heureka

Let \(a\) , \(b\) , and \(c\) be values chosen from \(\{~0,~1,~2,~3,~\ldots ~,100~\}\) .
They need not be distinct, but \(a\geq b\) .
How many ordered triples \((~a,~b,~c~)\) are there satisfying both \(|~a-b~|=c\) and \(|~b-c~|=a\) ?

\(\small{ \begin{array}{|lrcll|} \hline & |~a-b~| &=& c \qquad \text{square both sides} \\ (1) & (a-b)^2 &=& c^2 \\\\ & |~b-c~| &=& a \qquad \text{square both sides} \\ (2) & (b-c)^2 &=& a^2 \\ \hline \\ (1)+(2): & (a-b)^2 + (b-c)^2 &=& c^2 + a^2 \\ & a^2-2ab+b^2+b^2-2bc+c^2 &=& c^2 + a^2 \\ & -2ab+b^2+b^2-2bc &=& 0 \\ & 2b^2-2ab-2bc &=& 0 \quad | \quad : 2 \\ & b^2-ab-bc &=& 0 \\ &\mathbf{ b(b-a-c) } &=& \mathbf{0} \\ \hline \\ 1) & \mathbf{b} &=& \mathbf{0} \\ & |~a-b~| &=& c \quad | \quad b = 0 \\ & |~a-0~| &=& c \\ & |~a ~| &=& c \\ & \mathbf{a} &=& \mathbf{c} \\ \quad \text{List}~ 1: & \mathbf{(~a,~0,~a~)} && \begin{array}{|lcll|} \hline (~0,~0,~0~),\quad (~1,~0,~1~),\quad (~2,~0,~2~),\ \ldots (~99,~0,~99~),\ (~100,~0,~100~) \\ \mathbf{101} ~\text{ordered triples} \\ \hline \end{array} \\ \hline \\ 2) & \mathbf{b-a-c} &=& \mathbf{0} \\ & a+c &=& b \\ & a &\geq& b \quad | \quad b = a+c \\ & a&\geq & a+c \quad \text{possible if $c=0$} \\ & \mathbf{c} &=& \mathbf{0} \\ & |~b-c~| &=& a \quad | \quad c = 0 \\ & |~b-0~| &=& a \\ & |~b ~| &=& a \\ & \mathbf{b} &=& \mathbf{a} \\ \quad \text{List}~ 2: & \mathbf{(~a,~a,~0~)} && \begin{array}{|lcll|} \hline (~1,~1,~0~),\quad (~2,~2,~0~),\quad (~3,~3,~0~),\ \ldots (~99,~99,~0~),\ (~100,~100,~0~) \\ \mathbf{100} ~\text{ordered triples}\qquad (~0,~0,~0~)~\text{ is already included in the first list } \\ \hline \end{array} \\ \hline \end{array} }\)

\(\text{There are $101+100 = \mathbf{201}$ ordered triples}\)

Find a monic quadratic polynomial f(x)  such that the remainder when f(x) is divided by x-1 is 2
and the remainder when f(x) is divided by x-3 is 4.

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{q_1(x)(x-1) + 2} \quad &| \quad x = 1 \\ f(1) &=& q_1(x)*0 + 2 \\ \mathbf{f(1)} &=& \mathbf{2} \\\\ \mathbf{f(x)} &=& \mathbf{ q_2(x)(x-3) + 4 } \quad &| \quad x = 3 \\ f(3) &=& q_2(x)*0 + 4 \\ \mathbf{f(3)} &=& \mathbf{4} \\ \hline \end{array}\)

Monic quadratic polynomial \(f(x) = x^2+ax+b\)

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{x^2+ax+b} \\\\ f(1) &=& 1^2+a*1 + b \quad & | \quad f(1) = 2 \\ 2 &=& 1+a+b \\ 1 &=& a+b \\ \mathbf{b} &=& \mathbf{1-a} \\\\ f(3) &=& 3^2+a*3 + b \quad & | \quad f(3) = 4 \\ 4 &=& 9+3a + b \quad &| \quad b=1-a \\ 4 &=& 9+3a + 1-a \\ 4 &=& 10+2a \\ -6 &=& 2a \\ a &=& -\dfrac{6}{2} \\ \mathbf{a} &=& \mathbf{-3} \\\\ \mathbf{b} &=& \mathbf{1-a} \\ b &=& 1-(-3) \\ b &=& 1+3 \\ \mathbf{b} &=& \mathbf{4} \\ \hline \end{array} \)

\(\mathbf{f(x) = x^2-3x+4 }\)

Suppose g(x) is a polynomial of degree five for which g(1) = 2, g(2) = 3, g(3) = 4, g(4) = 5, g(5) = 6, and g(6) = -113.

