heureka

Suppose \(x\), \(y\), and \(z\) form a geometric sequence.
If you know that and \(x + y + z = 18\),
find the value of \(x^2+y^2+z^2=612\).

I assume:

Geometric sequence:
\(x = a \\ y = ar \\ z = ar^2\)

\(\begin{array}{|lrcll|} \hline & \mathbf{x+y+z} &=& \mathbf{18} \\ &a+ar+ar^2 &=& 18 \\ &a(1+r+r^2) &=& 18 \quad |\quad 1+r+r^2 = \dfrac{1-r^3}{1-r} \qquad {\displaystyle |r|<1} \\\\ (1) &\mathbf{a\left(\dfrac{1-r^3}{1-r}\right)} &=& \mathbf{18} \qquad \text{or} \quad \mathbf{a} = \mathbf{18\left(\dfrac{1-r}{1-r^3}\right)} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline & \mathbf{x^2+y^2+z^2} &=& \mathbf{612} \\ &a^2+a^2r^2+a^2r^4 &=& 612 \\ &a^2(1+r^2+r^4) &=& 612 \quad &|\quad 1+r^2+r^4 = \dfrac{1-r^6}{1-r^2} \qquad {\displaystyle |r|<1} \\ (2)&\mathbf{a^2\left(\dfrac{1-r^6}{1-r^2}\right)} &=& \mathbf{612} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \dfrac{(2)}{(1)}: & \dfrac{ a^2\left(\dfrac{1-r^6}{1-r^2}\right) } { a\left(\dfrac{1-r^3}{1-r}\right)} &=& \dfrac{612}{18} \\\\ & \dfrac{ a\left(\dfrac{1-r^6}{1-r^2}\right) } { \left(\dfrac{1-r^3}{1-r}\right)} &=& 34 \\\\ & a \dfrac{(1-r^6)(1-r)}{(1-r^3)(1-r^2)} &=& 34 \quad |\quad 1-r^6 = (1-r^3)(1+r^3) \\\\ & a \dfrac{(1-r^3)(1+r^3)(1-r)}{(1-r^3)(1-r^2)} &=& 34 \quad |\quad 1-r^2 = (1-r)(1+r) \\\\ & a \dfrac{(1-r^3)(1+r^3)(1-r)}{(1-r^3)(1-r)(1+r)} &=& 34 \\\\ (3)&\mathbf{a\left(\dfrac{1+r^3}{1+r}\right)} &=& \mathbf{34} \qquad \text{or} \quad \mathbf{a} = \mathbf{34\left(\dfrac{1+r}{1+r^3}\right)} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1)=(3): & a=18\left(\dfrac{1-r}{1-r^3}\right) &=& 34\left(\dfrac{1+r}{1+r^3}\right) \\\\ & 18\left(\dfrac{1-r}{1-r^3}\right) &=& 34\left(\dfrac{1+r}{1+r^3}\right) \\\\ & 18(1-r)(1+r^3) &=& 34(1+r)(1-r^3) \quad & | \quad : 2 \\ & 9(1-r)(1+r^3) &=& 17(1+r)(1-r^3) \\ & 9(1-r+r^3-r^4) &=& 17(1+r-r^3-r^4) \\ & 9-9r+9r^3-9r^4 &=& 17+17r-17r^3-17r^4 \\ & 8r^4+26r^3-26r-8 &=& 0 \quad & | \quad : 2 \\ & \mathbf{4r^4+13r^3-13r-4} &=& \mathbf{0} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \text{Solutions: }\\ r &=& -1 & \text{no solution} \quad |r|< 1! \\ r &=& 1 & \text{no solution} \quad |r|< 1! \\ \mathbf{ r } &=& \mathbf{ \dfrac{-13+\sqrt{105}}{8} }\ \checkmark & (r=-0.34413115426) \\ \mathbf{ r } &=& \mathbf{ \dfrac{-13-\sqrt{105}}{8} }\ \checkmark & (r=-2.90586884574) \\ \hline \end{array}\)

