Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn for all $i=1,2,3,\ldots, m$ and $j=1,\allowbreak 2,\allowbreak 3, \ldots, \allowbreak n$ , no point strictly between $\ell_A$ and $\ell_B$ lies on more than 1 of the segments. Find the number of bounded regions into which this figure divides the plane when $m=7$ and $n=5$ . The figure shows that there are 8 regions when $m=3$ and $n=2$ [asy] import geometry; size(10cm); draw((-2,0)--(13,0)); draw((0,4)--(10,4)); label("$\ell_A$",(-2,0),W); label("$\ell_B$",(0,4),W); point A1=(0,0),A2=(5,0),A3=(11,0),B1=(2,4),B2=(8,4),I1=extension(B1,A2,A1,B2),I2=extension(B1,A3,A1,B2),I3=extension(B1,A3,A2,B2); draw(B1--A1--B2); draw(B1--A2--B2); draw(B1--A3--B2); label("$A_1$",A1,S); label("$A_2$",A2,S); label("$A_3$",A3,S); label("$B_1$",B1,N); label("$B_2$",B2,N); label("1",centroid(A1,B1,I1)); label("2",centroid(B1,I1,I3)); label("3",centroid(B1,B2,I3)); label("4",centroid(A1,A2,I1)); label("5",(A2+I1+I2+I3)/4); label("6",centroid(B2,I2,I3)); label("7",centroid(A2,A3,I2)); label("8",centroid(A3,B2,I2)); dot(A1); dot(A2); dot(A3); dot(B1); dot(B2); [/asy]

Solution: We know the by Euler's Formula for planar graphs that $F-E+V=2$ , where $F$ is the number of bounded faces, plus the outer region, $E$ is the number of edges, and $V$ is the number of vertices. Temporarily disregarding the intersections between the lines, we can easily calculate that:
$V_{i}=7+5=12$
$E_{i}=6+4+7\cdot5=45$
However, the resulting graph is not planar, as the edges clearly intersect. To account for this, we must turn all intersection points into vertices, and update our values accordingly.
Observe that each intersection point can be mapped to two points on either line, and analogously, two points on either line can be mapped to one intersection point, uniquely. Thus, to count intersection points, we simply calculate:
${7 \choose 2}{5 \choose 2} = 210$
And thus,
$V=V_{i}+210=222$
We must also account for the edges. Observe that each intersection point turns the two edges that make it into four, that is, each intersection point adds $2$ to the number of edges. Therefore,
$E=E_{i}+2\cdot210=465$
Plugging these into Euler's Formula we get: $F-E+V=2$
$F-465+222=2$
$F=245$
Disregarding the outer region, we conclude that our answer is $F-1=245-1=\boxed{244}$