Find g(0).

\(\begin{array}{rcll} \mathbf{g(0)} &=& \dfrac{ \begin{array}{|rrrrrr|} ~1 & 1 & 1 & 1 & 1 & 2 ~ \\ ~2^5& 2^4& 2^3& 2^2& 2& 3 ~ \\ ~3^5& 3^4& 3^3& 3^2& 3& 4 ~ \\ ~4^5& 4^4& 4^3& 4^2& 4& 5 ~ \\ ~5^5& 5^4& 5^3& 5^2& 5& 6 ~ \\ ~6^5& 6^4& 6^3& 6^2& 6&-113 ~ \\ \end{array} }{ \begin{array}{|rrrrrr|} ~1 & 1 & 1 & 1 & 1 & 1 ~ \\ ~2^5& 2^4& 2^3& 2^2& 2& 1 ~ \\ ~3^5& 3^4& 3^3& 3^2& 3& 1 ~ \\ ~4^5& 4^4& 4^3& 4^2& 4& 1 ~ \\ ~5^5& 5^4& 5^3& 5^2& 5& 1 ~ \\ ~6^5& 6^4& 6^3& 6^2& 6& 1 ~ \\ \end{array} } = \dfrac{-4181760}{-34560} = \mathbf{121} \\ \end{array}\)

The first four terms in an arithmetic sequence are x + y, x - y, xy, and x/y, in that order.
What is the fifth term?
Express your answer as a common fraction.

\(\begin{array}{rclcl} a_1 &=& a &=& x+y \\ a_2 &=& a+d &=& x-y \\ a_3 &=& a+2d &=& xy \\ a_4 &=& a+3d &=& \dfrac{x}{y} \quad &| \quad y\neq 0 ~!\\ a_5 &=& a+4d &=& \ ? \\ \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{a_2} &=& \mathbf{a+d} \quad &| \quad a=x+y\\ a_2 &=& x+y+d \\ a_2-(x+y) &=&d \quad &| \quad a_2 = x-y \\ x-y-(x+y) &=&d \\ \mathbf{-2y} &=&\mathbf{d}\qquad (1) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{a_3} &=& \mathbf{a+2d} \quad &| \quad a=x+y \\ a_3 &=& x+y+2d \\ a_3-(x+y) &=& 2d \quad &| \quad a_3 = xy \\ \mathbf{ xy-(x+y)} &=& \mathbf{2d}\qquad (2) \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline \dfrac{(2)}{(1)}: & \mathbf{ \dfrac{xy-(x+y)}{-2y}} &=& \mathbf{ \dfrac{2d}{d}} \\\\ & \dfrac{xy-(x+y)}{-2y} &=& 2 \\ & xy-(x+y) &=& -4y \\ & xy-x-y &=& -4y \\ & xy-x+3y &=& 0 \\ &\mathbf{ x-xy } &=& \mathbf{3y} \qquad (3) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{a_4} &=& \mathbf{a+3d} \quad &| \quad a=x+y \\ a_4 &=& x+y+3d \\ a_4-(x+y) &=& 3d \quad &| \quad a_4 = \dfrac{x}{y} \\ \mathbf{\dfrac{x}{y}-(x+y)} &=& \mathbf{3d} \qquad (4) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \dfrac{(4)}{(1)}: & \mathbf{ \dfrac{\dfrac{x}{y}-(x+y)}{-2y}} &=& \mathbf{ \dfrac{3d}{d}} \\\\ & \dfrac{\dfrac{x}{y}-(x+y)}{-2y} &=& 3 \\ & \dfrac{x}{y}-(x+y) &=& -6y \quad &| \quad *y \\ & x-(x+y)y &=& -6y^2 \\ & x-xy-y^2 &=& -6y^2 \\ & x-xy+5y^2 &=& 0 \quad &| \quad \mathbf{ x-xy =3y} \qquad (3) \\ & 3y+5y^2 &=& 0 \\ & y(3+5y) &=& 0 \quad &| \quad y\neq 0 ~!\\ & 3+5y &=& 0 \\ & 5y &=& -3 \\ & \mathbf{y} &=& \mathbf{-\dfrac{3}{5}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{d} &=& \mathbf{-2y} \qquad (1) \\ d &=& -2\left(-\dfrac{3}{5}\right) \\ \mathbf{d} &=& \mathbf{ \dfrac{6}{5} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{ xy-(x+y)} &=& \mathbf{2d}\qquad (2) \\ \ldots \\ x &=& \dfrac{2d+y}{y-1} \quad & | \quad \mathbf{y=-\dfrac{3}{5}},\ \mathbf{d= \dfrac{6}{5} } \\ x &=& \dfrac{2 *\dfrac{6}{5}-\dfrac{3}{5}}{-\dfrac{3}{5}-1} \\ x &=& -\dfrac{9}{5}* \dfrac{5}{8} \\ \mathbf{x} &=& \mathbf{-\dfrac{9}{8}} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline a &=& x+y \\ a &=& -\dfrac{9}{8}-\dfrac{3}{5} \\ \mathbf{a} &=& \mathbf{-\dfrac{69}{40}} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline a_5 &=& a+4d \\ a_5 &=& -\dfrac{69}{40} + 4*\dfrac{6}{5} \\ a_5 &=& -\dfrac{69}{40} + \dfrac{24}{5} \\ a_5 &=& -\dfrac{69}{40} + \dfrac{24*8}{5*8} \\ \mathbf{a_5} &=& \mathbf{\dfrac{123}{40}} \\ \hline \end{array}\)

