Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$

Solution: Let $O_i$ and $r_i$ be the center and radius of $\omega_i$ , and let $O$ and $r$ be the center and radius of $\omega$
Since $\overline{AB}$ extends to an arc with arc $120^\circ$ , the distance from $O$ to $\overline{AB}$ is $r/2$ . Let $X=\overline{AB}\cap \overline{O_1O_2}$ . Consider $\triangle OO_1O_2$ . The line $\overline{AB}$ is perpendicular to $\overline{O_1O_2}$ and passes through $X$ . Let $H$ be the foot from $O$ to $\overline{O_1O_2}$ ; so $HX=r/2$ . We have by tangency $OO_1=r+r_1$ and $OO_2=r+r_2$ . Let $O_1O_2=d$ [asy] unitsize(3cm); pointpen=black; pointfontpen=fontsize(9);  pair A=dir(110), B=dir(230), C=dir(310);   DPA(A--B--C--A);    pair H = foot(A, B, C);  draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted);  pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X);   D("O",A,dir(A)); D("O_1",B,dir(B)); D("O_2",C,dir(C)); D("H",H,dir(270)); D("X",X,dir(225)); D("A",R1,dir(180)); D("B",R2,dir(180));  draw(rightanglemark(Y,X,C,3));   [/asy] Since $X$ is on the radical axis of $\omega_1$ and $\omega_2$ , it has equal power with respect to both circles, so \[O_1X^2 - r_1^2 = O_2X^2-r_2^2 \implies O_1X-O_2X = \frac{r_1^2-r_2^2}{d}\] since $O_1X+O_2X=d$ . Now we can solve for $O_1X$ and $O_2X$ , and in particular, \begin{align*} O_1H &= O_1X - HX = \frac{d+\frac{r_1^2-r_2^2}{d}}{2} - \frac{r}{2} \\ O_2H &= O_2X + HX = \frac{d-\frac{r_1^2-r_2^2}{d}}{2} + \frac{r}{2}.  \end{align*} We want to solve for $d$ . By the Pythagorean Theorem (twice): \begin{align*} &\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\ &\implies \left(d+r-\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_2)^2 = \left(d-r+\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_1)^2 \\ &\implies 2dr - 2(r_1^2-r_2^2)-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\ &\implies 4dr = 8rr_2-8rr_1 \\ &\implies d=2r_2-2r_1 \end{align*} Therefore, $d=2(r_2-r_1) = 2(961-625)=\boxed{672}$

Answer:  672
Problem: Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$

Solution: Let $O_{1}$ $O_{2}$ , and $O$ be the centers of $\omega_{1}$ $\omega_{2}$ , and $\omega$ with $r_{1}$ $r_{2}$ , and $r$ their radii, respectively. Then, the distance from $O$ to the radical axis $\ell\equiv\overline{AB}$ of $\omega_{1}, \omega_{2}$ is equal to $\frac{1}{2}r$ . Let $x=O_{1}O_{2}$ and $O^{\prime}$ the orthogonal projection of $O$ onto line $\ell$ . Define the function $f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ by \[f(X)=\text{Pow}_{\omega_{1}}(X)-\text{Pow}_{\omega_{2}}(X).\] Then \begin{align*} f(O_{1})=-r_{1}^{2}-(x-r_{2})(x+r_{2})&=-x^{2}+r_{2}^{2}-r_{1}^{2}, \\ f(O_{2})=(x-r_{1})(x+r_{1})-(-r_{2}^{2})&=x^{2}+r_{2}^{2}-r_{1}^{2}, \\ f(O)=r(r+2r_{1})-r(r+2r_{2})&=2r(r_{1}-r_{2}), \\ f(O^{\prime})&=0. \end{align*} By linearity \[\frac{f(O_{2})-f(O_{1})}{f(O)-f(O^{\prime})}=\frac{O_{1}O_{2}}{OO^{\prime}}=\frac{x}{\tfrac{1}{2}r}=\frac{2x}{r}.\] Notice that $f(O_{2})-f(O_{1})=x^{2}-(-x^{2})=2x^{2}$ and $f(O)-f(O^{\prime})=2r(r_{1}-r_{2})$ , thus \begin{align*}\frac{2x^{2}}{2r(r_{1}-r_{2})}&=\frac{2x}{r}\end{align*} Dividing both sides by $\frac{2x}{r}$ (which is obviously nonzero as $x$ is nonzero) gives us \begin{align*}\frac{x}{2(r_{1}-r_{2})}&=1\end{align*} so $x=2(r_{1}-r_{2})$ . Since $r_{1}=961$ and $r_{2}=625$ , the answer is $x=2\cdot(961-625)=\boxed{672}$