Answer:  244
Problem: Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$

Solution: [asy]  defaultpen(fontsize(12)+0.6); size(300);  pair A,B,C,D,M,H; real xb=71, xd=121; A=origin; D=(7,0); B=2*dir(xb); C=3*dir(xd)+D; M=(B+C)/2; H=foot(M,A,D); path c=CR(D,3); pair A1=bisectorpoint(D,A,B), D1=bisectorpoint(C,D,A), Bp=IP(CR(A,2),A--H), Cp=IP(CR(D,3),D--H);  draw(B--A--D--C--B); draw(A--M--D^^M--H^^Bp--M--Cp, gray+0.4); draw(rightanglemark(A,H,M,5));  dot("$A$",A,SW); dot("$D$",D,SE); dot("$B$",B,NW); dot("$C$",C,NE); dot("$M$",M,up); dot("$H$",H,down); dot("$B'$",Bp,down); dot("$C'$",Cp,down); [/asy]
According to the problem, we have $AB=AB'=2$ $DC=DC'=3$ $MB=MB'$ $MC=MC'$ , and $B'C'=7-2-3=2$
Because $M$ is the midpoint of $BC$ , we have $BM=MC$ , so: \[MB=MB'=MC'=MC.\]
Then, we can see that $\bigtriangleup{MB'C'}$ is an isosceles triangle with $MB'=MC'$
Therefore, we could start our angle chasing: $\angle{MB'C'}=\angle{MC'B'}=180^\circ-\angle{MC'D}=180^\circ-\angle{MCD}$
This is when we found that points $M$ $C$ $D$ , and $B'$ are on a circle. Thus, $\angle{BMB'}=\angle{CDC'} \Rightarrow \angle{B'MA}=\angle{C'DM}$ . This is the time we found that $\bigtriangleup{AB'M} \sim \bigtriangleup{MC'D}$
Thus, $\frac{AB'}{B'M}=\frac{MC'}{C'D} \Longrightarrow (B'M)^2=AB' \cdot C'D = 6$
Point $H$ is the midpoint of $B'C'$ , and $MH \perp AD$ $B'H=HC'=1 \Longrightarrow MH=\sqrt{B'M^2-B'H^2}=\sqrt{6-1}=\sqrt{5}$
The area of this quadrilateral is the sum of areas of triangles: \[S_{\bigtriangleup{ABM}}+S_{\bigtriangleup{AB'M}}+S_{\bigtriangleup{CDM}}+S_{\bigtriangleup{CD'M}}+S_{\bigtriangleup{B'C'M}}\] \[=S_{\bigtriangleup{AB'M}}\cdot 2 + S_{\bigtriangleup{B'C'M}} + S_{\bigtriangleup{C'DM}}\cdot 2\] \[=2 \cdot \frac{1}{2} \cdot AB' \cdot MH + \frac{1}{2} \cdot B'C' \cdot MH + 2 \cdot \frac{1}{2} \cdot C'D \cdot MH\] \[=2\sqrt{5}+\sqrt{5}+3\sqrt{5}=6\sqrt{5}\]
Finally, the square of the area is $(6\sqrt{5})^2=\boxed{180}$

Answer:  180
Problem: Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$

Solution: Denote by $M$ the midpoint of segment $BC$ .
Let points $P$ and $Q$ be on segment $AD$ , such that $AP = AB$ and $DQ = DC$
Denote $\angle DAM = \alpha$ $\angle BAD = \beta$ $\angle BMA = \theta$ $\angle CMD = \phi$
Denote $BM = x$ . Because $M$ is the midpoint of $BC$ $CM = x$
Because $AM$ is the angle bisector of $\angle BAD$ and $AB = AP$ $\triangle BAM \cong \triangle PAM$ .
Hence, $MP = MB$ and $\angle AMP = \theta$ .
Hence, $\angle MPD = \angle MAP + \angle PMA = \alpha + \theta$
Because $DM$ is the angle bisector of $\angle CDA$ and $DC = DQ$ $\triangle CDM \cong \triangle QDM$ .
Hence, $MQ = MC$ and $\angle DMQ = \phi$ .
Hence, $\angle MQA = \angle MDQ + \angle QMD = \beta + \phi$
Because $M$ is the midpoint of segment $BC$ $MB = MC$ .
Because $MP = MB$ and $MQ = MC$ $MP = MQ$
Thus, $\angle MPD = \angle MQA$
Thus, \[ \alpha + \theta = \beta + \phi . \hspace{1cm} (1) \]
In $\triangle AMD$ $\angle AMD = 180^\circ - \angle MAD - \angle MDA = 180^\circ - \alpha - \beta$ .
In addition, $\angle AMD = 180^\circ - \angle BMA - \angle CMD = 180^\circ - \theta - \phi$ .
Thus, \[ \alpha + \beta = \theta + \phi . \hspace{1cm} (2) \]
Taking $(1) + (2)$ , we get $\alpha = \phi$ .
Taking $(1) - (2)$ , we get $\beta = \theta$
Therefore, $\triangle ADM \sim \triangle AMB \sim \triangle MDC$
Hence, $\frac{AD}{AM} = \frac{AM}{AB}$ and $\frac{AD}{DM} = \frac{DM}{CD}$ .
Thus, $AM = \sqrt{AD \cdot AD} = \sqrt{14}$ and $DM = \sqrt{AD \cdot CD} = \sqrt{21}$
In $\triangle ADM$ , by applying the law of cosines, $\cos \angle AMD  = \frac{AM^2 + DM^2 - AD^2}{2 AM \cdot DM} = - \frac{1}{\sqrt{6}}$ .
Hence, $\sin \angle AMD = \sqrt{1 - \cos^2 \angle AMD} = \frac{\sqrt{5}}{\sqrt{6}}$ .
Hence, ${\rm Area} \ \triangle ADM = \frac{1}{2} AM \cdot DM \dot \sin \angle AMD = \frac{7 \sqrt{5}}{2}$
Therefore, \begin{align*} {\rm Area} \ ABCD & = {\rm Area} \ \triangle AMD + {\rm Area} \ \triangle ABM + {\rm Area} \ \triangle MCD \\ & = {\rm Area} \ \triangle AMD \left( 1 + \left( \frac{AM}{AD} \right)^2 + \left( \frac{MD}{AD} \right)^2 \right) \\ & = 6 \sqrt{5} . \end{align*}
Therefore, the square of ${\rm Area} \ ABCD$ is $\left( 6 \sqrt{5} \right)^2 = \boxed{180}$

