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A sequence satisifes \(a_1 = 3\) and \(a_{n + 1} = \dfrac{a_n}{a_n + 1}\) for \(n \geq 1\). 

What is \( a_{14}\)?

\(\begin{array}{|lrcll|} \hline & \mathbf{a_{n + 1}} &=& \mathbf{\dfrac{a_n}{a_n + 1}} \\\\ & \dfrac{1}{a_{n + 1}} &=& \dfrac{a_n + 1}{a_n} \\\\ & \mathbf{\dfrac{1}{a_{n + 1}}} &=& \mathbf{1+ \dfrac{1}{a_n}} \\\\ \hline \\ & \dfrac{1}{a_{n + 2}} &=& 1+ \dfrac{1}{a_{n+1}} \quad | \quad \mathbf{\dfrac{1}{a_{n + 1}}=1+ \dfrac{1}{a_n}} \\\\ & \dfrac{1}{a_{n + 2}} &=& 1+ 1+ \dfrac{1}{a_n} \\\\ & \mathbf{\dfrac{1}{a_{n + 2}}} &=& \mathbf{2+ \dfrac{1}{a_n}} \\\\ \hline \\ & \dfrac{1}{a_{n + 3}} &=& 1+ \dfrac{1}{a_{n+2}} \quad | \quad \mathbf{\dfrac{1}{a_{n + 2}}=2+ \dfrac{1}{a_n}} \\\\ & \dfrac{1}{a_{n + 3}} &=& 1+ 2+ \dfrac{1}{a_n} \\\\ & \mathbf{\dfrac{1}{a_{n + 3}}} &=& \mathbf{3+ \dfrac{1}{a_n}} \\\\ \hline \\ & \dfrac{1}{a_{n + 4}} &=& 1+ \dfrac{1}{a_{n+3}} \quad | \quad \mathbf{\dfrac{1}{a_{n + 3}}=3+ \dfrac{1}{a_n}} \\\\ & \dfrac{1}{a_{n + 4}} &=& 1+ 3+ \dfrac{1}{a_n} \\\\ & \mathbf{\dfrac{1}{a_{n + 4}}} &=& \mathbf{4+ \dfrac{1}{a_n}} \\ \hline & \ldots \\ \hline \\ & \mathbf{\dfrac{1}{a_{n + m}}} &=& \mathbf{m+ \dfrac{1}{a_n}} \\\\ n=1,\ m=13: & \mathbf{\dfrac{1}{a_{1 + 13}}} &=& \mathbf{13+ \dfrac{1}{a_1}} \\\\ & \dfrac{1}{a_{14}} &=& 13+ \dfrac{1}{a_1} \quad | \quad a_1 =3 \\\\ & \dfrac{1}{a_{14}} &=& 13+ \dfrac{1}{3} \\\\ & \dfrac{1}{a_{14}} &=& \dfrac{3*13+1}{3} \\\\ & \dfrac{1}{a_{14}} &=& \dfrac{40}{3} \\\\ & \mathbf{ a_{14} } &=& \mathbf{\dfrac{3}{40} } \\ \hline \end{array}\)

Given the functions \(f(x)=\dfrac{x+5}{3}\) and \(g(x)=\dfrac{1}{f^{-1}(x)+1}\), find the value of g(3).

Hello EpicWater!

\(f(x)=\dfrac{x+5}{3}\\\color{blue} f^{-1}(x)=\frac{3}{x+5}\) 

\(\color{BrickRed} g(x)=\dfrac{1}{f^{-1}(x)+1}\\ g(x)=\frac{1}{\frac{3}{x+5}+1}\\ \color{BrickRed}x=3\\ g(3)=\frac{1}{\frac{3}{3+5}+1}= \frac{1}{\frac{3}{8}+1}=\frac{1}{\frac{11}{8}}\)

\(g(3)=\frac{8}{11}\)

  !

The function \(f(x)=x^2+ax+2b-a+3\) has distinct real root when the discriminant of the quadratic expression \(x^2+ax+(2b-a+3)\) is positive. That is when \(a^2-4(2b-a+3)>0\); this translates into

\(a^2-4a-12-8b>0\). This inequality must hold for any value of a, and so it must hold for the minimum value of the quadratic expression \(a^2-4a-12\) , which is equal to - 16 attained at a = - 2 ( think of the vertex of of the parabola defined by the quadratic). So b has to be an integer that results in \(-16-8b>0\). Solving this inequality we get \(b<-2\) and the largest integer smaller than - 2 is - 3.