Answer:  672
Problem: Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$

Solution: Denote by $O_1$ $O_2$ , and $O$ the centers of $\omega_1$ $\omega_2$ , and $\omega$ , respectively. Let $R_1 = 961$ and $R_2 = 625$ denote the radii of $\omega_1$ and $\omega_2$ respectively, $r$ be the radius of $\omega$ , and $\ell$ the distance from $O$ to the line $AB$ . We claim that \[\dfrac{\ell}{r} = \dfrac{R_2-R_1}{d},\] where $d = O_1O_2$ . This solves the problem, for then the $\widehat{PQ} = 120^\circ$ condition implies $\tfrac{\ell}r = \cos 60^\circ = \tfrac{1}{2}$ , and then we can solve to get $d = \boxed{672}$ [asy] import olympiad; size(230pt); defaultpen(linewidth(0.8)+fontsize(10pt)); real r1 = 17, r2 = 27, d = 35, r = 18; pair O1 = origin, O2 = (d,0); path w1 = circle(origin,r1), w2 = circle((d,0),r2), w1p = circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label("$O_1$",origin,N); label("$O_2$",(d,0),N); label("$O$",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label("$X$",T,N,gray(0.6)); label("$Y$",foot(X[0],O1,O2),NE,gray(0.6)); label("$\ell$",(O+Tp)/2,S,gray(0.6)); [/asy]

Answer:  672
Problem: Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$

Solution: Suppose we label the points as shown below [asy] defaultpen(fontsize(12)+0.6); size(300); pen p=fontsize(10)+royalblue+0.4;  var r=1200; pair O1=origin, O2=(672,0), O=OP(CR(O1,961+r),CR(O2,625+r)); path c1=CR(O1,961), c2=CR(O2,625), c=CR(O,r); pair  A=IP(CR(O1,961),CR(O2,625)), B=OP(CR(O1,961),CR(O2,625)), P=IP(L(A,B,0,0.2),c),  Q=IP(L(A,B,0,200),c),  F=IP(CR(O,625+r),O--O1),  M=(F+O2)/2,  D=IP(CR(O,r),O--O1),  E=IP(CR(O,r),O--O2), X=extension(E,D,O,O+O1-O2), Y=extension(D,E,O1,O2);  draw(c1^^c2); draw(c,blue+0.6); draw(O1--O2--O--cycle,black+0.6); draw(O--X^^Y--O2,black+0.6); draw(X--Y,heavygreen+0.6); draw((X+O)/2--O,MidArrow); draw(O2--Y-(300,0),MidArrow);  dot("$A$",A,dir(A-O2/2)); dot("$B$",B,dir(B-O2/2)); dot("$O_2$",O2,right+up);  dot("$O_1$",O1,left+up);  dot("$O$",O,dir(O-O2));  dot("$D$",D,dir(170)); dot("$E$",E,dir(E-O1)); dot("$X$",X,dir(X-D)); dot("$Y$",Y,dir(Y-D));   label("$R$",O--E,right+up,p); label("$R$",O--D,left+down,p); label("$2R$",(X+O)/2-(150,0),down,p); label("$961$",O1--D,2*(left+down),p); label("$625$",O2--E,2*(right+up),p);  MA("",E,D,O1,100,fuchsia+linewidth(1)); MA("",X,D,O,100,fuchsia+linewidth(1)); MA("",Y,E,O2,100,orange+linewidth(1)); MA("",D,E,O,100,orange+linewidth(1)); [/asy] By radical axis, the tangents to $\omega$ at $D$ and $E$ intersect on $AB$ . Thus $PDQE$ is harmonic, so the tangents to $\omega$ at $P$ and $Q$ intersect at $X \in DE$ . Moreover, $OX \parallel O_1O_2$ because both $OX$ and $O_1O_2$ are perpendicular to $AB$ , and $OX = 2OP$ because $\angle POQ = 120^{\circ}$ . Thus \[O_1O_2 = O_1Y - O_2Y = 2 \cdot 961 - 2\cdot 625 = \boxed{672}\] by similar triangles.

