heureka

Find the value of:

\(x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\).

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}} } \quad &| \quad -1 \\ (x-1) &=& \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}} \\ (x-1) &=& \cfrac{1}{2 + (x-1)} \quad & | \quad x-1= \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}} \\ (x-1) &=& \cfrac{1}{1+x} \\ (x-1)(1+x) &=& 1 \\ x+x^2-1-x &=& 1 \\ x^2 &=& 2 \\ \mathbf{x} &=& \mathbf{\sqrt{2}} \\ \hline \end{array}\)

This number \(65^4\) has a long sequence of positive consecutive 4-digit integers.
The first term, the 3001st term and the last term are all 4-digit prime numbers.
What is the first term, the last term and the number of terms of this sequence?

\(\text{$a_1$ is a prime number hence odd } \\ \text{$a_n$ is a prime number hence odd } \)

\(\begin{array}{|rcll|} \hline && a_n = a_1+(n-1) \\ &or& n = a_n-a_1+1 \qquad \text{$a_n$ is odd and $a_1$ is odd hence $n$ is odd} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline 65^4 &=& a_1 + (a_1+1)+ (a_1+2) + \ldots + \Big(a_1+(n-1) \Big) \\ 65^4 &=& na_1 + \dfrac{ \Big(1+(n-1)\Big)}{2} (n-1 ) \\ 65^4 &=& na_1 + \dfrac{n(n-1)}{2} \\ 65^4 &=& n\left( a_1 + \dfrac{n-1}{2}\right) \\ 65^4 &=& n\left( \dfrac{2a_1+n-1}{2}\right) \quad | \quad n = a_n-a_1+1 \\ 65^4 &=& (a_n-a_1+1)\left( \dfrac{2a_1+a_n-a_1+1-1}{2}\right) \\ \mathbf{65^4} &=& \mathbf{(a_n-a_1+1)\left( \dfrac{ a_1+a_n }{2}\right)} \quad | \quad \text{$a_1+a_n$ is even! $\quad a_n-a_1+1=n$ is odd!} \\ \mathbf{65^4} &=& \mathbf{x\times y} \\ && \boxed{x = a_n-a_1+1},\ \boxed{y = \left( \dfrac{ a_1+a_n }{2}\right)} \quad \text{$x$ and $y$ are divisors of $65^4$ } \\\\ \hline x &=& a_n-a_1+1 \quad \text{ or } \\ \mathbf{ a_n } &=& \mathbf{x+p-1} \\\\ y &=& \left( \dfrac{ a_1+a_n }{2}\right) \\ y &=& \left( \dfrac{ a_1+x+p-1 }{2}\right)\quad \text{ or } \\ \mathbf{a_1} &=& \mathbf{y-\dfrac{x-1}{2}} \\ \hline \end{array}\)

The Divisors of \(65^4\) are:

\(\text{1 | 5 | 13 | 25 | 65 | 125 | 169 | 325 | 625 | 845 | 1625 | 2197 | 4225|$\\$ | 8125 | 10985 | 21125 | 28561 | 54925 | 105625 | 142805 | 274625 |$\\$ |714025 | 1373125 | 3570125 | 17850625 (25 divisors)}\)

\(\begin{array}{|rr|cc|cc|c|} \hline & & & \text{prime} &&\text{prime} \\ x & y & a_1 = y-\dfrac{x-1}{2} & \text{number} & a_n = x+a_1-1 & \text{number} & n =a_n-a_1+1 \\ \hline 1 & 17850625 & a_1 > 9999 \\ 5 & 3570125 & a_1 > 9999 \\ 13 &1373125 & a_1 > 9999 \\ 25 & 714025 & a_1 > 9999 \\ 65 & 274625 & a_1 > 9999 \\ 125 & 142805 & a_1 > 9999 \\ 169 & 105625 & a_1 > 9999 \\ 325 & 54925 & a_1 > 9999 \\ 625 & 28561 & a_1 > 9999 \\ 845 & 21125 & a_1 > 9999 \\ 1625 & 21125 & a_1 > 9999 \\ 2197 & 10985 & a_1=7027 & \text{yes} & a_n= 9223 & \text{no} \\ 4225 &4225& \color{red}a_1 = 2113 & \text{yes} & \color{red}a_n= 6337 & \text{yes} & \color{red}n= 4225 \\ \hline \end{array}\)

Here is only one solution so \(a_{3001} \)must be a prime number. \((a_{3001}=a_1+3000 = 2113+3000=5113\text{ prime number})\)

\(a_{3001} \) is irrelevant for the calculation!