First solution: \(\mathbf{ r =\dfrac{-13+\sqrt{105}}{8} }\)

\(\begin{array}{|rcll|} \hline \mathbf{a(1+r+r^2)} &=& \mathbf{18} \\\\ a &=& \dfrac{18}{1+r+r^2} \\\\ a &=& \dfrac{18}{ 1+ \dfrac{\sqrt{105}-13}{8}+ \left( \dfrac{\sqrt{105}-13}{8} \right)^2 } \\ \ldots \\ a &=& \dfrac{64}{13-\sqrt{105}} * \left( \dfrac{13+\sqrt{105}}{13+\sqrt{105}} \right) \\\\ \mathbf{a} &=& \mathbf{13+\sqrt{105}} & (a=23.2469507660) \\\\ x &=& a \\ \mathbf{x} &=& \mathbf{13+\sqrt{105}} \\\\ y &=& ar \\ y &=& \left(13+\sqrt{105}\right) \left(\dfrac{-13+\sqrt{105}}{8} \right) \\ y &=& -\left(13+\sqrt{105}\right) \left(\dfrac{13-\sqrt{105}}{8} \right) \\ y &=& \dfrac{-64}{8} \\ \mathbf{y} &=& \mathbf{-8} \\\\ z = ar^2 &=& yr \\ z &=& -8\left( \dfrac{-13+\sqrt{105}}{8} \right) \\ \mathbf{z} &=& \mathbf{13-\sqrt{105}} & (z=2.75304923404) \\ \\ \mathbf{x^2+y^2+z^2} &=& \mathbf{612} \\ \left(13+\sqrt{105}\right)^2 + (-8)^2 + \left(13-\sqrt{105}\right)^2 &=& 612 \ \checkmark \\ \hline \end{array} \)

Second solution: \(\mathbf{ r =\dfrac{-13-\sqrt{105}}{8} } \)

\(\begin{array}{|rcll|} \hline \mathbf{a(1+r+r^2)} &=& \mathbf{18} \\\\ a &=& \dfrac{18}{1+r+r^2} \\\\ a &=& \dfrac{18}{1+\dfrac{-13-\sqrt{105}}{8}+\left(\dfrac{-13-\sqrt{105}}{8}\right)^2} \\ \ldots \\ a &=& \dfrac{64}{13+\sqrt{105}} * \left( \dfrac{13-\sqrt{105}}{13-\sqrt{105}} \right) \\\\ \mathbf{a} &=& \mathbf{13-\sqrt{105}}& (a=2.75304923404) \\\\ x &=& a \\ \mathbf{x} &=& \mathbf{13-\sqrt{105}} \\\\ y &=& ar \\ y &=& \left(13-\sqrt{105}\right) \left(\dfrac{-13-\sqrt{105}}{8} \right) \\ y &=& -\left(13-\sqrt{105}\right) \left(\dfrac{13+\sqrt{105}}{8} \right) \\ y &=& \dfrac{-64}{8} \\ \mathbf{y} &=& \mathbf{-8} \\\\ z = ar^2 &=& yr \\ z &=& -8\left( \dfrac{-13-\sqrt{105}}{8} \right) \\ \mathbf{z} &=& \mathbf{13+\sqrt{105}} & (z=23.2469507660) \\ \\ \mathbf{x^2+y^2+z^2} &=& \mathbf{612} \\ \left(13-\sqrt{105}\right)^2 + (-8)^2 + \left(13+\sqrt{105}\right)^2 &=& 612 \ \checkmark \\ \hline \end{array}\)

For a certain value of k, the system has no solutions. What is the value of

\(k x + y + 3z = 10,\\ -4x + 2y + 5z = 7,\\ kx + z = 3 \)

There is no solution, if the determinant \(\begin{vmatrix} k&1&3\\-4&2&5\\k&0&1\end{vmatrix} = 0\)