\((x-b)(x-a)+(x-b)(x-c)=(x-b)(2x-a-c)=0\)gives you roots \(x=b\ and\ x= \frac{a+c}{2} \), so sum of the roots would be \(b+ \frac{a+c}{2} \). This is a linear function of a, b, and c that needs to be maximized under certain constraints that we can write as

\(-9\leq a \leq 9\),

\(-9 \leq b \leq 9\), and

\(-9 \leq c \leq 9\),

since a, b, and c are one-digit integers. This is essentially a linear programming problem and it is known that the maximum of a linear function of a, b, and c , would occur at one of the corner points of the feasible region, in this case described by the above inequalities. The region is a cube with corner points (-9, -9, -9), (-9, -9, 9), (-9, 9, -9), (9, -9, -9), (9, 9, -9), (9, -9, 9), (-9, 9, 9), and (9, 9, 9). It is clear that the max value would be at (9, 9, 9), but we are looking for distinct integers. So we take the next largest positive integers, that is 7, 8, and 9. The largest value of \(b+ \frac{a+c}{2} \) is attained for a=7, b=9 and c=8, or a = 8, b = 9, and c = 7 and it is \( \frac{15}{2} +9=16.5\).

In convex quadrialteral ABCD, the area of triangle DAB is 1, that of triangle ABC is 6, and that of triangle CDA is 2. 

Find the area of triangle BCD.

\(\begin{array}{|rcll|} \hline \mathbf{CDA + ABC} &=& \mathbf{DAB + BCD} \\ 2 + 6 &=& 1 + BCD \\ 8 &=& 1 + BCD \\ BCD &=& 8-1 \\ \mathbf{BCD} &=& \mathbf{7} \\ \hline \end{array}\)

\(\boxed{2nd}\quad \boxed{atan}\quad \text{your number }=\)

arctan is the same as inverse tan.

ALSO

How many ways are there to put 9 differently colored beads on a grid if the purple bead and the green bead cannot be adjacent (either horizontally, vertically, or diagonally), and rotations and reflections of the grid are considered the same?

A grid of 3 dots by 3 dots.      9 colours but pink and green cannot be next to each other.

I really do not know but here is my 'guess' 

So there are 7 colours that can go into the middle.

Rotations are the same just put a green in a corner

You cannot have a pink next to it but you cannot have a pink on the adjacent corners either because it it is rotated they will become diagonally adjacent,

So

I get   7*1*6*5*4*3 *3*2*1 = 15120

There are 4 axes of symmetry so divide by 2^4=16

15120/16 = 945

I doubt that it is correct though.

If you know what the number answer is then please share that knowledge with the forum.

For all real numbers x and y, the function f satisfies f(x + y) = f(x) + f(y).  If f(1) = 3, then find f(10).

\(\begin{array}{|lrcll|} \hline & \mathbf{f(x+y)} &=& \mathbf{f(x) + f(y)} \\\\ x=1,y=1: & f(1+1) &=& f(1) + f(1) \\ & \mathbf{f(2)} &=& \mathbf{2f(1)} \\\\ x=1,y=2: & f(1+2) &=& f(1) + f(2) \\ & f(3) &=& f(1) + 2f(1) \\ & \mathbf{f(3)} &=& \mathbf{3f(1)} \\\\ x=1,y=3: & f(1+3) &=& f(1) + f(3) \\ & f(4) &=& f(1) + 3f(1) \\ & \mathbf{f(4)} &=& \mathbf{4f(1)} \\\\ \ldots \\ & \mathbf{f(n)} &=& \mathbf{nf(1)} \\\\ \hline n=10: & f(10) &=& 10f(1) \quad | \quad f(1) =3 \\ & f(10) &=& 10*3 \\ & \mathbf{f(10)} &=& \mathbf{30} \\ \hline \end{array} \)

Find the value of x that satisfies \(\log_3 (x) =\Big(-2 + \log_2 (100) \Big)\Big(\log_3 (\sqrt{2}) \Big) \)