Answer:  180
Problem: Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$

Solution: Claim
In the triangle $ABC, AB = 2AC, M$ is the midpoint of $AB. D$ is the point of intersection of the circumcircle and the bisector of angle $A.$ Then $DM = BD.$
Proof
Let $A = 2\alpha.$ Then $\angle DBC = \angle DCB = \alpha.$
Let $E$ be the intersection point of the perpendicular dropped from $D$ to $AB$ with the circle.
Then the sum of arcs $\overset{\Large\frown} {BE} + \overset{\Large\frown}{AC} + \overset{\Large\frown}{CD} = 180^\circ.$ \[\overset{\Large\frown} {BE} = 180^\circ – 2\alpha – \overset{\Large\frown}{AC}.\]
Let $E'$ be the point of intersection of the line $CM$ with the circle. $CM$ is perpendicular to $AD, \angle AMC = 90^\circ – \alpha,$ the sum of arcs $\overset{\Large\frown}{A}C +  \overset{\Large\frown}{BE'} = 180^\circ – 2\alpha \implies E'$ coincides with $E.$
The inscribed angles $\angle DEM = \angle DEB, M$ is symmetric to $B$ with respect to $DE, DM = DB.$
Solution
Let $AB' = AB, DC' = DC, B'$ and $C'$ on $AD.$
Then $AB' = 2, DC' = 3, B'C' = 2 = AB'.$
Quadrilateral $ABMC'$ is cyclic.
Let $\angle A = 2\alpha.$ Then $\angle MBC' = \angle MC'B = \alpha.$
Circle $BB'C'C$ centered at $M, BC$ is its diameter, $\angle BC'C = 90^\circ.$ $\angle DMC' = \angle MC'B,$ since they both complete $\angle MC'C$ to $90^\circ.$
$\angle MB'A = \angle MC'D,$ since they are the exterior angles of an isosceles $\triangle MB'C'.$ $\triangle AMB' \sim \triangle MDC'$ by two angles. $\frac {AB'}{MC'} = \frac {MB'}{DC'}, MC' =\sqrt{AB' \cdot C'D} = \sqrt{6}.$
The height dropped from $M$ to $AD$ is $\sqrt{MB'^2 - (\frac{B'C'}{2})^2} =\sqrt{6 - 1} = \sqrt{5}.$
The areas of triangles $\triangle AMB'$ and $\triangle MC'B'$ are equal to $\sqrt{5},$ area of $\triangle MC'D$ is $\frac{3}{2} \sqrt{5}.$
\[\triangle AMB' = \triangle AMB, \triangle MC'D = \triangle MCD \implies\] The area of $ABCD$ is $(1 + 2 + 3) \sqrt{5} = 6\sqrt{5} \implies 6^2 \cdot 5 = \boxed{180}.$

