)+b=44$ , or $b^2-24b+44=0$ . Factoring this gives $(b-2)(b-22)=0$ , so $b=2$ or $b=22$ . Both of these yield integer solutions for $a$ , giving $a=21$ or $a=1$ , respectively.
Therefore, the only solutions are $(21, 2, 1)$ and $(1, 22, 1)$ , yielding $2$ solutions, $\boxed{2}$

Answer: C 2
Problem: If $a$ and $b$ are positive real numbers and each of the equations $x^2+ax+2b=0$ and $x^2+2bx+a=0$ has real roots , then the smallest possible value of $a+b$ is
$\mathrm{(A) \ }2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ }5 \qquad \mathrm{(E) \ } 6$

Solution: Since both of the equations have real roots, both of their discriminants are nonnegative. Therefore, we have
$a^2-4(1)(2b)=a^2-8b\geq0\implies a^2\geq8b$ from the first equation, and
$(2b)^2-4(1)(a)=4b^2-4a\geq0\implies 4b^2\geq4a\implies b^2\geq a$ from the second.
We can square the second equation to get $b^4\geq a^2$ , and combining this with the first one gives $b^4\geq a^2\geq8b$ , so $b^4\geq8b$ . We can divide both sides by $b$ , since it is positive, and take the cubed root of that to get $b\geq2$ . Therefore, we have $a^2\geq8b\geq8(2)=16$ , and since $a$ is positive, we can take the square root of this to get $a\geq4$ . Therefore, $a+b\geq 2+4=6$ , and the smallest possible value is $6, \boxed{6}$

Answer: E 6
Problem: The number of distinct pairs of integers $(x, y)$ such that $0<x<y$ and $\sqrt{1984}=\sqrt{x}+\sqrt{y}$ is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ }4 \qquad \mathrm{(E) \ } 7$

Solution: We can simplify $\sqrt{1984}$ to $8\sqrt{31}$ . Therefore, the only solutions are $a\sqrt{31}+b\sqrt{31}$ such that $a+b=8$ and $0<a<b$ . The only solutions to this are $a=1, b=7; a=2, b=6; a=3, b=5$ . Each of these gives distinct pairs of $(x, y)$ , so there are $3$ pairs, $\boxed{3}$

Answer: C 3
Problem: If $x \neq 0, \frac x{2} = y^2$ and $\frac{x}{4} = 4y$ , then $x$ equals
$\textbf{(A)}\ 8\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 32\qquad \textbf{(D)}\ 64\qquad \textbf{(E)}\ 128$

Solution: From $\frac{x}{4} = 4y$ , we get $x=16y$ . Plugging in the other equation, $\frac{16y}{2} = y^2$ , so $y^2-8y=0$ . Factoring, we get $y(y-8)=0$ , so the solutions are $0$ and $8$ . Since $x \neq 0$ , we also have $y \neq 0$ , so $y=8$ . Hence $x=16\cdot 8 = \boxed{128}$

Answer: E 128
Problem: Point $P$ is outside circle $C$ on the plane. At most how many points on $C$ are $3$ cm from $P$
$\textbf{(A)} \ 1 \qquad  \textbf{(B)} \ 2 \qquad  \textbf{(C)} \ 3 \qquad  \textbf{(D)} \ 4 \qquad  \textbf{(E)} \ 8$

Solution: The points $3$ cm away from $P$ can be represented as a circle centered at $P$ with radius $3$ cm. The maximum number of intersection points of two circles is $\boxed{2}$

Answer: B 2
Problem: Three primes $p,q$ , and $r$ satisfy $p+q = r$ and $1 < p < q$ . Then $p$ equals
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 13\qquad \textbf{(E)}\ 17$

Solution: We are given that $p,q$ and $r$ are primes. In order for $p$ and $q$ to sum to another prime, either $p$ or $q$ has to be even, because the sum of two odd numbers would be even, and the only even prime is $2$ (but $p + q = 2$ would have, as the only solution in positive integers, $p = q = 1$ , and $1$ is not prime). Thus, with one of either $p$ or $q$ being even, either $p$ or $q$ must be $2$ , and as $p < q$ , we deduce $p = 2$ (as $2$ is the smallest prime). This means the answer is $\boxed{2}$

Answer: A 2
Problem: When $x^5, x+\frac{1}{x}$ and $1+\frac{2}{x} + \frac{3}{x^2}$ are multiplied, the product is a polynomial of degree.
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8$

Solution: We have $x^5\left(x+\frac{1}{x}\right)\left(1+\frac{2}{x}+\frac{3}{x^2}\right) = (x^6+x^4)\left(1+\frac{2}{x}+\frac{3}{x^2}\right) = x^6 + \text{lower order terms}$ , where we know that the $x^6$ will not get cancelled out by e.g. a $-x^6$ term since all the terms inside the brackets are positive. Thus the degree is $6$ , which is choice $\boxed{6}$

Answer: C 6