heureka

1.
Find the greatest integer x such that \(|x^2 - 98x+1| = 98x-x^2-1\).

\(\begin{array}{|rclrll|} \hline \mathbf{|x^2 - 98x+1|} &=& \mathbf{98x-x^2-1} \\\\ |x^2 - 98x+1| &=& -\underbrace{(x^2 - 98x+1)}_{<0!} \\\\ && x^2 - 98x+1 &<& 0 \\ && \left(x-\dfrac{98}{2}\right)^2 -\dfrac{98^2}{4} + 1 &<& 0 \\ && \left(x-\dfrac{98}{2}\right)^2 &<& \dfrac{98^2}{4} - 1 \\ && \left(x-\dfrac{98}{2}\right)^2 &<& \dfrac{98^2-4}{4} \\ && x-\dfrac{98}{2} &<& \dfrac{\sqrt{98^2-4}}{2} \\ && x &<& \dfrac{98}{2} +\dfrac{\sqrt{98^2-4}}{2} \\ && x &<& \dfrac{98+\sqrt{98^2-4}}{2} \\ && x &<& 97.9897948557 \\ &&\mathbf{ x_{\text{max}}} &=& \mathbf{97} \\ \hline \end{array} \)

Find the sum of the roots, real and non-real, of the equation \(x^{2001}+\left(\dfrac 12-x\right)^{2001}=0\),
given that there are no multiple roots.

\(\begin{array}{|lcll|} \hline \mathbf{x^{2001}+\left(\dfrac 12-x\right)^{2001}} &=& {0} \\\\ x^{2001} + \dbinom{2001}{0}\left(\dfrac12\right)^{2001} - \dbinom{2001}{1} \left(\dfrac12\right)^{2000}x^1+\ldots + \\ - \dbinom{2001}{1999} \left(\dfrac12\right)^{2}x^{1999} + \dbinom{2001}{2000} \left(\dfrac12\right)^{1}x^{2000} - \dbinom{2001}{2001}x^{2001}&=& 0 \\\\ x^{2001} + \dbinom{2001}{0}\left(\dfrac12\right)^{2001} - \dbinom{2001}{1} \left(\dfrac12\right)^{2000}x^1+\ldots + \\ - \dbinom{2001}{1999} \left(\dfrac12\right)^{2}x^{1999} + \dbinom{2001}{2000} \left(\dfrac12\right)^{1}x^{2000} - x^{2001}&=& 0 \\\\ \dbinom{2001}{0}\left(\dfrac12\right)^{2001} - \dbinom{2001}{1} \left(\dfrac12\right)^{2000}x^1+\ldots + \\ - \dbinom{2001}{1999} \left(\dfrac12\right)^{2}x^{1999} + \dbinom{2001}{2000} \left(\dfrac12\right)^{1}x^{2000} &=& 0 \\ \hline \end{array} \)

\(\begin{array}{|lcll|} \hline \dbinom{2001}{2000} \left(\dfrac12\right)^{1}x^{2000} - \dbinom{2001}{1999} \left(\dfrac12\right)^{2}x^{1999}+- \ldots + \dbinom{2001}{0}\left(\dfrac12\right)^{2001} &=& 0 \\\\ 1000.5x^{2000} - 500250 x^{1999}+- \ldots + \left(\dfrac12\right)^{2001} &=& 0 \quad | \quad : 1000.5 \\\\ x^{2000} - \dfrac{500250}{1000.5}x^{1999}+- \ldots + \dfrac{\left(\dfrac12\right)^{2001}}{1000.5} &=& 0 \\\\ x^{2000} - 500x^{1999}+- \ldots + \dfrac{\left(\dfrac12\right)^{2001}}{1000.5} &=& 0 \\\\ x^{2000} \underbrace{- 500}_{=-\sum \limits_{k=1}^{2000} x_k}x^{1999}+- \ldots + \dfrac{\left(\dfrac12\right)^{2001}}{1000.5} &=& 0 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline -\sum \limits_{k=1}^{2000} x_k &=& -500 \\ \mathbf{\sum \limits_{k=1}^{2000} x_k} &=& \mathbf{500} \\ \hline \end{array}\)

The sum of the roots is 500

Thank you, CPhill !

Thank you, Melody !

Thank you NoobGuest !