Answer:  180
Problem: Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$

Solution: Extend $AB$ and $CD$ so they intersect at a point $X$ . Then note that $M$ is the incenter of $\triangle{XAD}$ , implying that $M$ is on the angle bisector of $X$ . Now because $XM$ is both an angle bisector and a median of $\triangle{XBC}$ $\triangle{XBC}$ is isosceles. Then we can start angle chasing:
Let $\angle{BAM}=a, \angle{CDM}=b,$ and $\angle{XBC}=c$ . Then $\angle{AMD}=\pi-(a+b), \angle{ABM}=\pi-c, \angle{DCM}=\pi-c$ , implying that $\angle{BMA}+\angle{CMD}=a+b$ , implying that $2c-(a+b)=(a+b)$ , or that $c=a+b$ . Substituting this into the rest of the diagram, we find that $\triangle{BMA} \sim \triangle{CDM} \sim \triangle{MDA}$
Then $\frac{AB}{BM}=\frac{MC}{CD}$ , or $BM=CM=\sqrt{6}$ . Moreover, $\frac{AB}{AM}=\frac{AM}{AD}$ , or $AM=\sqrt{14}$ . Similarly, $\frac{CD}{MD}=\frac{MD}{AD}$ , or $DM=\sqrt{21}$ . Then using Law of Cosines on $\triangle{AMD}$ , to get that $cos\angle{AMD}=-\frac{\sqrt{6}}{6}$ , or $sin\angle{AMD}=\frac{\sqrt{30}}{6}$
We finish by using the formula $K=\frac{1}{2}absinC$ , as follows:
$[ABCD]=[ABM]+[CDM]+[ADM]=\frac{\frac{\sqrt{30}}{6}(2\sqrt{6}+3\sqrt{6}+7\sqrt{6})}{2}=6\sqrt{5}$
$(6\sqrt{5})^2=\boxed{180}$

Answer:  180
Problem: Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$

Solution: As shown in paragraph one of solution 4, extending $AB$ and $CD$ to $X$ , we realize that $\triangle{XBC}$ is isosceles, thus $XM \perp BC$ . Let $XB = XC = x$ . And, midpoint $M$ is the incenter of $\triangle{XAD}$ . Construct perpendiculars $ME, MF, MG$ to sides $AD, AX, DX$ respectively (constructing the radii of the incircle). Let $EM = FM = GM = r$ . The semiperimeter $s = \frac{2x + 2 + 3 + 7}{2} = x+6$ . Since $FX$ is the tangent off the incircle, $FX = s - AD = x-1$ . So, $BF = BX - FX = 1$ . Because $\triangle{BFM} \sim \triangle{MFX}$
\[\frac{BF}{FM} = \frac{FM}{FX} \implies {FM}^2 = BF \cdot FX = x - 1 \implies r^2 = x - 1\]
By Heron's formula and the inradius area formula,
\[(x+6)r = \sqrt{(x+6)\cdot 4 \cdot 3 \cdot (x-1)} \implies (x+6)r^2 = 12(x-1) \implies (x+6)(x-1) = 12(x-1) \implies x=6\]
Then, $r^2 = x - 1 = 5 \implies r = \sqrt{5}$ . Finally,
\[[ABCD] = [ABM] + [CDM] + [AMD] = \frac{AB \cdot FM}{2} + \frac{CD \cdot GM}{2} + \frac{AD \cdot EM}{2} = \frac{2r}{2} + \frac{3r}{2} + \frac{7r}{2} = 6r = 6\sqrt{5}\]
Thus, our answer is $(6\sqrt{5})^{2} = \boxed{180}$

