heureka

Let r and s be the roots of $3x^2 + 4x + 12 = 2x^2 - 7x + 5$.
Find $r^2 + s^2$.

$\begin{array}{|rcll|} \hline \mathbf{ 3x^2 + 4x + 12 } &=& \mathbf{ 2x^2 - 7x + 5 } \\ 3x^2-2x^2 +4x+7x+12-5 &=& 0 \\ \mathbf{ x^2+11x+7 } &=& \mathbf{0} \\ \hline \end{array}$

Vieta: $rs=7,~ -(r+s) = 11$

$\begin{array}{|rcll|} \hline -(r+s) &=& 11 \\ r+s &=& -11 \\ (r+s)^2 &=& (-11)^2 \\ r^2+2rs+ s^2 &=& 121 \quad &|\quad rs &=& 7 \\ r^2 +2*7 + s^2 &=& 121 \\ r^2+s^2 &=& 121 - 14 \\ \mathbf{ r^2+s^2 } &=& \mathbf{ 107 } \\ \hline \end{array}$

The number $210$ can be written as the sum of consecutive integers in several ways.
(a) When Written as the sum of the greatest possible number of consecutive positive integers, what is the largest of these integers?
(b) What if we allow negative integerrs into the sum; then what is the greatest possible number of consecutive integers that sum to $210$?

Arithmetic progression:
$\begin{array}{|rcll|} \hline a_n &=& a_1+(n-1)*d \quad | \quad d=1 \\ a_n &=& a_1+(n-1)*1 \\ a_n &=& a_1+n-1 \\ \mathbf{n} &=& \mathbf{a_n-a_1+1} \\ \hline \end{array}$

Sum of the Arithmetic progression:
$\begin{array}{|rcll|} \hline \text{sum} &=& \dfrac{a_1+a_n}{2}*n \quad &|\quad \text{sum} = 210 \\\\ 210 &=& \dfrac{a_1+a_n}{2}*n \\\\ \left(\dfrac{a_1+a_n}{2}\right)*n &=& 210 \\\\ \mathbf{(a_1+a_n)*n} &=& \mathbf{420} \\ \hline \end{array}$

The divisors of $420$ are:
Divisors:
 1 |  2 |  3 |  4 |   5 |   6 |   7 | 10 |
12 | 14 | 15 | 20 |  21 |  28 |  30 | 35 |
42 | 60 | 70 | 84 | 105 | 140 | 210 | 420  (24 divisors)

$\text{Let $a_1+a_n=k\quad$or$\quad \mathbf{a_n=k-a_1}$}$

$\begin{array}{|rcll|} \hline (a_1+a_n)*n &=& 420 \\ k*n &=& 1*420 \\ &=& 2*210 \\ &=& 3*140 \\ &=& 4*105 \\ &=& 5*84 \\ &=& 6*70 \\ &=& 7*60 \\ &=& 10*42 \\ &=& 12*35 \\ &=& 14*30 \\ &=& 15*28 \\ &=& 20*21 \\ &=& 21*20 \\ &=& 28*15 \\ &=& 30*14 \\ &=& 35*12 \\ &=& 42*10 \\ &=& 60*7 \\ &=& 70*6 \\ &=& 84*5 \\ &=& 105*4 \\ &=& 140*3 \\ &=& 210*2 \\ &=& 420*1 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline n &=& a_n-a_1+1 \quad &| \quad a_n = k - a_1 \\ n &=& k-a_1-a_1+1 \\ n &=& k-2a_1+1 \\ 2a_1 &=& 1+k-n \\\\ \mathbf{a_1} &=& \mathbf{ \dfrac{1+k-n}{2} } \\ \hline \mathbf{a_n} &=& \mathbf{ k-a_1 } \\ \hline \end{array}$