Let \(f_1,f_2,f_3,\ldots\) be a sequence of numbers such that \(f_n = f_{n - 1} + f_{n - 2}\) for every integer \(n \ge 3\) .

If \(f_7 = 83\), what is the sum of the first 10 terms of the sequence?

\(\begin{array}{|rclcl|} \hline f_1 &=& 0\cdot f_2 &+& 1\cdot f_1 \\ f_2 &=& 1\cdot f_2 &+& 0\cdot f_1 \\ f_3 &=& 1\cdot f_2 &+& 1\cdot f_1 \\ f_4 &=& 2\cdot f_2 &+& 1\cdot f_1 \\ f_5 &=& 3\cdot f_2 &+& 2\cdot f_1 \\ f_6 &=& 5\cdot f_2 &+& 3\cdot f_1 \\ f_7 &=& 8\cdot f_2 &+& 5\cdot f_1 \\ \ldots \\ \mathbf{f_n} &=& \mathbf{ F_{n-1} \cdot f_2} &\mathbf{+}& \mathbf{F_{n-2}\cdot f_1 } \\\\ && && \text{Fibonacci Numbers:} \\ && && \ldots \\ && && F_{-1} = 1 \\ && && F_{0} = 0 \\ && && F_{1} = 1 \\ && && F_{2} = 1 \\ && && F_{3} = 2 \\ && && F_{4} = 3 \\ && && F_{5} = 5 \\ && && F_{6} = 8 \\ && && F_{7} = 13 \\ && && F_{8} =21 \\ && && F_{9} = 34 \\ && && F_{10} = 55 \\ && && F_{11} = 89 \\ && && \ldots \\ \hline \end{array} \)

\(\mathbf{f_1 = \ ?,\ f_2 = \ ?}\)

\(\begin{array}{|rcll|} \hline f_7 = 83 &=& F_6\cdot f_2 + F_5\cdot f_1 \quad | \quad F_6 = 8,\qquad F_5=5 \\ \hline \end{array}\)

\(\begin{array}{|rclrclrclrcl|} \hline 83 &=& 8f_2 + 5f_1 \\ 5f_1 &=&83- 8f_2 \\ f_1 &=& \dfrac{83- 8f_2}{5} \\ f_1 &=& \dfrac{80-5f_2-3f_2+3}{5} \\ f_1 &=& 16-f_2+\underbrace{\dfrac{3-3f_2}{5}}_{=a} \\ f_1 &=& 16-f_2+a & a &=& \dfrac{3-3f_2}{5} \\ & & & 5a &=& 3-3f_2 \\ & & & 3f_2 &=& 3-5a \\ & & & f_2 &=& \dfrac{3-5a}{3} \\ & & & f_2 &=& \dfrac{3-3a-2a}{3} \\ & & & f_2 &=& 1-a-\underbrace{\dfrac{2a}{3}}_{=b} \\ & & & f_2 &=& 1-a-b & b &=& \dfrac{2a}{3} \\ & & & & & & 3b &=& 2a \\ & & & & & & 2a &=& 3b \\ & & & & & & a &=& \dfrac{3b}{2} \\ & & & & & & a &=& \dfrac{2b+b}{2} \\ & & & & & & a &=& b +\underbrace{\dfrac{b}{2}}_{=c} \\ & & & & & & a &=& b +c & c &=& \dfrac{b}{2} \\ & & & & & & & & & \mathbf{b} &=& \mathbf{2c} \\ & & & & & & a &=& 2c +c \\ & & & & & & \mathbf{a} &=& \mathbf{3c} \\ & & & f_2 &=& 1-3c-2c \\ & & & \mathbf{f_2} &=& \mathbf{1-5c} \\ f_1 &=& 16-(1-5c)+3c \\ \mathbf{f_1} &=& \mathbf{15+8c} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{f_1} &=& \mathbf{15+8c} ,\ c\in \mathbb{Z}\\ \mathbf{f_2} &=& \mathbf{1-5c},\ c\in \mathbb{Z}\\ \hline \end{array}\)

sum:

\(\begin{array}{|rclcl|} \hline s_1 &=& 0\cdot f_2 &+& 1\cdot f_1 \\ s_2 &=& 1\cdot f_2 &+& 1\cdot f_1 \\ s_3 &=& 2\cdot f_2 &+& 2\cdot f_1 \\ s_4 &=& 4\cdot f_2 &+& 3\cdot f_1 \\ s_5 &=& 7\cdot f_2 &+& 5\cdot f_1 \\ s_6 &=& 20\cdot f_2 &+& 8\cdot f_1 \\ \ldots \\ \mathbf{s_n} &=& \mathbf{ (F_{n+1}-1) \cdot f_2} &\mathbf{+}& \mathbf{F_{n}\cdot f_1 } \\ \hline \end{array}\)

\(\mathbf{s_{10}=\ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{s_{10}} &=& \mathbf{(F_{11}-1)\cdot f_2 +F_{10}} \cdot f_1 \quad &| \quad F_{11}=89,\ F_{10}= 55 \\\\ &=& 88 f_2 +55f_1 \quad &| \quad f_1 = 15+8c,\ f_2 = 1-5c \\ &=& 88\cdot(1-5c) +55\cdot(15+8c) \\ &=& 88 -88\cdot 5c +55\cdot 15+ 55\cdot 8c \quad &| \quad 88\cdot 5 = 55\cdot 8 \\ &=& 88 +55\cdot 15 \\ &=& \mathbf{913} \\ \hline \end{array}\)

The sum of the first 10 terms of the sequence is 913

Compute \(\mathbf{i^{1234}}\).

\(\begin{array}{|rcll|} \hline && \mathbf{i^{1234}} \\ &=& \left(i^2\right)^{617} \quad | \quad i^2 = -1 \\ &=& \left(-1\right)^{617} \\ &=& \mathbf{ -1} \\ \hline \end{array} \)

The function f(x,y) accepts an ordered pair as input and gives another ordered pair as output.

It is defined according to the following rules: If x>4,f(x,y) = (x-4,y) . If x<= 4 but y>4, f(x,y) = (x,y-4). Otherwise,(x+5,y+6) . 

A robot starts by moving to the point (1,1). 

Every time it arrives at a point (x,y), it applies f to that point and then moves to f(x,y). 

If the robot runs forever, how many different points will it visit?

\( f(x,y) =\begin{cases} f(x-4,y), & \text{if } x>4 \\\\ f(x,y-4), & \text{if } x\le4 \text{ and } y > 4 \\\\ f(x+5,y+5), & \text{if } x\le4 \text{ and } y \le 4 \\ \end{cases}\)

\(\begin{array}{|rcrcrcrc|} \hline & &f(x+5,y+5) & & f(x-4,y) & & f(x,y-4) \\ \hline f(1,1) &\rightarrow & f(6,6) &\rightarrow & f(2,6) & \rightarrow & f(2,2) \\ &\rightarrow & f(7,7) &\rightarrow & f(3,7) & \rightarrow & f(3,3) & \\ &\rightarrow & f(8,8) &\rightarrow & f(4,8) & \rightarrow & f(4,4) & \\ &\rightarrow & f(9,9) &\rightarrow & f(5,9) \\ & & &\rightarrow & f(1,9) & \rightarrow & f(1,5) & \\ & & & & & \rightarrow & f(1,1) \\ \hline \end{array} \)

The robot will 14  different points visit.

3.
A sequence \((a_n)\) is defined as follows: \(a_1 = 1,\ a_2 = \dfrac{1}{2}\), and \(a_n = \dfrac{1 - a_{n - 1}}{2a_{n - 2}}\) for all \(n > 2\). Find \(a_{120}\).

\( a_n =\begin{cases} \dfrac{1}{2}, & \text{if }n\equiv 2 \pmod{3} \\\\ \dfrac{n}{2n+6}, & \text{if }n\equiv 0 \pmod{3} \\\\ \dfrac{n+5}{2n+4}, & \text{if }n\equiv 1 \pmod{3} \\ \end{cases}\)

\(\begin{array}{|rcll|} \hline n &=& 120 \quad \text{ and }\quad 120 &\equiv& 0 \pmod{3} \\ \\ a_{120} &=& \dfrac{n}{2n+6} \\\\ &=& \dfrac{120}{2\cdot 120+6} \\\\ &=& \dfrac{120}{246} \\\\ \mathbf{a_{120}} &=& \mathbf{\dfrac{20}{41}} \\ \hline \end{array}\)

3.
A sequence \((a_n)\) is defined as follows: \(a_1 = 1,\ a_2 = \dfrac{1}{2}\), and \(a_n = \dfrac{1 - a_{n - 1}}{2a_{n - 2}}\) for all \(n > 2\). Find \(a_{120}\).