Hello Guest,

yes,  your way is an acceptable way to do it.

math help

What is the soluton to the quadratic inequality?
\(6x^2 \geq 5-13x\)

Compute \(\sum \limits_{x=0^\circ}^{90^\circ} \cos^2(x)\)

\(\begin{array}{|rcll|} \hline &&\mathbf{ \sum \limits_{x=0^\circ}^{90^\circ} \left(\cos^2(x) \right) } \\ &=& \sum \limits_{x=0^\circ}^{44^\circ} \left(\cos^2(x)\right)+\cos^2(45^\circ)+\sum \limits_{x=46^\circ}^{90^\circ} \left(\cos^2(x)\right) \\ &=& \sum \limits_{x=0^\circ}^{44^\circ} \left(\cos^2(x)\right)+\sum \limits_{x=46^\circ}^{90^\circ} \left(\cos^2(x)\right)+\cos^2(45^\circ) \\ &=& \sum \limits_{x=0^\circ}^{44^\circ} \left(\cos^2(x)\right)+\sum \limits_{x=0^\circ}^{44^\circ} \left(\cos^2(90^\circ-x)\right)+\cos^2(45^\circ) \\ &=& \sum \limits_{x=0^\circ}^{44^\circ} \left(\cos^2(x)\right)+\sum \limits_{x=0^\circ}^{44^\circ} \left(\sin^2(x)\right)+\cos^2(45^\circ) \\ &=& \sum \limits_{x=0^\circ}^{44^\circ} \left(\cos^2(x)+\sin^2(x)\right)+\cos^2(45^\circ) \\ &=& \sum \limits_{x=0^\circ}^{44^\circ} \left(1\right)+\cos^2(45^\circ) \\ &=& 45+\cos^2(45^\circ) \quad | \quad \cos(45^\circ) = \dfrac{\sqrt{2}}{2} \\ &=& 45+ \left(\dfrac{\sqrt{2}}{2}\right)^2 \\ &=& 45+ \dfrac{2}{4} \\ &=& 45+ \dfrac{1}{2} \\ &=& \mathbf{45.5} \\ \hline \end{array}\)

Find all the Intersecting points of a circle and a line:
Circle equation   :  (\(x-1)^2+(y-1)^2=1\)
Line equation     :  \(y=2x+1\)

\(\begin{array}{|rclrcl|} \hline \mathbf{x_1} &=& \mathbf{0} \\ && & y_1 &=& 2x_1+1 \\ && & y_1 &=& 2\cdot 0 +1 \\ && &\mathbf{ y_1} &=& \mathbf{ 1} \\\\ \mathbf{x_2} &=& \mathbf{0.4} \\ && & y_2 &=& 2x_2+1 \\ && & y_2 &=& 2\cdot 0.4 +1 \\ && &\mathbf{ y_2} &=& \mathbf{ 1.8 } \\ \hline \end{array} \)

The two Intersecting points of a circle and a line are: \(\mathbf{(0,\ 1)}\) and \(\mathbf{(0.4,\ 1.8)}\)

Find

\(\left|\left(1+2i\right)^8\right|\).

\(\begin{array}{|rclrcl|} \hline && \left|\left(1+2i\right)^8\right| \\ &=& \left|\Big(\left(1+2i\right)^2\Big)^4\right| & \qquad \left(1+2i\right)^2 &=& 1+4i+4i^2 \quad | \quad i^2 = -1\\ & & & &=& 1+4i-4 \\ & & & &=& -3+4i \\ &=& \left|\left(-3+4i\right)^4\right| \\ &=& \left|\Big(\left(-3+4i\right)^2\Big)^2\right| & \qquad \left(-3+4i\right)^2 &=& 9-24i+16i^2 \quad | \quad i^2 = -1\\ & & & &=& 9-24i-16i \\ & & & &=& -7-24i \\ &=& \left|\left(-7-24i\right)^2\right| \\ &=& \left|\left(-1 \right)^2\left(7+24i\right)^2\right| \\ &=& \left| \left(7+24i\right)^2\right| & \qquad \left(7+24i\right)^2 &=& 49+336i+576i^2 \quad | \quad i^2 = -1\\ & & & &=& 49+336i-576i \\ & & & &=& -527+336i \\ &=& \left|-527+336i\right| \\ &=&\sqrt{\left(-527 \right)^2 + 336^2} \\ &=&\sqrt{390625} \\ &=&\mathbf{625} \\ \hline \end{array}\)