arithmetric sequence: \(\{-\dfrac{69}{40},~-\dfrac{21}{40},~\dfrac{27}{40},~\dfrac{75}{40},~\dfrac{123}{40},~\ldots\}\)

If

\(f(x)=\dfrac{x^5-1}3\),

find \(f^{-1} \left( \dfrac{-31}{96} \right)\).

\(\begin{array}{|rcll|} \hline \mathbf{f\Big( f^{-1} \left( x \right) \Big)} &=& \mathbf{x} \quad | \quad x = -\dfrac{31}{96} \\\\ f\Bigg( f^{-1} \left( -\dfrac{31}{96} \right) \Bigg) &=& -\dfrac{31}{96} \\ \hline \\ && \boxed{f(x)=\dfrac{x^5-1}3 \\ x=f^{-1}\left( -\dfrac{31}{96} \right)} \\ \dfrac{\left[ f^{-1}\left( -\dfrac{31}{96} \right) \right]^5 -1 }{3} &=& -\dfrac{31}{96} \\\\ \left[ f^{-1}\left( -\dfrac{31}{96} \right) \right]^5 -1 &=& -\dfrac{31}{32} \\\\ \left[ f^{-1}\left( -\dfrac{31}{96} \right) \right]^5 &=&1 -\dfrac{31}{32} \\\\ \left[ f^{-1}\left( -\dfrac{31}{96} \right) \right]^5 &=& \dfrac{1}{32} \\\\ \left[ f^{-1}\left( -\dfrac{31}{96} \right) \right]^5 &=& \dfrac{1}{2^5} \\\\ f^{-1}\left( -\dfrac{31}{96} \right) &=& \dfrac{1}{\sqrt[5]{2^5}} \\\\ \mathbf{ f^{-1}\left( -\dfrac{31}{96} \right) } &=& \mathbf{ \dfrac{1}{2} } \\ \hline \end{array}\)

Let \(x\) and \(y\) be vectors such that \(\operatorname{proj}_{\mathbf{x}}(\mathbf{y})=\dbinom{5}{3} \)and \(\operatorname{proj}_{\mathbf{x}}(\mathbf{y})=\dbinom{5}{3}\).
Compute the ratio \(\dfrac{\|\mathbf{x}\|}{\|\mathbf{y}\|}\).