Answer:  672
Problem: Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$

Solution: Like in other solutions, let $O$ be the center of $\omega$ with $r$ its radius; also, let $O_{1}$ and $O_{2}$ be the centers of $\omega_{1}$ and $\omega_{2}$ with $R_{1}$ and $R_{2}$ their radii, respectively. Let line $OP$ intersect line $O_{1}O_{2}$ at $T$ , and let $u=TO_{2}$ $v=TO_{1}$ $x=PT$ , where the length $O_{1}O_{2}$ splits as $u+v$ . Because the lines $PQ$ and $O_{1}O_{2}$ are perpendicular, lines $OT$ and $O_{1}O_{2}$ meet at a $60^{\circ}$ angle.
Applying the Law of Cosines to $\triangle O_{2}PT$ $\triangle O_{1}PT$ $\triangle O_{2}OT$ , and $\triangle O_{1}OT$ gives \begin{align*}\triangle O_{2}PT&:O_{2}P^{2}=u^{2}+x^{2}-ux \\ \triangle O_{1}PT&:O_{1}P^{2}=v^{2}+x^{2}+vx \\ \triangle O_{2}OT&:(r+R_{2})^{2}=u^{2}+(r+x)^{2}-u(r+x) \\ \triangle O_{1}OT&:(r+R_{1})^{2}=v^{2}+(r+x)^{2}+v(r+x)\end{align*}
Adding the first and fourth equations, then subtracting the second and third equations gives us \[\left(O_{2}P^{2}-O_{1}P^{2}\right)+\left(R_{1}^{2}-R_{2}^{2}\right)+2r(R_{1}-R_{2})=r(u+v)\]
Since $P$ lies on the radical axis of $\omega_{1}$ and $\omega_{2}$ , the power of point $P$ with respect to either circle is \[O_{2}P^{2}-R_{2}^{2}=O_{1}P^{2}-R_{1}^{2}.\]
Hence $2r(R_{1}-R_{2})=r(u+v)$ which simplifies to \[u+v=2(R_{1}-R_{2}).\]
The requested distance \[O_{1}O_{2}=O_{1}T+O_{2}T=u+v\] is therefore equal to $2\cdot(961-625)=\boxed{672}$