\(\begin{array}{|rcll|} \hline \begin{vmatrix} k&1&3\\-4&2&5\\k&0&1\end{vmatrix} &=& k\begin{vmatrix} 1&3\\2&5\end{vmatrix}+\begin{vmatrix} k&1\\ -4&2\end{vmatrix} \\ &=& k(5-6)+3k+4 \\ &=& -k+2k+4\\ &=& k+4 \quad &| \quad \begin{vmatrix} k&1&3\\-4&2&5\\k&0&1\end{vmatrix} = 0 \\ 0 &=& k+4\\ \mathbf{k} &=& \mathbf{-4} \\ \hline \end{array} \)

Thank you, Melody !

An infinite geometric series has first term -3/2 and sums to twice the common ratio.

Find the sum of all possible values for the common ratio.

\(\text{Let $a_1=a=-\dfrac{3}{2} $} \\ \text{Let the common ratio $=r$} \\ \text{Let the sum of the infinite geometric series $= \dfrac{a}{1-r} $ } \)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{a}{1-r}} &=& \mathbf{2r} \\\\ a &=& 2r(1-r) \\ a &=& 2r-2r^2 \\ 2r^2-2r+a &=& 0 \quad &|\quad a = -\dfrac{3}{2} \\ 2r^2-2r-\dfrac{3}{2} &=& 0 \quad &|\quad \cdot 2 \\ 4r^2-4r-3 &=& 0 \\\\ r &=& \dfrac{4\pm \sqrt{4^2-4\cdot 4\cdot (-3)} }{2\cdot 4} \\ &=& \dfrac{4\pm \sqrt{4^2+4^2\cdot 3} }{2\cdot 4} \\ &=& \dfrac{4\pm \sqrt{4^2\cdot 4} }{2\cdot 4} \\ &=& \dfrac{4\pm 4\cdot 2 }{2\cdot 4} \\ &=& \dfrac{1\pm 2 }{2} \\\\ r_1 &=& \dfrac{1+ 2 }{2} \\ \mathbf{ r_1 } &=& \mathbf{ \dfrac{3}{2} } \\\\ r_2 &=& \dfrac{1- 2 }{2} \\ \mathbf{ r_2 } &=& \mathbf{ -\dfrac{1}{2} } \\ \hline \end{array} \)

The sum of all possible values for the common ratio \(\dfrac{3}{2}-\dfrac{1}{2} = \mathbf{1}\)

Four positive integers \(w\), \(x\), \(y\),and \(z\) satisfy \(wx + w + x = 524\), \(xy + x + y = 146\), \(yz + y + z = 104\), and \(wxyz= 8!\).
What are \(w\), \(x\), \(y\), and \(z\) ?

Solution by substitution:

\(\begin{array}{|rcll|} \hline wx + w + x &=& 524 \\ x(w+1)+w &=& 524 \\ x(w+1) &=& 524-w \\ \mathbf{x} &=& \mathbf{\dfrac{524-w} {w+1}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline xy + x + y &=& 146 \\ y(x+1) + x &=& 146 \\ y(x+1) &=& 146-x \\ \mathbf{y} &=& \mathbf{\dfrac{146-x} {x+1}} \\\\ y &=& \dfrac{146-\dfrac{524-w} {w+1}} {\dfrac{524-w} {w+1}+1} \\ \ldots \\ \mathbf{y} &=& \mathbf{\dfrac{147w-378}{525} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline yz + y + z &=& 104 \\ z(y+1) + x &=& 146 \\ z(y+1) &=& 104-y \\ \mathbf{z} &=& \mathbf{\dfrac{104-y} {y+1}} \\ z &=& \dfrac{104-\dfrac{147w-378}{525}} {\dfrac{147w-378}{525}+1} \\ \ldots \\ \mathbf{z} &=& \mathbf{\dfrac{54978-147w}{147(w+1)} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{wxyz} &=& \mathbf{8!} \\ wxyz &=& 40320 \\ 40320 &=& w \left(\dfrac{524-w} {w+1}\right) \left(\dfrac{147w-378}{525} \right) \left(\dfrac{54978-147w}{147(w+1)} \right) \\ 40320 &=& \dfrac{w(524-w)(147w-378)(54978-147w)} {77175(w+1)^2} \\ 3111696000 (w+1)^2 &=& w(524-w)(147w-378 )(54978-147w ) \\ \ldots \\ \hline \end{array}\\ \begin{array}{|lrcll|} \hline & -21609 w^4 + 19460448 w^3 - 1173047652 w^2 + 17112994416 w + 3111696000 &=& 0 \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \text{Solutions: }\\ &\mathbf{ w } &=& \mathbf{24}\ \checkmark \qquad \text{integer} \\ & w&=&0.17961 \\ &w&=&39.918 \\ &w&=&36.83 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{\dfrac{524-w} {w+1}} \\ x &=& \dfrac{524-24} {24+1} \\ x &=& \dfrac{500} {25} \\ \mathbf{x} &=& \mathbf{20} \\\\ \mathbf{y} &=& \mathbf{\dfrac{146-x} {x+1}} \\ y &=& \dfrac{146-20} {20+1} \\ y &=& \dfrac{126} {21} \\ \mathbf{y} &=& \mathbf{6} \\\\ \mathbf{z} &=& \mathbf{\dfrac{104-y} {y+1}} \\ z &=& \dfrac{104-6} {6+1} \\ z &=& \dfrac{98} {7} \\ \mathbf{z} &=& \mathbf{14} \\ \hline \end{array}\)

Let n be a positive integer greater than or equal to 3.
Let a,b be integers such that is invertible modulo n and \((ab)^{-1}\equiv 2\pmod n\).
Given a+b is invertible, what is the remainder when \((a+b)^{-1}(a^{-1}+b^{-1})\) is divided by n?

\(\begin{array}{|rcll|} \hline (ab)^{-1} &\equiv& 2\pmod n \\ a^{-1}b^{-1} &\equiv& 2\pmod n \quad | \quad \cdot a \\ \mathbf{b^{-1}} &\equiv&\mathbf{ 2a\pmod n} \\\\ a^{-1}b^{-1} &\equiv& 2\pmod n \quad | \quad \cdot b \\ \mathbf{a^{-1}} &\equiv& \mathbf{2b\pmod n} \\\\ (a+b)^{-1}(a^{-1}+b^{-1}) &\equiv& x \pmod n \\ (a+b)^{-1}(2b+2a) &\equiv& x \pmod n \\ 2(a+b)^{-1}(ab) &\equiv& x \pmod n \\ 2 &\equiv& x \pmod n \\ x &\equiv& 2 \pmod n \\ \hline \end{array}\)

What is the largest integer \(n\) such that \(7^n\) divides \(1000!\)

\(\text{The largest integer $= \left[\dfrac{1000}{7}\right] + \left[\dfrac{1000}{7^2}\right] + \left[\dfrac{1000}{7^3}\right] \qquad [\ldots] =$ integer part. } \\ \text{The largest integer $= 142 + 20 + 2 $} \\ \text{The largest integer $= 164 $} \)

1.
What is the smallest distance between the origin and a point on the graph of \(y=\dfrac{1}{\sqrt{2}}\left(x^2-3\right)?\)

\(\text{Let the distance $=s$} \)