Answer:  180
Problem: Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$

Solution: Let the midpoint of $BC$ be $M$ . Angle-chase and observe that $\Delta AMD~\Delta ABM~\Delta MCD$ . Let $BM=CM=a$ and $AM=x$ and $DM=y$ . As a result of this similarity, we write
\[\dfrac2a=\dfrac a3,\]
which gives $a=\sqrt 6$ . Similarly, we write
\[\dfrac2x=\dfrac x7\]
and
\[\dfrac3y=\dfrac y7\]
to get $x=\sqrt{14}$ and $y=\sqrt{21}$
We now have all required side lengths; we can find the area of $\Delta AMD$ with Heron's formula. Doing so yields $\dfrac72\sqrt5$ . We could also bash out the areas of the other two triangles since we know all their side lengths (this is what I did :sob:), but a more intelligent method is to recall the triangles' similarity. The ratio of similarlity between $\Delta AMD$ and $\Delta ABM$ is $\dfrac{\sqrt{14}}7=\sqrt{\dfrac27}$ , and between $\Delta AMD$ and $\Delta MCD$ is $\dfrac{\sqrt{21}}7=\sqrt{\dfrac37}$ . Thus, the area ratios are $\dfrac27$ and $\dfrac37$ , respectively, so adding together we have $\dfrac27+\dfrac37=\dfrac57$ . Multiplying this by our $\dfrac72\sqrt5$ , we have $\dfrac52\sqrt5$ as their total area. Adding this to our original area, we have $\dfrac52\sqrt5+\dfrac72\sqrt5=\sqrt5\left(\dfrac52+\dfrac72\right)=\sqrt5\left(\dfrac{12}2\right)=6\sqrt5$
The square of this is $\boxed{180}$

Answer:  180
Problem: Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$

Solution: As in solution 4, let $X=AB\cap CD$ , so $M$ is the incenter of $ADX$ and $XB=XC$ . Let $XB=XC=x$ . Then the normalized barycentric coordinates of $B$ $C$ , and $M$ with respect to $ADX$ are $\left[\frac{x}{x+2}:0:\frac{2}{x+2}\right]$ $\left[0:\frac{x}{x+3}:\frac{3}{x+3}\right]$ , and $\left[\frac{x+3}{2x+12}:\frac{x+2}{2x+12}:\frac{7}{2x+12}\right]$ . So we have $\frac{1}{2}\frac{x}{x+2}=\frac{x+3}{2x+12}$ giving $x=6$ . The sidelengths of $ADX$ are thus $AD=7$ $AX=8$ , and $DX=9$ giving $[ADX]=12\sqrt 5$ . Also, we have $[BCX]=\frac{6}{8}\cdot\frac{6}{9}[ADX]=6\sqrt 5$ so that $[ABCD]=[ADX]-[BCX]=6\sqrt 5$ . The area squared is thus $\boxed{180}$ .
~~ golue3120

Answer:  180
Problem: Let $a, b, x,$ and $y$ be real numbers with $a>4$ and $b>1$ such that \[\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1.\] Find the least possible value of $a+b.$

Solution: Denote $P = \left( x , y \right)$
Because $\frac{x^2}{a^2}+\frac{y^2}{a^2-16} = 1$ $P$ is on an ellipse whose center is $\left( 0 , 0 \right)$ and foci are $\left( - 4 , 0 \right)$ and $\left( 4 , 0 \right)$
Hence, the sum of distance from $P$ to $\left( - 4 , 0 \right)$ and $\left( 4 , 0 \right)$ is equal to twice the major axis of this ellipse, $2a$
Because $\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2} = 1$ $P$ is on an ellipse whose center is $\left( 20 , 11 \right)$ and foci are $\left( 20 , 10 \right)$ and $\left( 20 , 12 \right)$
Hence, the sum of distance from $P$ to $\left( 20 , 10 \right)$ and $\left( 20 , 12 \right)$ is equal to twice the major axis of this ellipse, $2b$
Therefore, $2a + 2b$ is the sum of the distance from $P$ to four foci of these two ellipses.
To make this minimized, $P$ is the intersection point of the line that passes through $\left( - 4 , 0 \right)$ and $\left( 20 , 10 \right)$ , and the line that passes through $\left( 4 , 0 \right)$ and $\left( 20 , 12 \right)$
The distance between $\left( - 4 , 0 \right)$ and $\left( 20 , 10 \right)$ is $\sqrt{\left( 20 + 4 \right)^2 + \left( 10 - 0 \right)^2} = 26$
The distance between $\left( 4 , 0 \right)$ and $\left( 20 , 12 \right)$ is $\sqrt{\left( 20 - 4 \right)^2 + \left( 12 - 0 \right)^2} = 20$
Hence, $2 a + 2 b = 26 + 20 = 46$
Therefore, $a + b = \boxed{023}.$