Answer:  672
Problem: Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$

Solution: Let circle $\omega$ tangent circles $\omega_1$ and $\omega_2,$ respectively at distinct points $C$ and $D$ . Let $O, O_1, O_2 (r, r_1, r_2)$ be the centers (the radii) of $\omega, \omega_1$ and $\omega_2,$ respectively. WLOG $r_1 < r_2.$ Let $F$ be the point of $OO_2$ such, that $OO_1 =OF.$ Let $M$ be the midpoint $FO_1, OE \perp AB, CT$ be the radical axes of $\omega_1$ and $\omega, T \in AB.$
Then $T$ is radical center of $\omega, \omega_1$ and $\omega_2, TD = CT.$
In $\triangle OFO_1 (OF = OO_1) OT$ is bisector of $\angle O, OM$ is median
$\hspace{10mm} \implies O,T,$ and $M$ are collinear.
\[\angle OCT = \angle ODT = \angle OET = 90^\circ \implies\]
$OCTDE$ is cyclic (in $\Omega), OT$ is diameter $\Omega.$ $O_1O_2 \perp AB, OM \perp FO_1 \implies \angle FO_1O_2 = \angle OTE$ $\angle OTE = \angle ODE$ as they intercept the same arc $\overset{\Large\frown}{OE}$ in $\Omega.$ \[OE \perp AB, O_1O_2 \perp AB \implies O_1 O_2 || OE \implies\] \[\angle OO_2O_1 = \angle O_2 OE \implies \triangle ODE \sim \triangle O_2 O_1 F \implies\] \[\frac {OE}{OD} = \frac {O_2F}{O_1O_2} \implies \cos \frac {120^\circ}{2} = \frac{r_2 + r - r_1 -r} {O_1O_2}\implies {O_1O_2}= 2|r_2 – r_1|.\]
Since $r_{1}=625$ and $r_{2}=961$ , the answer is $2\cdot|961-625|=\boxed{672}$

Answer:  672
Problem: Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$

Solution: We are not given the radius of circle $w$ , but based on the problem statement, that radius isn't important. We can set $w$ to have radius infinity (solution 8), but if you didn't observe that, you could also set the radius to be $2r$ so that the line containing the center of $w$ , call it $W$ , and $w_2$ , call it $W_2$ , is perpendicular to the line containing the center of $w_1$ , call it $W_1$ and $w_2$ . Let $AB = 2h$ and $W_1W_2 = x$ . Also, let the projections of $W$ and $W_1$ onto line $AB$ be $X$ and $Y$ , respectively.
By Pythagorean Theorem on $\triangle WW_1W_2$ , we get \[x^2+(625+2x)^2=(961+2r)^2 \;(1)\] Note that since $\angle PWQ = 120$ $\angle PWX = 60$ . So, $WX = 2r/2 = r = W_1Y$ . We now get two more equations from Pythag: \[h^2+r^2 = 625^2 \; (2)\] \[h^2+(x-r)^2 = 961^2 \; (3)\] From subtracting $(2)$ and $(3)$ $x^2-2rx=961^2-625^2 \; (4)$ . Rearranging $(3)$ yields $x^2-1344r = 961^2+625^2$ . Plugging in our result from $(4)$ $x^2-1344r= x^2-2rx \implies 1344r = 2rx \implies x=\boxed{672}$

Answer:  672
Problem: Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$

Solution: Let the circle $\omega$ be infinitely big (a line). Then for it to be split into an arc of $120^{\circ}$ $\overline{PQ}$ must intersect at a $60^{\circ}$ with line $\omega$
Notice the 30-60-90 triangle in the image. $O_1R = 961 - 625$
Thus, the distance between the centers of $\omega_1$ and $\omega_2$ is $2(961-625)=\boxed{672}$

Answer:  672
Problem: problem_id
891fbd11f453d2b468075929a7f4cfd8    For any positive integer $a, \sigma(a)$ denote...
891fbd11f453d2b468075929a7f4cfd8    Warning: This solution doesn't explain why $43...
Name: Text, dtype: object