$\begin{array}{|r|r|r|r|r|l|} \hline k*n & k & n & a_1 & a_n & \text{Arithmetic} \\ &&&& & \text{progression} \\ \hline 1*420 & 1 &\color{red}420 & -209 & 210 &-209,-208,\dots,209,210 \\ 2*210 & 2 & 210 & (-103.5)& &\text{$a_1$ no integer } \\ 3*140 & 3 & 140 & -68 & 71 &-68,-67,\dots,70,71 \\ 4*105 & 4 & 105 & -50 & 54 &-50,-49,\dots,53,54 \\ 5*84 & 5 & 84 & -39 & 44 &-39,-38,\dots,43,44 \\ 6*70 & 6 & 70 & (-31.5)& &\text{$a_1$ no integer } \\ 7*60 & 7 & 60 & -26 & 33 &-26,-25,\dots,32,33 \\ 10*42 & 10 & 42 & (-15.5)& &\text{$a_1$ no integer } \\ 12*35 & 12 & 35 & -11 & 23 &-11,-10,\dots,22,23\\ 14*30 & 14 & 30 & (-7.5)& &\text{$a_1$ no integer } \\ 15*28 & 15 & 28 & -6 & 21 &-6,-5,\dots,20,21\\ \hline 20*21 & 20 & 21 & (0) & 20 &\text{$a_1$ no integer } \\ \hline 21*20 & 21 & \color{red}20 & 1 & 20 & 1,2,\dots,19,20 \\ 28*15 & 28 & 15 & 7 & 21 & 7,8,\dots,20,21\\ 30*14 & 30 & 14 & (8.5)& &\text{$a_1$ no integer } \\ 35*12 & 35 & 12 & 12 & 23 & 12,13,\dots,22,23\\ 42*10 & 42 & 10 & (16.5)& &\text{$a_1$ no integer } \\ 60*7 & 60 & 7 & 27 & 33 & 27,28,\dots,32,33\\ 70*6 & 70 & 6 & (32.5)& &\text{$a_1$ no integer } \\ 84*5 & 84 & 5 & 40 & 44 & 40,41,42,43,44 \\ 105*4 &105 & 4 & 51 & 54 & 51,52,53,54\\ 140*3 &140 & 3 & 69 & 71 & 69,70,71\\ 210*2 &210 & 2 & (104.5)& &\text{$a_1$ no integer } \\ 420*1 &420 & 1 & 210 & 210 & 210 \\ \hline \end{array}$

(a) 20
(b) 420

For how many integer values of $\color{red}a$ does the equation
$x^2 + ax + 28a = 0$
have integer solutions for x?

$\text{Let the roots are $r_1$ and $r_2$}:$

Vieta:

$\begin{array}{|rcll|} \hline x^2 \underbrace{+a}_{=-(r_1+r_2)}x \underbrace{+ 28a}_{=r_1r_2} \\ \hline \mathbf{a} &=& \mathbf{-(r_1+r_2)} \\ 28a &=& r_1r_2 \\ 28\left(-(r_1+r_2)\right) &=& r_1r_2 \\ -28r_1-28r_2 &=& r_1r_2 \\ r_1r_2+28r_1+28r_2 &=& 0 \\ (r_1+28)(r_2+28) -28*28 &=& 0 \\ (r_1+28)(r_2+28)&=& 28*28 \\ \mathbf{(r_1+28)(r_2+28)}&=& \mathbf{784} \\ \hline \end{array}$

The divisors of 784 are:
  1 |  2 |  4 |   7 |   8 |  14 |  16 | 28 |
 49 | 56 | 98 | 112 | 196 | 392 | 784  (15 divisors)

$\begin{array}{|rcll|} \hline (r_1+28)(r_2+28)&=& 1*784 \\ &=& 2*392 \\ &=& 4*196 \\ &=& 7*112 \\ &=& 8*98 \\ &=& 14*56 \\ &=& 16*49 \\ &=& 28*28 \\ \hline \end{array}$