\( a(1) = 1 \\ a(2) = 1/2 \\ a(3) = 1/4 \\ a(4) = 3/4 \\ a(5) = 1/2 \\ a(6) = 1/3 \\ a(7) = 2/3 \\ a(8) = 1/2 \\ a(9) = 3/8 \\ a(10) = 5/8 \\ a(11) = 1/2 \\ a(12) = 2/5 \\ a(13) = 3/5 \\ a(14) = 1/2 \\ a(15) = 5/12 \\ a(16) = 7/12 \\ a(17) = 1/2 \\ a(18) = 3/7 \\ a(19) = 4/7 \\ a(20) = 1/2 \\ a(21) = 7/16 \\ a(22) = 9/16 \\ a(23) = 1/2 \\ a(24) = 4/9 \\ a(25) = 5/9 \\ a(26) = 1/2 \\ a(27) = 9/20 \\ a(28) = 11/20 \\ a(29) = 1/2 \\ a(30) = 5/11 \\ a(31) = 6/11 \\ a(32) = 1/2 \\ a(33) = 11/24 \\ a(34) = 13/24 \\ a(35) = 1/2 \\ a(36) = 6/13 \\ a(37) = 7/13 \\ a(38) = 1/2 \\ a(39) = 13/28 \\ a(40) = 15/28 \\ a(41) = 1/2 \\ a(42) = 7/15 \\ a(43) = 8/15 \\ a(44) = 1/2 \\ a(45) = 15/32 \\ a(46) = 17/32 \\ a(47) = 1/2 \\ a(48) = 8/17 \\ a(49) = 9/17 \\ a(50) = 1/2 \\ a(51) = 17/36 \\ a(52) = 19/36 \\ a(53) = 1/2 \\ a(54) = 9/19 \\ a(55) = 10/19 \\ a(56) = 1/2 \\ a(57) = 19/40 \\ a(58) = 21/40 \\ a(59) = 1/2 \\ a(60) = 10/21 \\ a(61) = 11/21 \\ a(62) = 1/2 \\ a(63) = 21/44 \\ a(64) = 23/44 \\ a(65) = 1/2 \\ a(66) = 11/23 \\ a(67) = 12/23 \\ a(68) = 1/2 \\ a(69) = 23/48 \\ a(70) = 25/48 \\ a(71) = 1/2 \\ a(72) = 12/25 \\ a(73) = 13/25 \\ a(74) = 1/2 \\ a(75) = 25/52 \\ a(76) = 27/52 \\ a(77) = 1/2 \\ a(78) = 13/27 \\ a(79) = 14/27 \\ a(80) = 1/2 \\ a(81) = 27/56 \\ a(82) = 29/56 \\ a(83) = 1/2 \\ a(84) = 14/29 \\ a(85) = 15/29 \\ a(86) = 1/2 \\ a(87) = 29/60 \\ a(88) = 31/60 \\ a(89) = 1/2 \\ a(90) = 15/31 \\ a(91) = 16/31 \\ a(92) = 1/2 \\ a(93) = 31/64 \\ a(94) = 33/64 \\ a(95) = 1/2 \\ a(96) = 16/33 \\ a(97) = 17/33 \\ a(98) = 1/2 \\ a(99) = 33/68 \\ a(100) = 35/68 \\ a(101) = 1/2 \\ a(102) = 17/35 \\ a(103) = 18/35 \\ a(104) = 1/2 \\ a(105) = 35/72 \\ a(106) = 37/72 \\ a(107) = 1/2 \\ a(108) = 18/37 \\ a(109) = 19/37 \\ a(110) = 1/2 \\ a(111) = 37/76 \\ a(112) = 39/76 \\ a(113) = 1/2 \\ a(114) = 19/39 \\ a(115) = 20/39 \\ a(116) = 1/2 \\ a(117) = 39/80 \\ a(118) = 41/80 \\ a(119) = 1/2 \\ a(120) = 20/41 \)