heureka

The roots of \(7x^2 + x - 5 = 0\) are \(a\) and \(b\).
Compute \(\dfrac{1}{a} + \dfrac{1}{b}\).

By Vieta:

\(\begin{array}{|rcll|} \hline 7x^2 + x - 5 &=& 0 \quad | \quad : 7 \\ x^2 + \underbrace{\frac{1}{7}}_{=-(a+b)}x \underbrace{-\frac{5}{7}}_{=ab} &=& 0 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline a+b &=& -\dfrac{1}{7} \\\\ ab &=& -\dfrac{5}{7} \\\\ \hline \dfrac{a+b}{ab} &=& -\dfrac{1}{7} \over -\dfrac{5}{7} \\\\ \dfrac{a}{ab}+\dfrac{b}{ab} &=& \dfrac{1}{7} \over \dfrac{5}{7} \\\\ \dfrac{1}{a}+\dfrac{1}{b} &=& \dfrac{1}{7} \times \dfrac{7}{5} \\\\ \mathbf{\dfrac{1}{a}+\dfrac{1}{b}} &=& \mathbf{\dfrac{1}{5}} \\ \hline \end{array}\)

When the positive integer x is divided by each of 4, 5, 6, and 7,

it has a remainder of 3.
What is the sum of the three smallest possible values of x?

\(\begin{array}{|rcll|} \hline x&\equiv& 3 \pmod{4} \\ x&\equiv& 3 \pmod{5} \\ x&\equiv& 3 \pmod{6} \\ x&\equiv& 3 \pmod{7} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{sum} &=& \Big( 3+0\times \text{lcm}(4,~5,~6,~7) \Big) \\ && +\Big( 3+1\times \text{lcm}(4,~5,~6,~7) \Big) \\ && +\Big( 3+2\times \text{lcm}(4,~5,~6,~7) \Big) \\\\ \text{sum} &=& 3 + ( 3+420 )+( 3+2\times420 ) \\\\ \text{sum} &=& 3 + 423+843 \\ \text{sum} &=& 1269 \\ \hline \end{array}\)

What is the largest number that will always divide the sum of any 8 consecutive positive integers?

My attempt:

\(\begin{array}{|rcll|} \hline \text{sum} &=& \dfrac{(a_1+a_8)}{2}\times 8 \\\\ \text{sum} &=& (a_1+a_8)\times \color{red}4 \\ \hline \end{array} \)

The largest number is \(\color{red} 4\)

Find length BC.

\(\text{Let $BC = \color{red}x$}\)

\(\begin{array}{|lrcll|} \hline (1): & AE^2+DE^2 &=& 4^2 \\ (2): & DE^2+CE^2 &=& 6^2 \\ (3): & AE^2+BE^2 &=& 3^2 \\ (4): & BE^2+CE^2 &=& x^2 \\ \hline \end{array}\)

\(\small{ \begin{array}{|lrcll|} \hline (3)-(1)+(2): & (AE^2+BE^2)-(AE^2+DE^2)+ (DE^2+CE^2) &=& 3^2-4^2+6^2 \\ & BE^2+CE^2 &=& 3^2-4^2+6^2 \\ & BE^2+CE^2 &=& 29 \quad | \quad BE^2+CE^2 = x^2 \quad (4) \\ & x^2 &=& 29 \\ & \mathbf{x} &=& \mathbf{\sqrt{29}} \\ \hline \end{array} }\)

\(BC = \sqrt{29}\)

Find the least positive four-digit solution to the following system of congruences.
\(7x \equiv 21 \pmod{14} \\ 2x + 18 \equiv 16 \pmod{9}\)

\(\begin{array}{|rcll|} \hline 7x &\equiv& 21 \pmod{14} \\ 7x &\equiv& 21-14 \pmod{14} \\ 7x &\equiv& 7 \pmod{14} \\ 7x &=& 7 +14m \quad m\in \mathbb{Z} \\ 7x &=& 7 +14m \quad | \quad :7 \\ x &=& 1 +2m \quad | \quad \text{$x$ is an odd number!} \\ x &\equiv& 1 \pmod{2} \\ \hline \end{array}\)

...so \(7x \equiv 21 \pmod{14}\) or \(x \equiv 1 \pmod{2}\)