\(\begin{array}{|rcll|} \hline \cos(\varphi) = \dfrac{\sqrt{8}}{\|\mathbf{x}\|} &=& \dfrac{\sqrt{34}}{\|\mathbf{y}\|} \\\\ \dfrac{\sqrt{8}}{\|\mathbf{x}\|} &=& \dfrac{\sqrt{34}}{\|\mathbf{y}\|} \\\\ \dfrac{\|\mathbf{x}\|}{\|\mathbf{y}\|} &=& \dfrac{\sqrt{8}}{\sqrt{34}} \\ \hline \end{array}\)

Triangles ABC and ADE have areas \(2007\) and \(7002\) respectively, with
\(B=(0,0)\), \(C=(223,0)\), \(D=(680,380)\), and \(E=(689,389)\).
What is the sum of all possible x-coordinates of \(A\)?

(The answer is not -1238)

First solution for \(y_A\):

\(\begin{array}{rcl} \mathbf{\text{2[ABC]} } = \begin{array}{|rr|} x_A & y_A \\ 0 & 0 \\ 223 & 0 & \\ x_A & y_A \\ \end{array} &=& 2* 2007 \\\\ \begin{array}{|rr|} x_A & y_A \\ 0 & 0 \\ 223 & 0 & \\ x_A & y_A \\ \end{array} &=& 2* 2007 \\ \begin{array}{|rr|} x_A*0-0*y_A + 0*0-223*0+223*y_A-x_A*0 \end{array} &=& 2* 2007 \\ 223*y_A&=& 2* 2007 \quad | \quad : 223 \\ \mathbf{y_A} &=& \mathbf{18} \\ \end{array} \)

First solution for \(x_A\):

\(\begin{array}{rcl} \mathbf{\text{2[ADE]} } = \begin{array}{|rr|} x_A & 18 \\ 680 & 380 \\ 689 & 389 & \\ x_A & 18 \\ \end{array} &=& 2* 7002 \\\\ \begin{array}{|rr|} x_A & 18 \\ 680 & 380 \\ 689 & 389 & \\ x_A & 18 \\ \end{array} &=& 2* 7002 \\ \begin{array}{|rr|} 380x_A-680*18+680*389-689*380+689*18-389x_A \end{array} &=& 2* 2007 \\ -9x_A+680*371-689*362 &=& 2* 7002 \\ -9x_A+2862 &=& 2* 7002 \quad | \quad : (-9) \\ x_A - 318 &=& -1556\\ \mathbf{x_A} &=& \mathbf{-1238} \\ \end{array}\)

Second solution for \(x_A\):

\(\begin{array}{rcl} \mathbf{\text{2[ADE]} } = \begin{array}{|rr|} x_A & 18 \\ 689 & 389 \\ 680 & 380 \\ x_A & 18 \\ \end{array} &=& 2* 7002 \\\\ \begin{array}{|rr|} x_A & 18 \\ 689 & 389 \\ 680 & 380 \\ x_A & 18 \\ \end{array} &=& 2* 7002 \\ \begin{array}{|rr|} 389x_A-689*18+689*380-680*389+680*18-380x_A \end{array} &=& 2* 2007 \\ 9x_A+689*362-689*371 &=& 2* 7002 \\ 9x_A-2862 &=& 2* 7002 \quad | \quad : 9 \\ x_A - 318 &=& 1556\\ \mathbf{x_A} &=& \mathbf{1874} \\ \end{array}\)

Second solution for \(y_A\):

\(\begin{array}{rcl} \mathbf{\text{2[ABC]} } = \begin{array}{|rr|} x_A & y_A \\ 223 & 0 & \\ 0 & 0 \\ x_A & y_A \\ \end{array} &=& 2* 2007 \\\\ \begin{array}{|rr|} x_A & y_A \\ 223 & 0 & \\ 0 & 0 \\ x_A & y_A \\ \end{array} &=& 2* 2007 \\ \begin{array}{|rr|} x_A*0-223*y_A +223*0 - 0*0 + 0*y_A -x_A*0 \end{array} &=& 2* 2007 \\ -223*y_A&=& 2* 2007 \quad | \quad : (-223) \\ \mathbf{y_A} &=& \mathbf{-18} \\ \end{array}\)

Third Solution for \(x_A\):

