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Hectictar just nominated this GP question.

Hi Heureka,

I have been learning from some of your modular arithemtic answers. I thank you for providing them.  

I looked at this one and thought there should be an easier method.

This is what I came up with.  Perhaps you could take a look and if you find fault with my logic then let me know.

\(\begin{align*} 3+x &\equiv 2^2 \pmod{3^3} \\ 5+x &\equiv 3^2 \pmod{5^3} \\ 7+x &\equiv 5^2 \pmod{7^3} \end{align*}\)

\(3+x\equiv 4 \quad \pmod{3^3}\\ x\equiv 1 \quad \pmod{3^3}\\ so\\ x\equiv 1 \quad \pmod{3}\\ x=1+3m\qquad \qquad (1) \)

\(5+x\equiv 9 \quad \pmod{5^3}\\ x\equiv 4 \quad \pmod{5^3}\\ x\equiv 4 \quad \pmod{5}\\ x\equiv -1 \quad \pmod{5}\\ x=-1+5n\qquad \qquad (2)\)

\(7+x\equiv 25 \quad \pmod{7^3}\\ x\equiv 18 \quad \pmod{7^3}\\ x\equiv 18 \quad \pmod{7}\\ x\equiv 4 \quad \pmod{7}\\ x=4+7k\qquad \qquad (3)\)

Solve the first 2 simultaneously

\(1+3m=-1+5n\\ 2=5n-3m\\ \text{One obvious solution for this is n=1 and m=1}\\ 2=5(1)-3(1)\\ \text{now I can add 15f and take it away again to get the general diophantine solution.}\\ 2=5(1+3f)-3(1+5f)\\ n=1+3f\\ x=-1+5n\\ x=-1+5(1+3f)\\ x=4+15f\\ \)

I could work it out with m instead of n but x in terms of f will be the same.

So now I have

\(x=4+7k \quad (3) \qquad and \qquad x=4+15f\\ \text{since }\\x\equiv 4 \pmod{7}\qquad and\\ x\equiv4\pmod{15}\\ \text{It follows that}\\ x\equiv 4 \pmod{7*15}\\ x\equiv 4 \pmod{105}\\ \)

So when x is divided by 105 the remainder is 4

Figure the area of the triangle again

What do we need to do after multiplying 10 x 8   ????   [ hint :    (1/2) base * height ]

That's a very interesting approach, heureka  !!!!

I like it  !!!!

Thankyou Cphill yes I was not even sure how to find that so thanks it makes more sense now! 

So first

8*6*1/2= 24 

10*8=80 

so 

80+24+24= 128

or 

80+24=104 

Not totally sure about this one....but....

The mean = 7

The first quartile is from  2 to 7

The second quartile is from 7 to 12

So   [ 1/4  + 1/4 ] *80  =   [1/2] * 80  =   40  days

OK, Nick....you can do this one

We have 1/2 the area of a 8 by 6 rectangle....what is its area ????

And we have a triangle with a base od 10 and a height of 8......what is its area  ???

Add these together.....

Thanks 

Your answer is correct, Nick....!!!!

Thanks C-phill! 

9 7 6 8 7 5

To find the mean absolute deviation of the data, start by finding the mean of the data set.

[ 9 + 7 + 6 + 8 + 7 + 5 ]  /  6  =    42 / 6  = 7

Find the absolute value of the difference between each data value and the mean: |data value – mean|.

l9 -7 l  = 2       l 6 - 7 l = 1     l 7 - 7 l = 0

l 7 - 7l = 0       l 8 - 7 l = 1     l 5  - 7 l = 2

Find the sum of the absolute values of the differences.

2 + 0 + 1 + 1 + 0 + 2  =   6

Divide the sum of the absolute values of the differences by the number of data values.

6 /6  =   

1  =   mean absolute deviation

(1/3)x = 6     multiply both sides by the reciprocal of 1/3   ⇒  3/1....this will "cancel" the fraction on the left

(3/1)(1/3)x =(3/1) (6)

x  =  18 / 1

x = 18

2 - ∑[(6*n + 2) / 4^(100 - n +1), n , 1, 100] = 200

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 \(a_{120}\).

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

solve 180 = F×(1 - 1.75^10)/(1 - 1.75) for F
First term=0.503......
∑[0.503 * 1.75^(2n - 1), n , 1, 5] =1260 / 11

63 is the answer! 

2.
Compute \(\dfrac{2 + 6}{4^{100}} + \dfrac{2 + 2 \cdot 6}{4^{99}} + \dfrac{2 + 3 \cdot 6}{4^{98}} + \dots + \dfrac{2 + 98 \cdot 6}{4^3} + \dfrac{2 + 99 \cdot 6}{4^2} + \dfrac{2 + 100 \cdot 6}{4}\)

\(\text{AP: $\quad a_n = a_1+(n-1)d,\qquad a_1=2,\ d=6 $} \)

\(\begin{array}{|rcll|} \hline a_n&=&2+(n-1)\cdot6\\\\ a_2 &=& 2+1\cdot 6 = 8 \\ a_3 &=& 2+2\cdot 6 \\ \ldots \\ a_{101} &=& 2+100\cdot 6 =602\\ \hline \end{array}\)

\(\text{GP: $\quad b_n = ar^{n-1},\qquad a=1,\ r=\dfrac{1}{4} $} \)