\(\begin{array}{|rcll|} \hline s^2 &=& x^2+y^2 \quad &|\quad y=\dfrac{1}{\sqrt{2}}\left(x^2-3\right) \\ s^2 &=& x^2+\left(\dfrac{1}{\sqrt{2}}\left(x^2-3\right)\right)^2 \\ s^2 &=& x^2+\dfrac{1}{2}\left(x^2-3\right)^2 \\ \dfrac{d\ s^2}{dx} &=& 2x_{\text{min}}+\dfrac{2}{2} (x_{\text{min}}^2-3)2x_{\text{min}} \\ &=& 2x_{\text{min}}+\dfrac{2}{2} (x_{\text{min}}^2-3)2x_{\text{min}} \\ &=& 2x_{\text{min}}(1+x_{\text{min}}^2-3) \\ &=& 2x_{\text{min}}(x_{\text{min}}^2-2) \\\\ \dfrac{d\ s^2}{dx} &=& 0 \\ 2x_{\text{min}}(x_{\text{min}}^2-2) &=& 0 \\ x_{\text{min}}^2-2 &=& 0 \\ \mathbf{ x_{\text{min}}^2 } &=& \mathbf{2} \\\\ y_{\text{min}} &=& \dfrac{1}{\sqrt{2}}\left(x_{\text{min}}^2-3\right) \\ y_{\text{min}} &=& \dfrac{1}{\sqrt{2}}\left(2-3\right) \\ \mathbf{ y_{\text{min}} } &=& -\dfrac{1}{\sqrt{2}} \\\\ s_{\text{min}} &=& \sqrt{x_{\text{min}}^2+y_{\text{min}}^2} \\ s_{\text{min}} &=& \sqrt{2+\left(-\dfrac{1}{\sqrt{2}}\right)^2} \\ s_{\text{min}} &=& \sqrt{ 2+ \dfrac{1}{2} } \\ s_{\text{min}} &=& \sqrt{ \dfrac{5}{2} } \\ \mathbf{ s_{\text{min}}} &=& \mathbf{1.58113883008\ldots} \\ \hline \end{array}\)

Suppose that x is an integer that satisfies the following congruences:
\(\begin{align*} 3+x &\equiv 2^2 \pmod{3^3} \\ 5+x &\equiv 3^2 \pmod{5^3} \\ 7+x &\equiv 5^2 \pmod{7^3} \end{align*}\)
What is the remainder when x is divided by 105?

idk how. can u like, explain how this can be done easily?

As n  ranges over the positive integers, what is the maximum possible value that the greatest common divisor of  13n + 8 and  5n+3 can take?

Formula: \(\boxed{\gcd(a,b)=gcd(a-b,b)}\)

\(\begin{array}{|rcll|} \hline && \mathbf{gcd(13n + 8, 5n+3)} \\ &=& gcd\left(13n+8-(5n+3), 5n+3\right) \\ &=& gcd\left(13n+8-5n-3, 5n+3\right) \\ &=& gcd\left(8n+5, 5n+3\right) \\ &=& gcd\left(8n+5-(5n+3), 5n+3\right) \\ &=& gcd\left(8n+5-5n-3, 5n+3\right) \\ &=& gcd\left(3n+2, 5n+3\right) \\ &=& gcd\left(5n+3,3n+2\right) \\ &=& gcd\left(5n+3-(3n+2),3n+2\right) \\ &=& gcd\left(5n+3-3n-2,3n+2\right) \\ &=& gcd\left(2n+1,3n+2\right) \\ &=& gcd\left(3n+2,2n+1\right) \\ &=& gcd\left(3n+2-(2n+1),2n+1\right) \\ &=& gcd\left(3n+2-2n-1,2n+1\right) \\ &=& gcd\left(n+1,2n+1\right) \\ &=& gcd\left(2n+1,n+1\right) \\ &=& gcd\left(2n+1-(n+1),n+1\right) \\ &=& gcd\left(2n+1-n-1,n+1\right) \\ &=& gcd\left(n,n+1\right) \\ &=& gcd\left(n+1,n\right) \\ &=& gcd\left(n+1-(n),n\right) \\ &=& gcd\left(n+1-n,n\right) \\ &=& gcd\left(1,n\right) \\ &=& gcd\left(n,1\right) \\ &=& \mathbf{1} \\ \hline \end{array}\)

The maximum possible value is \(\mathbf{1}\)