$\begin{array}{|r|r|r|r|r|} \hline (r_1+28) & r_1 & (r_2+28) &r_2 & a=-r_1-r_2 \\ \hline 1 & 1-28 = -27 & 784 & 784-28=756 & 27-756=\color{red}-729 \\ 2 & 2-28 = -26 & 392 & 392-28=364 & 26-364=\color{red}-338 \\ 4 & 4-28 = -24 & 196 & 196-28=168 & 24-168=\color{red}-144 \\ 7 & 7-28 = -21 & 112 & 112-28= 74 & 21-74=\color{red}-53 \\ 8 & 8-28 = -20 & 98 & 98-28= 70 & 20-70=\color{red}-50 \\ 14 & 14-28 = -14 & 56 & 56-28= 28 & 14-28=\color{red}-14 \\ 16 & 16-28 = -12 & 49 & 49-28= 21 & 12-21=\color{red}-9 \\ 28 & 28-28 = 0 & 28 & 28-28= 0 & -0-0=\color{red}0 \\ \hline 49 & 49-28= 21 & 16 & 16-28 = -12 & -21+12=\color{red}-9 \\ 56 & 56-28= 28 & 14 & 14-28 = -14 & -28+14=\color{red}-14 \\ 98 & 98-28= 70 & 8 & 8-28 = -20 & -70+20=\color{red}-50 \\ 112 & 112-28= 74 & 7 & 7-28 = -21 & -74+21=\color{red}-53 \\ 196 & 196-28=168 & 4 & 4-28 = -24 & -168+24=\color{red}-144 \\ 392 & 392-28=364 & 2 & 2-28 = -26 & -364+26=\color{red}-338 \\ 784 & 784-28=756 & 1 & 1-28 = -27 & -756+27=\color{red}-729 \\ \hline \end{array}$

Distinct Integer values of $\color{red} a$ are 7 without $a= 0$: $\mathbf{\{ -9,~-14,~-50,~-53~-144,~-338,~-729\}}$

Let a, b, and c be positive real numbers.

Prove that $a^2 + b^2 + c^2 >= ab + ac + bc$.

$\large{AM\ge GM}$

$\begin{array}{|rcll|} \hline \mathbf{\dfrac{a+b}{2}} &\ge& \mathbf{\sqrt{ab}} \\ a+b &\ge& 2\sqrt{ab} \quad &| \quad \text{square both sides} \\ (a+b)^2 &\ge& 4ab \\ a^2+2ab+b^2 &\ge& 4ab \quad &| \quad -2ab \\ a^2+b^2 &\ge& 4ab-2ab \\ \mathbf{a^2+b^2} &\ge& \mathbf{2ab} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{\dfrac{b+c}{2}} &\ge& \mathbf{\sqrt{bc}} \\ b+c &\ge& 2\sqrt{bc} \quad &| \quad \text{square both sides} \\ (b+c)^2 &\ge& 4bc \\ b^2+2bc+c^2 &\ge& 4bc \quad &| \quad -2bc \\ b^2+c^2 &\ge& 4bc-2bc \\ \mathbf{b^2+c^2} &\ge& \mathbf{2bc} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{\dfrac{a+c}{2}} &\ge& \mathbf{\sqrt{ac}} \\ a+c &\ge& 2\sqrt{ac} \quad &| \quad \text{square both sides} \\ (a+c)^2 &\ge& 4ac \\ a^2+2ac+c^2 &\ge& 4ac \quad &| \quad -2ac \\ a^2+c^2 &\ge& 4ac-2ac \\ \mathbf{a^2+c^2} &\ge& \mathbf{2ac} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{a^2+b^2} &\ge& \mathbf{2ab} \\ \mathbf{b^2+c^2} &\ge& \mathbf{2bc} \\ \mathbf{a^2+c^2} &\ge& \mathbf{2ac} \\ \hline (a^2+b^2)+(b^2+c^2) + (a^2+c^2) &\ge& 2ab + 2bc +2ac \\ 2a^2+2b^2+2c^2 &\ge& 2(ab +bc +ac)\\ 2(a^2+b^2+c^2) &\ge& 2(ab +bc +ac) \quad &| \quad : 2\\ \mathbf{a^2+b^2+c^2} &\ge& \mathbf{ab +bc +ac} \\ \hline \end{array}$

Under what conditions does equality occur? 