Answer:  023
Problem: There is a polynomial $P(x)$ with integer coefficients such that \[P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}\] holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$

Solution: Because $0 < x < 1$ , we have \begin{align*} P \left( x \right) & = \sum_{a=0}^6  \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \binom{6}{a} x^{2310a} \left( - 1 \right)^{6-a} x^{105b} x^{70c} x^{42d} x^{30e} \\ & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-a} x^{2310 a + 105 b + 70 c + 42 d + 30 e} . \end{align*}
Denote by $c_{2022}$ the coefficient of $P \left( x \right)$ .
Thus, \begin{align*} c_{2022} & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-a} \Bbb I \left\{ 2310 a + 105 b + 70 c + 42 d + 30 e = 2022 \right\} \\ & =  \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-0} \Bbb I \left\{ 2310 \cdot 0 + 105 b + 70 c + 42 d + 30 e = 2022 \right\} \\ & = \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \Bbb I \left\{ 105 b + 70 c + 42 d + 30 e = 2022 \right\} . \end{align*}
Now, we need to find the number of nonnegative integer tuples $\left( b , c , d , e \right)$ that satisfy \[ 105 b + 70 c + 42 d + 30 e = 2022 . \hspace{1cm} (1) \]
Modulo 2 on Equation (1), we have $b \equiv 0 \pmod{2}$ .
Hence, we can write $b = 2 b'$ . Plugging this into (1), the problem reduces to finding the number of 
nonnegative integer tuples $\left( b' , c , d , e \right)$ that satisfy \[ 105 b' + 35 c + 21 d + 15 e = 1011 . \hspace{1cm} (2) \]
Modulo 3 on Equation (2), we have $2 c \equiv 0 \pmod{3}$ .
Hence, we can write $c = 3 c'$ . Plugging this into (2), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c' , d , e \right)$ that satisfy \[ 35 b' + 35 c' + 7 d + 5 e = 337 . \hspace{1cm} (3) \]
Modulo 5 on Equation (3), we have $2 d \equiv 2 \pmod{5}$ .
Hence, we can write $d = 5 d' + 1$ . Plugging this into (3), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c' , d' , e \right)$ that satisfy \[ 7 b' + 7 c' + 7 d' + e = 66 . \hspace{1cm} (4) \]
Modulo 7 on Equation (4), we have $e \equiv 3 \pmod{7}$ .
Hence, we can write $e = 7 e' + 3$ . Plugging this into (4), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c' , d' , e' \right)$ that satisfy \[ b' + c' + d' + e' = 9 . \hspace{1cm} (5) \]
The number of nonnegative integer solutions to Equation (5) is $\binom{9 + 4 - 1}{4 - 1} = \binom{12}{3} =  \boxed{220}$

Answer:  220
Problem: There is a polynomial $P(x)$ with integer coefficients such that \[P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}\] holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$

Solution: Note that $2022 = 210\cdot 9 +132$ . Since the only way to express $132$ in terms of $105$ $70$ $42$ , or $30$ is $132 = 30+30+30+42$ , we are essentially just counting the number of ways to express $210*9$ in terms of these numbers. Since $210 = 2*105=3*70=5*42=7*30$ , it can only be expressed as a sum in terms of only one of the numbers ( $105$ $70$ $42$ , or $30$ ). Thus, the answer is (by sticks and stones) \[\binom{12}{3} = \boxed{220}\]