\( \begin{array}{rcl} \mathbf{\text{2[ADE]} } = \begin{array}{|rr|} x_A & -18 \\ 680 & 380 \\ 689 & 389 & \\ x_A & -18 \\ \end{array} &=& 2* 7002 \\\\ \begin{array}{|rr|} x_A & -18 \\ 680 & 380 \\ 689 & 389 & \\ x_A & -18 \\ \end{array} &=& 2* 7002 \\ \begin{array}{|rr|} 380x_A+680*18+680*389-689*380-689*18-389x_A \end{array} &=& 2* 2007 \\ -9x_A+680*407-689*398 &=& 2* 7002 \\ -9x_A+2538 &=& 2* 7002 \quad | \quad : (-9) \\ x_A - 282 &=& -1556\\ \mathbf{x_A} &=& \mathbf{-1274} \\ \end{array} \)

Fourth Solution for \(x_A\):

\(\begin{array}{rcl} \mathbf{\text{2[ADE]} } = \begin{array}{|rr|} x_A & -18 \\ 689 & 389 \\ 680 & 380 \\ x_A & -18 \\ \end{array} &=& 2* 7002 \\\\ \begin{array}{|rr|} x_A & -18 \\ 689 & 389 \\ 680 & 380 \\ x_A & -18 \\ \end{array} &=& 2* 7002 \\ \begin{array}{|rr|} 389x_A+689*18+689*380-680*389-680*18-380x_A \end{array} &=& 2* 2007 \\ 9x_A+689*398-680*407 &=& 2* 7002 \\ 9x_A-2538 &=& 2* 7002 \quad | \quad : 9 \\ x_A - 282 &=& 1556\\ \mathbf{x_A} &=& \mathbf{1838} \\ \end{array}\)

The sum of all possible x-coordinates of \(A\) is \(-1238+1874-1274+1838 = \mathbf{1200}\)

Rectangle ABCD contains a point such that AX = 1, BX = 7 and CX = 8 Find DX.

\(\begin{array}{|rcll|} \hline (w-p)^2+(h-q)^2 &=& 8^2 \quad | \quad (w-p)^2+q^2 =7^2 \rightarrow (w-p)^2 = 7^2-q^2 \\ 7^2-q^2+(h-q)^2 &=& 8^2 \quad | \quad (h-q)^2+p^2 =x^2 \rightarrow (h-q)^2 = x^2-p^2 \\ 7^2-q^2+x^2-p^2 &=& 8^2 \\ 7^2+x^2-(p^2+q^2) &=& 8^2 \quad | \quad p^2+q^2 = 1^2 \\ 7^2+x^2-1^2 &=& 8^2 \\ \mathbf{ x^2-1^2 }&=& \mathbf{8^2 -7^2} \\ x^2&=& 8^2 -7^2+1^2 \\ x^2&=& 64 - 49 +1 \\ x^2&=& 16 \\ \mathbf{ x }&=& \mathbf{4} \\ \hline \end{array}\)

Triangles ABC and ADE have areas \(2007 \)and \(7002 \)respectively, with
\(B=(0,0)\),
\(C=(223,0)\),
\(D=(680,380)\), and
\(E=(689,389)\).
What is the sum of all possible x-coordinates of A?

My attempt:

I use the Gauss's shoelace area formula.