\(\begin{array}{|rcll|} \hline b_n &=& \left(\dfrac{1}{4}\right)^{n-1} \\\\ b_1 &=& \left(\dfrac{1}{4}\right)^{0} = 1 \\ b_2 &=& \dfrac{1}{4} \\ b_3 &=& \left(\dfrac{1}{4}\right)^{2} \\ \ldots \\ b_{101} &=& \left(\dfrac{1}{4}\right)^{100} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline s &=& \dfrac{2 + 6}{4^{100}} + \dfrac{2 + 2 \cdot 6}{4^{99}} + \dfrac{2 + 3 \cdot 6}{4^{98}} + \dots + \dfrac{2 + 98 \cdot 6}{4^3} + \dfrac{2 + 99 \cdot 6}{4^2} + \dfrac{2 + 100 \cdot 6}{4} \\ \hline \end{array}\)

\(\begin{array}{|rclll|} \hline s &=& a_2b_{101}+& a_3b_{100}+a_4b_{99}+\ldots+ a_{101}b_{2} \\ \dfrac{s}{\frac{1}{4}} &=& & a_2b_{100}+ a_3b_{99}+\ldots+ a_{100}b_{2}+a_{101}b_1 \quad | \quad b_1 = 1,\ a_{n+1}-a_n = d \\ \hline s- \dfrac{s}{\frac{1}{4}} &=& a_2b_{101}+& d(b_2+b_3+\ldots + b_{100})-a_{101} \quad | \quad a_2 = 8,\ a_{101} = 602,\ d = 6 \\ -3s &=& 8b_{101}+& 6(\underbrace{b_2+b_3+\ldots + b_{100}}_{=S~ (GP)})-602 \\ -3s &=& 8b_{101}+& 6S-602 \\\\ &&&\begin{array}{|rclll|} \hline S &=& b_2+&b_3+\ldots + b_{100} \\ \dfrac{1}{4}S &=& & b_3+\ldots + b_{100}+b_{101} \\ \hline S - \dfrac{1}{4}S &=& b_2-& b_{101} \\ \dfrac{3}{4}S &=& b_2-& b_{101} \\ S &=& \dfrac{4}{3}b_2-& \dfrac{4}{3}b_{101} \\ \hline \end{array} \\\\ -3s &=& 8b_{101}+& 6\left(\dfrac{4}{3}b_2- \dfrac{4}{3}b_{101} \right)-602 \\ -3s &=& 8b_{101}+& 8b_2- 8b_{101} -602 \\ -3s &=& & 8b_2 -602 \quad | \quad b_2 = \dfrac{1}{4} \\ -3s &=& & 2 -602 \\ -3s &=& & -600 \quad | \quad : (-3) \\ \mathbf{ s} &=& &\mathbf{ 200 } \\ \hline \end{array}\)

\(\mathbf{\dfrac{2 + 6}{4^{100}} + \dfrac{2 + 2 \cdot 6}{4^{99}} + \dfrac{2 + 3 \cdot 6}{4^{98}} + \dots + \dfrac{2 + 98 \cdot 6}{4^3} + \dfrac{2 + 99 \cdot 6}{4^2} + \dfrac{2 + 100 \cdot 6}{4} = 200}\)

Thank You heureka!

1.
The graphs of \(y = x^3 - 3x + 2 \text{ and } x + 4y = 4\) intersect in the points \((x_1,y_1), (x_2,y_2), \text{ and } (x_3,y_3)\).
If \(x_1 + x_2 + x_3 = A \text{ and } y_1 + y_2 + y_3 = B\), compute the ordered pair \((A,B)\).

\(\mathbf{\text{Vieta:}} \)

\(\begin{array}{rcll} x^3+ a_2x^2+a_1x+a_0 &=& 0 \qquad \mathbf{ a_2 =-(x_1+x_2+x_3)} \\ \end{array}\)

\(\begin{array}{|rcll|} \hline x + 4y &=& 4 \quad |\quad y = x^3 - 3x + 2 \\ x + 4(x^3 - 3x + 2) &=& 4 \\ 4x^3-11x+4 &=& 0 \quad |\quad : 4 \\ x^3-\dfrac{11}{4}x+1 &=& 0 \quad |\quad a_2 = 0~ !\\\\ \mathbf{x_1+x_2+x_3 } &=& \mathbf{0} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline x + 4y &=& 4 \\ x &=& 4-4y \\ \mathbf{x} &=& \mathbf{4(1-y)} \\\\ y &=& x^3 - 3x + 2 \quad |\quad \mathbf{x = 4(1-y)}\\ y &=& 64(1-y)^3 -3\cdot 4(1-y)+2 \\ \ldots \\ 64y^3-129y^2+181y-54 &=& 0 \quad |\quad : 64 \\ y^3-3y^2+\dfrac{181}{64}y-\dfrac{54}{64} &=& 0 \quad |\quad a_2 = -3~ !\\\\ -(y_1+y_2+y_3) &=& a_2 \\ -(y_1+y_2+y_3) &=& -3 \\ \mathbf{y_1+y_2+y_3 } &=& \mathbf{3} \\ \hline \end{array} \)

\(\mathbf{(A,B) = (0,3)}\)

What will be the discriminant for this equation 1x + 11x + 36 = 0?

Hi Guest!

\( 1x+11x+36=0\\ 12x+36=0\\ 12x=-36\\ x=-36/12\\ \color{blue}x=-3 \)

The discriminant indicates how many real solutions have an equation.

The discriminant is 1.

  !

Different approach, same result:

.