\(\mathbf{\left|\left(1+2i\right)^8\right| = 625}\)

What is the soluton to the quadratic inequality?
\(6x^2 \geq 5-13x\)

\(\begin{array}{|rcll|} \hline \mathbf{6x^2} &\geq& \mathbf{5-13x} \quad | \quad -5+13x \\ 6x^2 +13x -5 &\geq& 0 \\ \overset{\text{quadratic formula}}{\ldots} \\ 6\left(x-\dfrac{1}{3}\right) \left(x+\dfrac{5}{2} \right) &\geq& 0 \quad | \quad : 6 \\\\ \mathbf{\left(x-\dfrac{1}{3}\right) \left(x+\dfrac{5}{2} \right)} &\geq&\mathbf{ 0 } \\ \hline \end{array} \)

We create a sign table:

\(\begin{array}{|l|c|c|c|c|c|c|c|} \hline \text{Interval or position} : & \left(-\infty,-\dfrac{5}{2}\right) & -\dfrac{5}{2} & \left(-\dfrac{5}{2},\dfrac{1}{3}\right) & \dfrac{1}{3} & \left(\dfrac{1}{3},\infty\right) \\ \hline \text{sign of } \left(x+\dfrac{5}{2} \right): & - & 0 & + & + & + \\ \hline \text{sign of } \left(x-\dfrac{1}{3}\right): & - & - & - & 0 & + \\ \hline \text{sign of }\left(x-\dfrac{1}{3}\right) \left(x+\dfrac{5}{2} \right): & \color{red}+ & \color{red}0 & - & \color{red}0 & \color{red}+ \\ \hline \end{array} \)

We can read the result:
in the interval \(\left(-\infty,-\dfrac{5}{2}\right]\) and \(\left[\dfrac{1}{3},\infty\right)\) the left side of the inequality is positive or zero, and thus the inequality is true there.

The solution is:  \(\mathbf{ x\leq -\dfrac{5}{2}}\)  and  \(\mathbf{x \geq \dfrac{1}{3}}\).

Sequence problem
The sequence \(\{a_n\}\) is defined by
\(a_0 = 1,a_1 = 1, \text{ and } a_n = a_{n - 1} + \dfrac {a_{n - 1}^2}{a_{n - 2}}\text{ for }n\ge2\).

The sequence \{b_n\} is defined by
\(b_0 = 1,b_1 = 3, \text{ and } b_n = b_{n - 1} + \dfrac {b_{n - 1}^2}{b_{n - 2}}\text{ for }n\ge2\).

Find \(\dfrac {b_{32}}{a_{32}}\).

