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FV = PV x [1 + R]^N

FV = 500 x [1 + 0.03]^10

FV = 500 x [  1.03  ]^10

FV = 500 x     1.34391637934412192049

FV = $672 - what Alan will have in his account after 10 years.

  3^{x + y} = 81, 81^{x - y} = 3, solve for x, y

3^(x +y) =3^4.................(1)

3^(4x - 4y) =3^1.............(2)

Equate the exponents on both equations:

(x + y) =4.......................(3)

(4x - 4y) = 1...................(4)

From (3) above:

x = 4 - y   sub this into (4) above

4(4 - y) - 4y = 1

16 - 4y - 4y = 1

16 - 8y = 1

15 - 8y = 0

15 = 8y

y = 15/8 =1 7/8, and x =17/8 =2 1/8

x*y =17/8 * 15/8 =255/64

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(a)

For what positive integers \(n\) does \(\displaystyle \left(x^2+\frac{1}{x}\right)^n\) have a nonzero constant term?

\(\begin{array}{|rcll|} \hline && \left(x^2+\dfrac{1}{x}\right)^n \\\\ &=& \left(\dfrac{x^3+1}{x}\right)^n \\\\ &=& \dfrac{1}{x^n} \cdot \left(1+ x^3 \right)^n \\\\ &=& \dfrac{1}{x^n} \cdot \left(\dbinom{n}{0} + \dbinom{n}{1}\left(x^3 \right)^1 + \dbinom{n}{2}\left(x^3 \right)^2 + \dbinom{n}{3}\left(x^3 \right)^3 + \ldots + \dbinom{n}{n}\left(x^3 \right)^n \right) \\\\ &=& \dfrac{1}{x^n} \cdot \sum \limits_{i=0}^{n} \dbinom{n}{i} \left(x^3 \right)^i \\ \hline \end{array}\)

\(\text{We have a nonzero constant, if $\\ \dfrac{1}{x^n} \cdot \dbinom{n}{i} \left(x^3 \right)^i = \dbinom{n}{i}$, $\ $ } \text{so $\left(x^3 \right)^i = x^{3i} = x^n $,$\ $ } \text{so $ \mathbf{3i = n}$ } \text{and $ i \in N $ }\)

The positive integers \(n\), which are 3 or multiples of 3 have a nonzero constant term.

\(n=3,6,9,12,\ldots.\)

 (b)

For the values of \(n\) that you found in part (a), what is that constant term? (You can leave your answer in the form of a combination.)

\(\begin{array}{|r|r|l|} \hline i & n=3i & \dbinom{n}{i} \\ \hline 1 & 3 & \dbinom{3}{1} = 3 \\ \hline 2 & 6 & \dbinom{6}{2} = 15 \\ \hline 3 & 9 & \dbinom{9}{3} = 84 \\ \hline 4 & 12 &\dbinom{12}{4} = 495 \\ \hline \ldots & \ldots & \ldots \\ \hline \end{array} \)

Determine the number of solutions in \(x\) of the congruence \(64x\equiv 2\pmod {66}\) such that \(0< x\le 100\).

\(\text{Determine the number of solutions in $x$ of the congruence $64x\equiv 2\pmod {66} \\$such that $0< x\le 100$.}\)

\(\begin{array}{|rcll|} \hline 64x &\equiv& 2\pmod {66} \quad \text{or} \quad 64x= 2 + 66n \\\\ 64x&=& 2 + 66n \quad | \quad : 2 \\ 32x&=& 1 + 33n \\\\ \mathbf{x}& \mathbf{=}& \mathbf{\dfrac{1 + 33n}{32}} \\\\ x &=& \dfrac{1 + 32n + n }{32} \\\\ x &=& \dfrac{32n+ (1 + n) }{32} \\\\ x &=& n+ \underbrace{\dfrac{1 + n}{32}}_{=a} \\\\ a &=& \dfrac{1 + n}{32} \\\\ 32a &=& 1+n \\ \mathbf{n}& \mathbf{=}& \mathbf{32a-1} \\\\ \hline \mathbf{x}& \mathbf{=}& \mathbf{\dfrac{1 + 33n}{32}} \quad | \quad \mathbf{n=32a-1} \\\\ x &=& \dfrac{1 + 33(32a-1)}{32} \\\\ x &=& \dfrac{1 + 33\cdot 32a- 33}{32} \\\\ x &=& \dfrac{-32 + 33\cdot 32a}{32} \\\\ x &=& -1 +33a \\ \mathbf{x}& \mathbf{=}& \mathbf{-1 +33a} \quad a\in N \\ \hline \end{array}\)