Solution: We first claim that $n$ must be divisible by $42$ . Since $\sigma(a^n)-1$ is divisible by $2021$ for all positive integers $a$ , we can first consider the special case where $a$ is prime and $a \neq 0,1 \pmod{43}$ . By Dirichlet's Theorem (Refer to the Remark section.), such $a$ always exists.
Then $\sigma(a^n)-1 = \sum_{i=1}^n a^i = a\left(\frac{a^n - 1}{a-1}\right)$ . In order for this expression to be divisible by $2021=43\cdot 47$ , a necessary condition is $a^n - 1 \equiv 0 \pmod{43}$ . By Fermat's Little Theorem $a^{42} \equiv 1 \pmod{43}$ . Moreover, if $a$ is a primitive root modulo $43$ , then $\text{ord}_{43}(a) = 42$ , so $n$ must be divisible by $42$
By similar reasoning, $n$ must be divisible by $46$ , by considering $a \not\equiv 0,1 \pmod{47}$
We next claim that $n$ must be divisible by $43$ . By Dirichlet, let $a$ be a prime that is congruent to $1 \pmod{43}$ . Then $\sigma(a^n) \equiv n+1 \pmod{43}$ , so since $\sigma(a^n)-1$ is divisible by $43$ $n$ is a multiple of $43$
Alternatively, since $\left(\frac{a(a^n - 1^n)}{a-1}\right)$ must be divisible by $43,$ by LTE, we have $v_{43}(a)+v_{43}{(a-1)}+v_{43}{(n)}-v_{43}{(a-1)} \geq 1,$ which simplifies to $v_{43}(n) \geq 1,$ which implies the desired result.
Similarly, $n$ is a multiple of $47$
Lastly, we claim that if $n = \text{lcm}(42, 46, 43, 47)$ , then $\sigma(a^n) - 1$ is divisible by $2021$ for all positive integers $a$ . The claim is trivially true for $a=1$ so suppose $a>1$ . Let $a = p_1^{e_1}\ldots p_k^{e_k}$ be the prime factorization of $a$ . Since $\sigma(n)$ is multiplicative , we have \[\sigma(a^n) - 1 = \prod_{i=1}^k \sigma(p_i^{e_in}) - 1.\] We can show that $\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}$ for all primes $p_i$ and integers $e_i \ge 1$ , so \[\sigma(p_i^{e_in}) = 1 + (p_i + p_i^2 + \ldots + p_i^n) + (p_i^{n+1} + \ldots + p_i^{2n}) + \ldots + (p_i^{n(e_i-1)+1} + \ldots + p_i^{e_in}),\] where each expression in parentheses contains $n$ terms. It is easy to verify that if $p_i = 43$ or $p_i = 47$ then $\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}$ for this choice of $n$ , so suppose $p_i \not\equiv 0 \pmod{43}$ and $p_i \not\equiv 0 \pmod{47}$ . Each expression in parentheses equals $\frac{p_i^n - 1}{p_i - 1}$ multiplied by some power of $p_i$ . If $p_i \not\equiv 1 \pmod{43}$ , then FLT implies $p_i^n - 1 \equiv 0 \pmod{43}$ , and if $p_i \equiv 1 \pmod{43}$ , then $p_i + p_i^2 + \ldots + p_i^n \equiv 1 + 1 + \ldots + 1 \equiv 0 \pmod{43}$ (since $n$ is also a multiple of $43$ , by definition). Similarly, we can show $\sigma(p_i^{e_in}) \equiv 1 \pmod{47}$ , and a simple CRT argument shows $\sigma(p_i^{e_in}) \equiv 1 \pmod{2021}$ . Then $\sigma(a^n) \equiv 1^k \equiv 1 \pmod{2021}$
Then the prime factors of $n$ are $2,3,7,23,43,$ and $47,$ and the answer is $2+3+7+23+43+47 = \boxed{125}$

Answer:  125
Problem: problem_id
891fbd11f453d2b468075929a7f4cfd8    For any positive integer $a, \sigma(a)$ denote...
891fbd11f453d2b468075929a7f4cfd8    Warning: This solution doesn't explain why $43...
Name: Text, dtype: object

Solution: $n$ only needs to satisfy $\sigma(a^n)\equiv 1 \pmod{43}$ and $\sigma(a^n)\equiv 1 \pmod{47}$ for all $a$ . Let's work on the first requirement (mod 43) first. All $n$ works for $a=1$ . If $a>1$ , let $a$ 's prime factorization be $a=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ . The following three statements are the same:
We can show this by casework on $p$
Similar arguments for modulo $47$ lead to $46|n$ and $47|n$ . Therefore, we get $n=\operatorname{lcm}[42,43,46,47]$ . Following the last paragraph of Solution 1 gives the answer $\boxed{125}$