Answer:  220
Problem: For positive integers $a$ $b$ , and $c$ with $a < b < c$ , consider collections of postage stamps in denominations $a$ $b$ , and $c$ cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to $1000$ cents, let $f(a, b, c)$ be the minimum number of stamps in such a collection. Find the sum of the three least values of $c$ such that $f(a, b, c) = 97$ for some choice of $a$ and $b$

Solution: This should be $\lfloor \frac{1000}{c} \rfloor$ . The current function breaks when $c \mid 1000$ and $b \mid c$ . Take $c = 200$ and $b = 20$ . Then, we have $\lfloor \frac{999}{200} \rfloor = 4$ stamps of value 200, $\lfloor \frac{199}{20} \rfloor = 9$ stamps of value b, and 19 stamps of value 1. The maximum such a collection can give is $200 \cdot 4 + 20 \cdot 9 +19 \cdot 1 = 999$ , just shy of the needed 1000. As for the rest of solution, proceed similarly, except use $1000$ instead of $999$
Also, some explanation: $b-1$ one cent stamps cover all residues module $b$ . Having $\lfloor \frac{c-1}{b} \rfloor$ stamps of value b covers all residue classes modulo $c$ . Finally, we just need $\lfloor \frac{1000}{c} \rfloor$ to cover everything up to 1000.
In addition, note that this function sometimes may not always minimize the number of stamps required. This is due to the fact that the stamps of value $b$ and of value $1$ have the capacity to cover values greater than or equal to $c$ (which occurs when $c-1$ has a remainder less than $b-1$ when divided by $b$ ). Thus, in certain cases, not all $\lfloor \frac{1000}{c} \rfloor$ stamps of value c may be necessary, because the stamps of value $b$ and 1 can replace one $c$
CrazyVideoGamez
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Therefore using $\lfloor \frac{999}{c} \rfloor$ stamps of value $c$ $\lfloor \frac{c-1}{b} \rfloor$ stamps of value $b$ , and $b-1$ stamps of value $1$ , all values up to $1000$ can be represented in sub-collections, while minimizing the number of stamps.
So, $f(a, b, c) = \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1$
$\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97$ . We can get the answer by solving this equation.
$c > \lfloor \frac{c-1}{b} \rfloor + b-1$
$\frac{999}{c} + c > \lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 = 97$
$c^2 - 97c + 999 > 0$ $c > 85.3$ or $c < 11.7$
$\lfloor \frac{999}{c} \rfloor + \lfloor \frac{c-1}{b} \rfloor + b-1 > \frac{999}{c}$
$97 > \frac{999}{c}$ $c>10.3$
The $3$ least values of $c$ are $11$ $88$ $89$ $11 + 88+ 89 = \boxed{188}$

Answer:  188
Problem: Zou and Chou are practicing their $100$ -meter sprints by running $6$ races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is $\frac23$ if they won the previous race but only $\frac13$ if they lost the previous race. The probability that Zou will win exactly $5$ of the $6$ races is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: For the next five races, Zou wins four and loses one. Let $W$ and $L$ denote a win and a loss, respectively. There are five possible outcome sequences for Zou:
We proceed with casework:
Case (1): Sequences #1-4, in which Zou does not lose the last race.
The probability that Zou loses a race is $\frac13,$ and the probability that Zou wins the next race is $\frac13.$ For each of the three other races, the probability that Zou wins is $\frac23.$
There are four sequences in this case. The probability of one such sequence is $\left(\frac13\right)^2\left(\frac23\right)^3.$
Case (2): Sequence #5, in which Zou loses the last race.
The probability that Zou loses a race is $\frac13.$ For each of the four other races, the probability that Zou wins is $\frac23.$
There is one sequence in this case. The probability is $\left(\frac13\right)^1\left(\frac23\right)^4.$
Answer
The requested probability is \[4\left(\frac13\right)^2\left(\frac23\right)^3+\left(\frac13\right)^1\left(\frac23\right)^4=\frac{32}{243}+\frac{16}{243}=\frac{48}{243}=\frac{16}{81},\] from which the answer is $16+81=\boxed{097}.$

Answer:  097