\(\begin{array}{|r|r|c|} \hline a & x = -1 +33a \\ \hline 1 & 32 & 0< x\le 100 & \checkmark \\ \hline 2 & 65 & 0< x\le 100 & \checkmark \\ \hline 3 & 98 & 0< x\le 100 & \checkmark \\ \hline 4 & 131 && \text{no solution } \\ \hline \gt 3 & && \text{no solution } \\ \hline \end{array} \)

\(x=32,65,98\)

Post your answer. Then someone can tell you that you are wrong.

Solved it

Solved it

If $p(x) = x^4 - 3x + 2$, then find the coefficient of the $x^3$ term in the polynomial $(p(x))^3$.

[ p(x] ^2  =   (x^4 - 3x + 2) ( x^4 - 3x + 2)  =

x^8 - 3x^5 + 2x^4

                  -3x^4  + 9x^2 - 6x

                 +2x^4              -6x    + 4

_____________________________

x^8  - 3x^5  + x^4  + 9x^2 - 12x  + 4

And

[p(x)]^2  * [p(x) ] =  [p(x)] ^3  =

 [x^8  - 3x^5  + x^4  + 9x^2 - 12x  + 4]  * [ x^4 - 3x + 2 ]

The x^3  term  will be    9x^2 * -3x   =   - 27x^3

So.....the coefficient is   -27

If $f(x)$ and $g(x)$ are polynomials such that $f(x) + g(x) = -2 + x,$ then what is $g(x)$ if $f(x) = x^3 - 2x - 2$?

We have that

f(x)  + g(x)  = x - 2     and      so

[ x^3 - 2x - 2 ]  +  g(x)   = x - 2      subtract  x^3 from both sides...add  2x, 2 to both sides

g(x)  = -x^3 + 3x 

 -4/x+1 = -1/x+5      cross  multiply

-4( x + 5)  = -1 ( x + 1)     simplify

-4x - 20  = -1x  - 1      add 1x, 20  to both sides

-3x = 19     divide both sides by -3

x  = -19 / 3

Check solution

-4/[ -19/3 + 1 ]  = -1 / [ -19/3 + 5]

-4/ [ -19/3 + 3/3]  = -1 / [ -19/3 + 15/3]

-4/ [ -16/3]  = -1 / [ -4/3]

-4*3 / -16  = -1 * 3 / -4

-12/ - 16  = -3 / -4

3/4  = 3/4

(x diamond y )  = ( 9 diamond 1)  =  √[9*1] / l 9 - 1 l   = √9 / l 8 l   =  3/8

(x diamond y )  = (4 diamond 3/8)  =  

√[ 4 * (3/8) ] / [ 4 - 3/8 l  = 

√ [12/8 ] / l 32/8 - 3/8 l = 

√ [3/2] / l 29/8 l  =

(8/29)* √ 1.5   ≈   0.3 

Thanks EP, guest and Heureka,

Here is another take, it is similar to EPs answer only with no trial and error.

It does not use trial and error.

We know that

\(7x+53=8y\)    where x is the number in each original pile and y is the number in the end piles. So obviously x and y are positive integers.

Rearranging I get

\(8y-7x=53\)

One  obvious solution is   \(x=53\;\; and \;\;y=53\)

so I can say

\(8(53 ) - 7(53 ) = 53\)

\(8(53+7t ) - 7(53 +8t) = 53 \)     this is true becasue I have simply added 56t and then taken it away again.

so

\(y=53+7t  \;\;  and\;\;   x=53+8t\)

Now you are also told that at the end you have more than 200 coins 

so\(8y>200\\ y>25\)

this means that

\(53+7t>25\\ 7t>25-53\\ 7t>-28\\ t>-4\\ \text{So lets try} \;\;  t=-3\\~\\ y=53-21=32\\ x=53-24=29\)

8*32-7*29 = 53  excellent

Initially you had \(7x\) coins.  the smallest number this could have been was \(7*29 = 203\) coins

Go online and use this Inverse Calculator: https://planetcalc.com/3311/

Enter "a" as an "integer". No matter what "n" you choose, it will always be an EVEN power of 3 and will not make any difference in the answer. Enter 9 under "modulo".