Answer:  125
Problem: problem_id
891fbd11f453d2b468075929a7f4cfd8    For any positive integer $a, \sigma(a)$ denote...
891fbd11f453d2b468075929a7f4cfd8    Warning: This solution doesn't explain why $43...
Name: Text, dtype: object

Solution: We perform casework on $a:$
Finally, the least such positive integer $n$ for all cases is \begin{align*} n&=\operatorname{lcm}(42,43,46,47) \\ &=\operatorname{lcm}(2\cdot3\cdot7,43,2\cdot23,47) \\ &=2\cdot3\cdot7\cdot23\cdot43\cdot47, \end{align*} so the sum of its prime factors is $2+3+7+23+43+47=\boxed{125}.$

Answer:  125
Problem: problem_id
891fbd11f453d2b468075929a7f4cfd8    For any positive integer $a, \sigma(a)$ denote...
891fbd11f453d2b468075929a7f4cfd8    Warning: This solution doesn't explain why $43...
Name: Text, dtype: object

Solution: Since the problem works for all positive integers $a$ , let's plug in $a=2$ and see what we get. Since $\sigma({2^n}) = 2^{n+1}-1,$ we have $2^{n+1} \equiv 2 \pmod{2021}.$ Simplifying using CRT and Fermat's Little Theorem , we get that $n \equiv 0 \pmod{42}$ and $n \equiv 0 \pmod{46}.$ Then, we can look at $a$ being a $1\pmod{43}$ prime and a $1\pmod{47}$ prime, just like in Solution 1, to find that $43$ and $47$ also divide $n$ . There don't seem to be any other odd "numbers" to check, so we can hopefully assume that the answer is the sum of the prime factors of $\text{lcm(42, 43, 46, 47)}.$ From here, follow solution 1 to get the final answer $\boxed{125}$

Answer:  125
Problem: Let $S$ be the set of positive integers $k$ such that the two parabolas \[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\] intersect in four distinct points, and these four points lie on a circle with radius at most $21$ . Find the sum of the least element of $S$ and the greatest element of $S$

Solution: Note that $y=x^2-k$ is an upward-opening parabola with the vertex at $(0,-k),$ and $x=2(y-20)^2-k$ is a rightward-opening parabola with the vertex at $(-k,20).$ We consider each condition separately:
Taking the intersection of Conditions 1 and 2 produces $5\leq k\leq280.$ Therefore, the answer is $5+280=\boxed{285}.$

Answer:  285
Problem: Let $S$ be the set of positive integers $k$ such that the two parabolas \[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\] intersect in four distinct points, and these four points lie on a circle with radius at most $21$ . Find the sum of the least element of $S$ and the greatest element of $S$

Solution: Make the translation $y \rightarrow y+20$ to obtain $20+y=x^2-k$ and $x=2y^2-k$ . Multiply the first equation by $2$ and sum, we see that $2(x^2+y^2)=3k+40+2y+x$ . Completing the square gives us $\left(y- \frac{1}{2}\right)^2+\left(x - \frac{1}{4}\right)^2 = \frac{325+24k}{16}$ ; this explains why the two parabolas intersect at four points that lie on a circle*. For the upper bound, observe that $LHS \leq 21^2=441 \rightarrow 24k \leq 6731$ , so $k \leq 280$
For the lower bound, we need to ensure there are $4$ intersections to begin with. (Here I'm using the un-translated coordinates.) Draw up a graph, and realize that two intersections are guaranteed, on the so called "right branch" of $y=x^2-k$ . As we increase the value of $k$ , two more intersections appear on the "left branch":
$k=4$ does not work because the "leftmost" point of $x=2(y-20)^2-4$ is $(-4,20)$ which lies to the right of $\left(-\sqrt{24}, 20\right)$ , which is on the graph $y=x^2-4$ . While technically speaking this doesn't prove that there are no intersections (why?), drawing the graph should convince you that this is the case. Clearly, $k<4$ does not work.
$k=5$ does work because the two graphs intersect at $(-5,20)$ , and by drawing the graph, you realize this is not a tangent point and there is in fact another intersection nearby, due to slope. Therefore, the answer is $5+280=\boxed{285}$