\(\begin{array}{|rcl|rcl|} \hline 2x + 18 &\equiv& 16 \pmod{9} \\ 2x &\equiv& 16-18 \pmod{9} \\ 2x &\equiv& -2 \pmod{9} \\ 2x &=& -2 +9n \quad n\in \mathbb{Z} \\ 2x &=& -2 +9n \quad | \quad x=1+2m \\ 2(1+2m) &=& -2 +9n \\ 2+4m &=& -2 +9n \\ 4m &=& -4 +9n \\ m &=& \frac{-4 +9n}{4} \\ m &=& \frac{-4 +8n+n}{4} \\ m &=& -1+2n+ \underbrace{ \frac{n}{4} }_{=r} \quad r \in \mathbb{Z} \\ m &=& -1+2n+ r & r &=& \frac{n}{4} \\ & & & 4r &=& n \\ & & & \mathbf{n} &=& \mathbf{4r} \\ m &=& -1+2(4r)+ r \\ \mathbf{m} &=& \mathbf{-1+9r} \\ \hline x &=& 1+2m \quad | \quad m=-1+9r \\ x &=& 1+2(-1+9r) \\ x &=& 1-2+18r \\ \mathbf{x} &=& \mathbf{-1+18r} \quad r \in \mathbb{Z} \\ \hline \end{array}\)

Find the least positive four-digit solution:

\(\begin{array}{|rcll|} \hline -1+18r &>& 999 \\ 18r &>& 1000 \\ r &>& \frac{1000}{18} \\ r &>& 55.\bar{5} \\ r &=& 56 \\ \hline x &=& -1+18*56 \\ \mathbf{ x} &=& \mathbf{1007} \\ \hline \end{array}\)

The least positive four-digit solution is 1007

Find the remainder when \(3*3^2*3^3*\ldots*3^{43}\) is divided by \(11\).

\(\begin{array}{|rcll|} \hline 3*3^2*3^3*\ldots*3^{43} &=& 3^{1+2+3+\ldots 42+43} \\ &=& \left.3\right.^{\left(\frac{1+43}{2}*43\right)} \\ &=& \left.3\right.^{\left(\frac{44}{2}*43\right)} \\ &=& 3^{22*43} \\ &=& 3^{946} \\ \hline \end{array}\)

\(\begin{array}{lrcll} \text{Euler:}\\ \boxed{3^{\phi(11)}\equiv 1 \pmod{11} \quad | \quad \phi(11)=10,~ \phi(p) = p-1\\3^{10}=1 \pmod{11} } \\ \end{array}\)

\(\begin{array}{|rcll|} \hline 3^{946} \pmod{11} &\equiv& 3^{10*94+6}\pmod{11} \\ &\equiv& \left(3^{10} \right)^{94}3^6\pmod{11} \quad | \quad 3^{10} \equiv 1\pmod{11} \\ &\equiv& 1^{94}3^6\pmod{11} \\ &\equiv& 3^6\pmod{11} \\ &\equiv& 729\pmod{11} \\ \mathbf{3^{946} \pmod{11}} &\equiv& \mathbf{ {\color{red}3}\pmod{11}} \\ \hline \end{array}\)

Find the remainder when \(5^{67}\) is divided by \(11\).

\(\begin{array}{|rcll|} \hline 5^{67} \pmod{11} &\equiv& 5^{5*13+2}\pmod{11} \\ &\equiv& \left(5^5 \right)^{13}5^2\pmod{11} \quad | \quad 5^5 \equiv 1\pmod{11} \\ &\equiv& 1^{13}5^2\pmod{11} \\ &\equiv& 5^2\pmod{11} \\ &\equiv& 25\pmod{11} \\ \mathbf{5^{67} \pmod{11}} &\equiv& \mathbf{ {\color{red}3}\pmod{11}} \\ \hline \end{array}\)

The remainder is 3.

For every value of a, the graph of each of these three equations is a line:
\(ax + y = 1 \\ x + ay = 1\\ x + y = 9\)
For what value of a do all three of these lines intersect at the same point?

My attempt:

\(\begin{array}{|lrcll|} \hline (1): & ax + y &=& 1 \\ (2): & x + ay &=& 1 \\ (3): & x + y &=& 9 \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1): & ax + y &=& 1 \\ & y &=& 1-ax \\\\ (3): & x + y &=& 9 \\ & x+1-ax &=& 9 \\ & x(1-a) &=& 8 \\ & \mathbf{x} &=& \mathbf{\dfrac{8}{1-a}} \\\\ & y &=& 9-x \\ & \mathbf{y} &=& \mathbf{9-\dfrac{8}{1-a}} \qquad (4) \\\\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (3): & x + y &=& 9 \\ & x &=& 9-y \\\\ (2): & x + ay &=& 1 \\ & 9-y + ay &=& 1 \\ & y(1-a) &=& 8 \\ & \mathbf{y} &=& \mathbf{\dfrac{8}{1-a}} \qquad (5) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (5)=(4):& y = \dfrac{8}{1-a} &=& 9-\dfrac{8}{1-a} \\\\ & \dfrac{8}{1-a} &=& 9-\dfrac{8}{1-a} \\\\ & \dfrac{2*8}{1-a} &=& 9 \\\\ & \dfrac{16}{1-a} &=& 9 \\\\ & 16 &=& 9(1-a) \\ & 16 &=& 9-9a \\ & 9a &=& 9-16 \\ & 9a &=& -7 \\\\ & \mathbf{a} &=& \mathbf{ -\dfrac{7}{9} } \\ \hline \end{array}\)

Assuming \(x\), \(y\), and \(z\) are positive real numbers satisfying
\(xy - z = 15\\ xz - y = 15\\ yz - x = 15\)
then, what is the value of \(xyz\)?