For example: a == (3^{2n}+4)^-1=(3^{2*3}+4 =3^6+4 =729 + 4 =733. Enter this under "integer" and press "calculate".

Note that we  have an adjacent side to the 31°  angle...and we are looking for the opposite side

So...the tangent seems the natural choice, here....so...

tan (31°)  = x  / 22         multiply both sides by 22

22 *  tan (31°)  = x  ≈  13.2  ft

Here's another way using similar triangles

Call the center of the larger circle, C   and the center  of the smaller circle, D

And call the point formed by the tangent and the radius at the  "top" of the smaller circle, E

Note that the vertical angles formed by the intersection of the tangent lines are equal...and if we connect the  centers of each circle, these  angles will be bisected and equal

So...we have triangles ABC  and  EBD, angle BAC  = angle BED

And angles ABC and  EBD  are also equal

Therefore.by  AA congruency...triangle  ABC  is similar to triangle EBD

Then AC / ED  = BC/BD

Then   24 / 18  = BC /BD

Then  4/3  = BC/ BD

Then CD has 7 equal parts

And each equal part is  70/7   = 10

So...BC  is 4 of these   = 40

So...by the Pythagorean Theorem, √ [ BC^2  - AC^2 ] = BC

So....√[40^2  - 24^2 ]  = √1024   = 32 cm   = BC

You forgot to multiply by 7 (the question asks for the minimal number of coins you had before you found the eighth bag, not for the minimal number of coins in each of the seven bags.

7*29=203

2)
You have seven bags of gold coins.
Each bag has the same number of gold coins.
One day, you find a bag of 53 coins.
You decide to redistribute the number of coins you have so that all eight bags you hold have the same number of coins.
You successfully manage to redistribute all the coins, and you also note that you have more than 200 coins.
What is the smallest number of coins you could have had before finding the bag of 53 coins?

Let n = number of gold coins in every of the seven bags.
Let m = number of gold coins in every of the eight bags.
Let 7n = sum of gold coins in the seven bags.
Let 8m = sum of gold coins in the eight bags.

\(\begin{array}{|rcl|cl|} \hline 7n+53 &=& 8m \\ 7n &=& -53+8m \\\\ 7n &\equiv& -53 \pmod{8} \\ 7n &\equiv& -53+7\cdot 8 \pmod{8} \\ 7n &\equiv& -53+56 \pmod{8} \\ \mathbf{7n} & \mathbf{\equiv} & \mathbf{3 \pmod{8}} \quad | \quad : 7 \\ n & \equiv & 3\cdot 7^{-1} \pmod{8} && 7^{-1} \pmod{8} \quad | \quad \gcd(7,8)=1 \\\\ && &\equiv & 7^{\phi(8)-1}\pmod{8} \quad | \quad \phi(8)= 8\cdot ( 1-\dfrac{1}{2} ) = 4 \\\\ && &\equiv & 7^{4-1} \pmod{8} \\\\ && &\equiv & 7^{3} \pmod{8} \\\\ && &\equiv & 343 \pmod{8} \\\\ && &\equiv & 7 \pmod{8} \\\\ n & \equiv & 3\cdot 7 \pmod{8} \\ n & \equiv & 21 \pmod{8} \\ n & = & 21 +z\cdot 8 \qquad z \in N \\ \hline \end{array} \)

\(\mathbf{n_{min} =\ ?}\)

\(\begin{array}{|rcll|} \hline 7n + 53 &\gt& 200 \\ 7(21 +z\cdot 8) + 53 &\gt& 200 \quad & | \quad -53\\ 7(21 +z\cdot 8) &\gt& 147 \quad & | \quad :7 \\ 21 +z\cdot 8 &\gt& 21 \quad & | \quad -21 \\ z\cdot 8 &\gt& 0 \quad & | \quad :8 \\ \mathbf{z} &\mathbf{\gt}& \mathbf{0} \\ \hline \end{array} \)

\(\text{So $z = 1$:}\)

\(\begin{array}{|rcll|} \hline n &=& 21 + 1\cdot 8 \\ n &=& 21 + 8 \\ n&=& 29 \\ \mathbf{n} &\mathbf{=}& \mathbf{29} \\ \hline \end{array}\)

The smallest number of coins you could have had before finding the bag of 53 coins is 7*29 = 203

2)
Three points are on the same line if the slope of the line through the first two points is the same as
the slope of the line through the second two points.
For what value of \(c\) are the three points
\((2,4), (6,3),\)
and
\((-5,c)\)
on the same line?