Answer:  285
Problem: Let $S$ be the set of positive integers $k$ such that the two parabolas \[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\] intersect in four distinct points, and these four points lie on a circle with radius at most $21$ . Find the sum of the least element of $S$ and the greatest element of $S$

Solution: Claim
Let the axes of two parabolas be perpendicular, their focal parameters be $p_1$ and $p_2$ and the distances from the foci to the point of intersection of the axes be $x_2$ and $y_1$ . Suppose that these parabolas intersect at four points.
Then these points lie on the circle centered at point $(p_2, p_1)$ with radius $r = \sqrt{2(p_1^2 + p_2^2 + p_1 y_1 + p_2 x_2)}.$
Proof
Let's introduce a coordinate system with the center at the point of intersection of the axes. Let the first (red) parabola have axis $x = 0,$ focal parameter $p_1$ and focus at point $A(0, –y_1), y_1 > 0.$ Let second (blue) parabola have axis $y = 0,$ focal parameter $p_2$ and focus at point $B(–x_2,0), x_2  > 0.$ Let us denote the angle between the vector connecting the focus of the first parabola and its point and the positive direction of the ordinate axis $2\theta,$ its length $\rho_1(\theta),$ the angle between the vector connecting the focus of the second parabola and its point and the positive direction of the abscissa axis $2\phi,$ its length $\rho_2(\phi).$ Then \[\rho_1(\theta) = \frac{p_1}{1 - \cos(2\theta)}, \rho_2(\phi) = \frac{p_2}{1 - \cos(2\phi)}.\]
Abscissa of the point of intersection is \begin{align*} x =\rho_1 \sin(2\theta) = p_1\cot\theta = \rho_2 \cos (2\phi) - x_2 = \frac{p_2}{2} (\cot^2\phi - 1)- x_2,\end{align*} \begin{align*} x^2 = p_1^2 \cot ^2 \theta , 2 p_1\cot\theta = p_2 \cos^2 \phi - p_2 - 2x_2 .\end{align*} Ordinate of the point of intersection is \begin{align*} y =\rho_2 \sin 2\phi = p_2\cot\phi = \rho_1 \cos 2\theta - y_1 = \frac{p_1}{2} (\cot^2\theta - 1)- y_1,\end{align*} \begin{align*} y^2 = p_2^2 \cot ^2 \phi , 2 p_2\cot\phi = p_1 \cos^2 \theta - p_1 - 2y_1 .\end{align*} The square of the distance from point of intersection to the point $(p_2, p_1)$ is \begin{align*} r^2 = (x-p_2)^2 + (y-p_1)^2 = x^2 + y^2 - 2 p_1 y - 2 p_2 x + p_1^2 + p_2^2 .\end{align*} After simple transformations, we get $r^2 = 2(p_1^2 + p_2^2 + p_1 y_1 + p_2 x_2).$
Hence, any intersection point has the same distance $r$ from the point $(p_2, p_1).$
Solution
Parameters of the parabola $y = x^2 – k$ are $p_1 = \frac{1}{2}, y_1 = 20 + k – \frac{1}{2}.$
Parameters of the parabola $\frac{x}{2} = (y – 20)^2 – \frac{k}{2}$ are $p_2 = \frac{1}{4}, x_2 = k – \frac{1}{4}  \implies r^2 = 20 + \frac{3k}{2}.$
If $r \le 21, k \le \frac{842}{3},$ then integer $k \le 280.$
The vertex of the second parabola is point $(– k,20)$ can be on the parabola $y = x^2 – k$ or below the point of the parabola with the same abscissa. So \[20 \ge (– k)^2 – k  \implies   5 \le k \le 280.\] Therefore, the answer is $5+280=\boxed{285}$

Answer:  285