\(\text{First solution for $y_A$} :\\ \begin{array}{rcl} \mathbf{\text{2[ABC]} } = \begin{array}{|rr|} x_A & y_A \\ 0 & 0 \\ 223 & 0 & \\ x_A & y_A \\ \end{array} &=& 2* 2007 \\\\ \begin{array}{|rr|} x_A & y_A \\ 0 & 0 \\ 223 & 0 & \\ x_A & y_A \\ \end{array} &=& 2* 2007 \\ \begin{array}{|rr|} x_A*0-0*y_A + 0*0-223*0+223*y_A-x_A*0 \end{array} &=& 2* 2007 \\ 223*y_A&=& 2* 2007 \quad | \quad : 223 \\ \mathbf{y_A} &=& \mathbf{18} \\ \end{array} \\ \text{First solution for $x_A$}: \\ \begin{array}{rcl} \mathbf{\text{2[ADE]} } = \begin{array}{|rr|} x_A & 18 \\ 680 & 380 \\ 689 & 389 & \\ x_A & 18 \\ \end{array} &=& 2* 7002 \\\\ \begin{array}{|rr|} x_A & 18 \\ 680 & 380 \\ 689 & 389 & \\ x_A & 18 \\ \end{array} &=& 2* 7002 \\ \begin{array}{|rr|} 380x_A-680*18+680*389-689*380+689*18-389x_A \end{array} &=& 2* 2007 \\ -9x_A+680*371-689*362 &=& 2* 7002 \\ -9x_A+2862 &=& 2* 7002 \quad | \quad : (-9) \\ x_A - 318 &=& -1556\\ \mathbf{x_A} &=& \mathbf{-1238} \\ \end{array} \\ \text{Second solution for $x_A$}: \\ \begin{array}{rcl} \mathbf{\text{2[ADE]} } = \begin{array}{|rr|} x_A & 18 \\ 689 & 389 \\ 680 & 380 \\ x_A & 18 \\ \end{array} &=& 2* 7002 \\\\ \begin{array}{|rr|} x_A & 18 \\ 689 & 389 \\ 680 & 380 \\ x_A & 18 \\ \end{array} &=& 2* 7002 \\ \begin{array}{|rr|} 389x_A-689*18+689*380-680*389+680*18-380x_A \end{array} &=& 2* 2007 \\ 9x_A+689*362-689*371 &=& 2* 7002 \\ 9x_A-2862 &=& 2* 7002 \quad | \quad : 9 \\ x_A - 318 &=& 1556\\ \mathbf{x_A} &=& \mathbf{1874} \\ \end{array} \\ \text{Second solution for $y_A$}: \\ \begin{array}{rcl} \mathbf{\text{2[ABC]} } = \begin{array}{|rr|} x_A & y_A \\ 223 & 0 & \\ 0 & 0 \\ x_A & y_A \\ \end{array} &=& 2* 2007 \\\\ \begin{array}{|rr|} x_A & y_A \\ 223 & 0 & \\ 0 & 0 \\ x_A & y_A \\ \end{array} &=& 2* 2007 \\ \begin{array}{|rr|} x_A*0-223*y_A +223*0 - 0*0 + 0*y_A -x_A*0 \end{array} &=& 2* 2007 \\ -223*y_A&=& 2* 2007 \quad | \quad : (-223) \\ \mathbf{y_A} &=& \mathbf{-18} \\ \end{array} \\ \text{Third Solution for $x_A$}: \\ \begin{array}{rcl} \mathbf{\text{2[ADE]} } = \begin{array}{|rr|} x_A & -18 \\ 680 & 380 \\ 689 & 389 \\ x_A & -18 \\ \end{array} &=& 2* 7002 \\\\ \begin{array}{|rr|} x_A & -18 \\ 680 & 380 \\ 689 & 389 \\ x_A & -18 \\ \end{array} &=& 2* 7002 \\ \begin{array}{|rr|} 380x_A+680*18+680*389-689*380-689*18-389x_A \end{array} &=& 2* 2007 \\ -9x_A+680*407-689*398 &=& 2* 7002 \\ -9x_A+2538 &=& 2* 7002 \quad | \quad : (-9) \\ x_A - 282 &=& -1556\\ \mathbf{x_A} &=& \mathbf{-1274} \\ \end{array} \\ \text{Fourth Solution for $x_A$}: \\ \begin{array}{rcl} \mathbf{\text{2[ADE]} } = \begin{array}{|rr|} x_A & -18 \\ 689 & 389 \\ 680 & 380 \\ x_A & -18 \\ \end{array} &=& 2* 7002 \\\\ \begin{array}{|rr|} x_A & -18 \\ 689 & 389 \\ 680 & 380 \\ x_A & -18 \\ \end{array} &=& 2* 7002 \\ \begin{array}{|rr|} 389x_A+689*18+689*380-680*389-680*18-380x_A \end{array} &=& 2* 2007 \\ 9x_A+689*398-680*407 &=& 2* 7002 \\ 9x_A-2538 &=& 2* 7002 \quad | \quad : 9 \\ x_A - 282 &=& 1556\\ \mathbf{x_A} &=& \mathbf{1838} \\ \end{array} \\ \text{The sum of all possible x-coordinates of A is $-1238+1874-1274+1838 = \mathbf{1200}$} \)

What is the smallest positive integer that has exactly 40 divisors?

\(\sigma_0(1680) = 40\)