$\begin{array}{|rcll|} \hline 2(a^2+b^2+c^2) &=& 2(ab +bc +ac) \\ 2(a^2+b^2+c^2)- 2(ab +bc +ac) &=& 0 \\ \hline \end{array}\\ 2(a^2+b^2+c^2)- 2(ab +bc +ac) =(a-b)^2+(b-c)^2+(c-a)^2 \\ \begin{array}{|lcll|} \hline (\underbrace{a-b}_{=0})^2+(\underbrace{b-c}_{=0})^2+(\underbrace{c-a}_{=0})^2&=& 0 \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline a-b&=& 0 \\ a&=& b \\\\ b-c&=& 0 \\ b&=& c \\\\ c-a &=& 0 \\ c&=& a \\ \hline \end{array}$

equality occur when $a=b=c$

how you got to the rule in the first place:

Arithmetic progression: $1,~ 3,~ 5,~ 7,~\ldots$
$\begin{array}{|rcll|} \hline a_n &=& a_1+(n-1)*d \quad | \quad a_1 = 1,~ d=2 \\ a_n &=& 1+(n-1)*2 \\ a_n &=& 1+2n -2 \\ a_n &=& 2n-1 \\ 2n &=& a_n + 1 \\ n &=& \dfrac{a_n+1}{2} \quad \text{or} \quad n = \dfrac{x+1}{2}\\ \hline \end{array}$

Hi friends, please help me with finding the rUle for these inputs aand outputs:

Inputs are 1; 3; 5 ;7 (X values)

Outputs are 5; 10; 15; 20 (Y values)

$\begin{array}{|r|r|r|r|} \hline x & \dfrac{x+1}{2} & \left(\dfrac{x+1}{2}\right) \times 5 \\ \hline 1 & \dfrac{1+1}{2}= 1 & 1*5 = 5 \\ \hline 3 & \dfrac{3+1}{2}= 2 & 2*5 = 10\\ \hline 5 & \dfrac{5+1}{2}= 3 & 3*5 = 15\\ \hline 7 & \dfrac{7+1}{2}= 4 & 4*5 = 20 \\ \hline \end{array}$

The rule is $y= \left(\dfrac{x+1}{2}\right)\times 5$

The Lucas sequence is the sequence $1,~ 3,~ 4,~ 7,~ 11,~\dots$
where the first term is 1,
the second term is 3 and each term after that is the sum of the previous two terms.
What is the remainder when the 100th term of the sequence is divided by 20?

In number theory, the nth Pisano period, written p(n), is the period with which
the sequence taken modulo n repeats.

Suppose $173 * 927 n \pmod {50}$ where $0\le n < 50$.
What is the value of $n$ ?

$\begin{array}{|rcll|} \hline && \mathbf{ 173 * 927\pmod {50}} \\ &\equiv& 160371 \pmod {50} \\ &\equiv& 160350 + 21 \pmod {50} \\ &\equiv& 3201*50 + 21 \pmod {50} \quad | \quad 3201*50 \equiv 0 \mod{50}\\ &\equiv& 0 + 21 \pmod {50} \\ &\equiv& \mathbf{21 \pmod {50}} \\ \hline \end{array}$

n = 21

Alex collected 3 times as many seashells as Clay.
When Alex gave Clay some of his seashells, they each had 14.
How many seashells had Alex given to Clay?

$\begin{array}{|lrcll|} \hline (1) & A &=& 3C \\ (2) & A-x &=& 14 \\ (3) & C+x &=& 14 \\ \hline (2)+(3) : & A+C &=& 28 \\ & 3C +C &=& 28 \\ & 4C &=& 28 \\ & \mathbf{C} &=& \mathbf{7} \\ \hline & A &=& 3C \\ & A &=& 3*7 \\ & \mathbf{A} &=& \mathbf{21} \\ \hline & C+x &=& 14 \\ & 7+x &=& 14 \\ & \mathbf{x} &=& \mathbf{7} \\ \hline \end{array}$

7 seashells had Alex given to Clay

The roots of  $x^2 + 8x + 4$ are the same as
the roots of $Ax^2 + 16x + B$ .
What is $A + B$?

$\begin{array}{|rcll|} \hline -(x_1+x_2) = 8 &=& \dfrac{16}{A} \\\\ 8A &=& 16 \\ \mathbf{A} &=& \mathbf{2} \\ \hline x_1*x_2 =4 &=& \dfrac{B}{A} \\\\ B &=& 4A \\ B &=& 4*2 \\ \mathbf{B} &=& \mathbf{8} \\ \hline A+ B &=& 2+8 \\ \mathbf{A+B} &=& \mathbf{10} \\ \hline \end{array}$