\(\text{Let $P_1=(2,4)$ }\\ \text{Let $P_2=(6,3)$ }\\ \text{Let $P_3=(-5,c)$ }\)

\(\begin{array}{|rcll|} \hline \dfrac{y_2-y_1}{x_2-x_1} &=& \dfrac{y_3-y_2}{x_3-x_2} \\\\ \dfrac{3-4}{6-2} &=& \dfrac{c-3}{-5-6} \\\ -\dfrac{1}{4} &=& -\dfrac{c-3}{11} \\\\ \dfrac{1}{4} &=& \dfrac{c-3}{11} \\\\ c-3 &=& \dfrac{11}{4} \\\\ c &=& 3+ \dfrac{11}{4} \\\\ c &=& \dfrac{12+11}{4} \\\\ c &=& \dfrac{23}{4} \\\\ \mathbf{c} &\mathbf{=}& \mathbf{5.75} \\ \hline \end{array}\)

1)

For a certain value of , the system 

\(\begin{align*} x + y + 3z &= 10, \\ -4x + 2y + 5z &= 7, \\ kx + z &= 3 \end{align*}\)

\begin{align*} x + y + 3z &= 10, \\ -4x + 2y + 5z &= 7, \\ kx + z &= 3 \end{align*}

has no solutions. What is this value of \( k \) ?

\(\text{Let $\det(A) = \begin{vmatrix} 1 & 1 & 3 \\ -4& 2& 5 & \\ k& 0& 1 \end{vmatrix}$ }\\ \text{Let $b = \begin{pmatrix} 10 \\ 7 \\ 3 \end{pmatrix}$ }\\ \text{If $\det(A)=0$, then the linear system $Ax=b, b\ne0$, has no solution. } \)

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

a)

Compute \(10^{-1} \pmod {1001}\). 

Express your answer as a residue from  0 to 1000,inclusive.

1. Solution

\(\small{ \begin{array}{|rcll|} \hline && 10^{-1} \pmod {1001} \quad | \quad \gcd(10,1001)=1 \\\\ &\equiv & 10^{\phi(1001)-1} \pmod {1001} \quad | \quad \phi(1001)= 1001\cdot(1-\dfrac17)\cdot(1-\dfrac{1}{11})\cdot(1-\dfrac{1}{13})=720 \\\\ &\equiv & 10^{720-1} \pmod{1001} \\ &\equiv & 10^{719} \pmod{1001} \\ &\equiv & 10^{3*239+2} \pmod{1001} \\ &\equiv & 10^{3*239}10^2 \pmod{1001} \\ &\equiv & \left(10^{3}\right)^{239}10^2 \pmod{1001} \quad | \quad 10^3 \equiv -1 \pmod{1001} \\ &\equiv & (-1)^{239}10^2 \pmod{1001} \\ &\equiv & -10^2 \pmod{1001} \\ &\equiv & -100 \pmod{1001} \\ &\equiv & -100+1001 \pmod{1001} \\ & \mathbf{\equiv} & \mathbf{901 \pmod{1001}} \\ \hline \end{array} }\)

2. Solution

\(\begin{array}{rcll} \text{Let} \\ & 10\cdot 10^{-1} \equiv 1 \pmod{1001} \\ \end{array}\)

\(\begin{array}{rcll} \text{Let} \\ & 10\cdot 100 = 1000 \equiv -1 \pmod {1001} \\ \end{array}\)

\(\begin{array}{llcll} \text{square this equation: } \\ & (10\cdot 100)(10\cdot 100) &\equiv& (-1)(-1) \pmod{1001} \\ & (10) (100^2\cdot 10) &\equiv& 1 \pmod{1001} \\ & (10) (100000) &\equiv& 1 \pmod{1001} \quad | \quad 100000 \equiv 901 \pmod{1001} \\ & (10)\underbrace{(901)}_{=(10)^{-1}} &\equiv& 1 \pmod{1001} \\ \end{array}\)

\(\text{So $10^{-1}=\boxed{901}$ is the multiplicative inverse to $10$ modulo $1001$}.\)