\(\begin{array}{|rcll|} \hline a_n &=& a_{n - 1} + \dfrac { a_{n - 1}^2 } { a_{n - 2} } \\ \mathbf{ a_n } &=& \mathbf{ a_{n - 1} \left( 1 + \dfrac{a_{n-1}}{a_{n-2}} \right) } \\ \hline a_0 &=& 1 \text{ or } a_0 = 0! \\ a_1 &=& 1 \text{ or } a_1 = 1! \\ \hline a_2 &=& 1!\left(1+\dfrac{1}{1}\right) \\ &=& 1!\cdot 2 \\ &=& \mathbf{2!} \\\\ a_3 &=& 2!\left(1+\dfrac{1!\cdot 2}{1!}\right) \\ &=& 2!\cdot 3 \\ &=& \mathbf{3!} \\\\ a_4 &=& 3!\left(1+\dfrac{2!\cdot 3}{2!}\right) \\ &=& 3!\cdot 4 \\ &=& \mathbf{4!} \\\\ a_5 &=& 4!\left(1+\dfrac{3!\cdot 4}{3!}\right) \\ &=& 4!\cdot 5 \\ &=& \mathbf{5!} \\ \ldots \\ \mathbf{a_n} &=& \mathbf{n!} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline b_n &=& b_{n-1} + \dfrac{ b_{n-1}^2 } { b_{n-2} } \\ \mathbf{ b_n } &=& \mathbf{ b_{n-1} \left( 1 + \dfrac{b_{n-1}}{b_{n-2}} \right) } \\ \hline b_0 &=& 1 \text{ or } b_0 = \dfrac{2!}{2!} \\ b_1 &=& 3 \text{ or } b_1 = \dfrac{3!}{2!} \\ \hline b_2 &=& 3\left(1+\dfrac{3}{1}\right) \\ &=& 3\cdot 4 \\\\ b_3 &=& 3\cdot 4\left(1+\dfrac{\not{3}\cdot 4}{\not{3}}\right) \\ &=& 3\cdot 4\cdot 5 \\\\ b_4 &=& 3\cdot 4\cdot 5\left(1+\dfrac{\not{3}\cdot \not{4}\cdot 5}{\not{3}\cdot\not{4}}\right) \\ &=& 3\cdot 4\cdot 5\cdot 6 \\\\ b_5 &=& 3\cdot 4\cdot 5\cdot 6\left(1+\dfrac{\not{3}\cdot \not{4}\cdot \not{5}\cdot 6}{\not{3}\cdot\not{4}\cdot \not{5}}\right) \\ &=& 3\cdot 4\cdot 5\cdot 6\cdot 7 \\ \ldots \\ \mathbf{b_n} &=& \mathbf{\dfrac{(n+2)!}{2!} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{b_n}{a_n} &=& \dfrac{\dfrac{(n+2)!}{2!}} {n!} \\\\ &=& \dfrac{ (n+2)! } {2!n!} \\\\ &=& \dfrac{ n!(n+1)(n+2) } {2!n!} \\\\ &=& \dfrac{ (n+1)(n+2) } {2!} \\\\ \mathbf{ \dfrac{b_n}{a_n} } &=& \mathbf{ \dfrac{(n+1)(n+2)} {2}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{b_{32}}{a_{32}} &=& \dfrac{(32+1)(32+2)} {2} \\\\ &=& \dfrac{33\cdot 34}{2} \\\\ &=& 33\cdot 17 \\\\ \mathbf{ \dfrac{b_{32}} {a_{32}} } &=&\mathbf{561} \\ \hline \end{array} \)

The American Mathematics College is holding its orientation for incoming freshmen.
The incoming freshman class contains fewer than \(500\) people.
When the freshmen are told to line up in columns of \(23\), \(22\) people are in the last column.
When the freshmen are told to line up in columns of \(21\), \(14\) people are in the last column.
How many people are in the incoming freshman class?

\(\begin{array}{|rclrclr|} \hline x &\equiv& 22 \pmod{23} \\ \text{or } \quad x &=& 22+23n \\ x &\equiv& 14 \pmod{21} \\ \text{or}\quad \mathbf{ x } &=& \mathbf{14+21m} \\ \hline x = 14+21m &=& 22+23n \\ 21m &=& 23n + 8 \\ m &=& \dfrac{23n + 8}{21} \\ m &=& \dfrac{21n +2n + 8}{21} \\ m &=& n+ \underbrace{\dfrac{2n + 8}{21}}_{=a} \\ \mathbf{ m } &=& \mathbf{n+a} & a &=& \dfrac{2n + 8}{21} \\ & & & 21a &=& 2n + 8 \\ &&& 2n &=& 21a-8 \\ &&& n &=& \dfrac{21a-8}{2} \\ &&& n &=& \dfrac{20a+a-8}{2} \\ &&& n &=& 10a+\underbrace{\dfrac{a-8}{2}}_{=b} \\ &&& \mathbf{ n } &=& \mathbf{ 10a+b} & b=\dfrac{a-8}{2} \\ &&& & & & 2b= a-8 \\ &&& & & & \mathbf {a= 2b+8} \\ &&& n &=& 10(2b+8)+b \\ &&& \mathbf{n} &=& \mathbf{21b+80} \\ m &=& 21b+80+2b+8 \\ \mathbf{m} &=& \mathbf{23b+88} \\ \hline x &=& 14+21(23b+88) \\ x &=& 1862+483b \quad | \quad 1863 \mod{483} \equiv 413 \\ \mathbf{x} &=& \mathbf{413+483b} \\ \hline \end{array}\)

In the incoming freshman class are 413 people