\(\begin{array}{|lrcll|} \hline (1) & xy - z &=& 15 \\ (2) & xz - y &=& 15 \\ (3) & yz - x &=& 15 \\ \hline (1)+(2)+(3):& (xy - z)+(xz - y)+(yz - x) &=& 45 \\ & \ldots \\ &\mathbf{ x+y+z } &=& \mathbf{ xy+xz+yz-45 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline yz - x &=& 15 \\ \mathbf{x} &=& yz-15 \\ \hline xz-y &=& 15 \\ (yz-15)z-y &=& 15 \\ yz^2-15z-y &=& 15 \\ yz^2-y &=& 15z+15 \\ y(z^2-1) &=& 15(z+1) \\ y(z-1)(z+1) &=& 15(z+1) \quad | \quad : (z+1),~ z>0!\\ y(z-1)&=& 15\\ \mathbf{y} &=& \mathbf{\dfrac{15} {z-1} }\qquad (4) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline xz - y &=& 15 \\ \mathbf{y} &=& xz-15 \\ \hline xy-z &=& 15 \\ (xz-15)x-z &=& 15 \\ zx^2-15x-z &=& 15 \\ zx^2-z &=& 15x+15 \\ z(x^2-1) &=& 15(x+1) \\ z(x-1)(x+1) &=& 15(x+1) \quad | \quad : (x+1),~ x>0!\\ z(x-1)&=& 15\\ \mathbf{z} &=& \mathbf{\dfrac{15} {x-1} }\qquad (5) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline xy - z &=& 15 \\ \mathbf{z} &=& xy-15 \\ \hline yz-x &=& 15 \\ (xy-15)y-x &=& 15 \\ xy^2-15y-x &=& 15 \\ xy^2-x &=& 15y+15 \\ x(y^2-1) &=& 15(y+1) \\ x(y-1)(y+1) &=& 15(y+1) \quad | \quad : (y+1),~ y>0!\\ x(y-1)&=& 15\\ \mathbf{x} &=& \mathbf{\dfrac{15} {y-1} }\qquad (6) \\ \hline \end{array}\)

\( \small{ \begin{array}{|lrcll|} \hline (4)\times(5)\times(6):& xyz &=& \dfrac{15} {y-1} \times \dfrac{15} {z-1} \times \dfrac{15} {x-1} \\\\ & xyz &=& \dfrac{15^3} {(x-1)(y-1)(z-1)} \\\\ &&& \boxed{\mathbf{(x-1)(y-1)(z-1)}\\=xyz-(xy+xz+yz)+(x+y+z) - 1} \\\\ & xyz &=& \dfrac{15^3} {xyz-(xy+xz+yz)+(x+y+z) - 1} \\\\ &&& \mathbf{ x+y+z=xy+xz+yz-45 } \\\\ & xyz &=& \dfrac{15^3} {xyz-(xy+xz+yz)+(xy+xz+yz-45) - 1} \\\\ & xyz &=& \dfrac{15^3} {xyz-46} \\\\ & xyz(xyz-46) &=& 15^3 \\\\ & (xyz)^2-46xyz- 15^3 &=& 0 \\ \hline & xyz &=& \dfrac{46\pm\sqrt{46^2-4*(-15^3)}}{2} \\\\ & xyz &=& \dfrac{46\pm\sqrt{46^2+4*15^3}}{2} \\\\ & xyz &=& \dfrac{46\pm\sqrt{15616}}{2} \\\\ & xyz &=& \dfrac{46\pm\sqrt{16^2*61}}{2} \\\\ & xyz &=& \dfrac{46\pm 16\sqrt{61}}{2} \\\\ & \mathbf{ xyz } &=& \mathbf{ 23 +8\sqrt{61} } \qquad xyz >0! \\ \hline \end{array} }\)

The letters A, B, C, D, E, and F represent digits and 

ABC,DEF represents a positive six-digit integer.

What is the number ABC,DEF if 4(ABC,DEF) = 9(DEF,ABC)?