\(\begin{array}{|rcll|} \hline \log_3 (x) &=& \Big(-2 + \log_2 (100) \Big)\Big(\log_3 (\sqrt{2}) \Big) \\ \log_3 (x) &=& \Big(\log_2 (100) -2 \Big)\Big(\log_3 (\sqrt{2}) \Big) \\ \log_3 (x) &=& \Big(\log_2 (10^{2}) -2 \Big)\Big(\log_3 (2^{\frac{1}{2}}) \Big) \\ \log_3 (x) &=& \Big(2\log_2 (10) -2 \Big)\Big(\dfrac{1}{2}\log_3 (2) \Big) \\ \log_3 (x) &=& \Big(\log_2 (10) -1 \Big)\dfrac{2}{2}\Big(\log_3 (2) \Big) \\ \log_3 (x) &=& \Big(\log_2 (10) -1 \Big)\Big(\log_3 (2) \Big) \quad | \quad \log_2(2) = 1 \\ \log_3 (x) &=& \Big(\log_2 (10) -\log_2(2) \Big)\Big(\log_3 (2) \Big) \\ && \boxed{\log\left( \dfrac{a}{b} \right) = \log(a)-\log(b) } \\ \log_3 (x) &=& \Big(\log_2\left( \dfrac{10}{2} \right) \Big)\Big(\log_3 (2) \Big) \\ \log_3 (x) &=& \Big(\log_2(5)\Big)\Big(\log_3 (2) \Big) \quad | \quad \\ && \boxed{\log_a(x)=\dfrac{\log_b(x)} {\log_b(a)}\qquad \log_2(5)=\dfrac{\log_3(5)} {\log_3(2)} } \\ \log_3 (x) &=& \dfrac{\log_3(5)} {\log_3(2)} \log_3 (2) \\ \log_3 (x) &=& \log_3(5) \\ \mathbf{x} &=& \mathbf{5} \\ \hline \end{array}\)

I assume you do not know how to do it but do you know what the answer is? 

Or do you have any information that could help answerers.

Please answer either way.   

A sequence goes 1,2,2,3,3,3,4,4,4,4,....   What is the 254th term?

\(\begin{array}{|rcll|} \hline \dfrac{(1+n)n}{2} &=& 254 \\\\ n^2+n &=& 508 \\\\ n^2+n-508 &=& 0 \\ n &=& \dfrac{-1 \pm \sqrt{1-4*(-508)}}{2} \\ n &=& \dfrac{-1 \pm \sqrt{1+2032}}{2} \\ n &=& \dfrac{-1 \pm \sqrt{2033}}{2} \\ n &=& \dfrac{-1 + \sqrt{2033}}{2} \\ n &=& \dfrac{-1 + 45.0888012704}{2} \\ n &=& \dfrac{44.0888012704}{2} \\ n &=& 22.0444006352 \\ \hline \end{array}\)

The 254th term is 23

1. The statement of this problem is not very clear. Is k a constant? or k is another function and g(x) =(kf)(x), the product of the functions k and f? If k is a constant, then the only value of k that makes g(x)=kf(x) not invertible is k = 0, and it makes no sense to ask for the sum of all possible values of k since there is only one value, that is 0. If k is a nonzero constant, then g always has an inverse; in fact \(g^{-1}(x)=f^{-1}(\frac{x}{k})\). For instance, we know that if \(f(x)=x^3\), then \(f^{-1}(x)=\sqrt[3] {x}\). The inverse of \(g(x)=kx^3\) is just \(g^{-1}(x)= \sqrt [3]{\frac{x}{k} }\), assuming k is not equal to zero.

If on the other hand k is a function and g(x) is the product of f and k, again the question makes no sense since there are lots of functions that when multiplied by f give a function that is not invertible. Two examples: \(g(x)=x^3\cdot \frac{1}{x} \), where \(f(x)=x^3\ and \ k(x)= \frac{1}{x} \); and \(g(x)=x^3 \cdot \frac{x+1}{x} \), where \(f(x)=x^3\) and \(k(x)= \frac{x+1}{x} \). In both cases g is not invertible because it is not 1-1.

2. Here I will assume you mean \(f(x)= \frac{ax+b}{cx+d} \). The answer to this question is: \(a+d=0\). There is quite a bit of Algebra involved in showing this and it starts by setting \(f(f(x))=x\), which means f is its own inverse. We have

\(f(f(x))=f(\frac{ax+b}{cx+d} )= \frac{a(\frac{ax+b}{cx+d})+b }{c(\frac{ax+b}{cx+d})+d } =\frac{a(ax+b)+b(cx+d)}{c(ax+b)+d(cx+d)} =\frac{x}{1} \);

If you cross-multiply and combine terms you get,

\((a^2+bc)x+(ab+bd)=(ca+dc)x^2+(cb+d^2)x\).

Setting the coefficients of the polynomials on the two sides of the equation equal, we get

\(ab+bd=0\), or \(b(a+d)=0\),

\(ca+dc=0\), or \(c(a+d)=0\)

\(a^2+bc=bc+d^2\), or \(a^2-d^2=0\).

The first two equations tell us that \(a+d=0\), since neither b nor c could be zero (one of the assumptions is that \(abcd \neq 0\))

I did it in a circle and I realize now that this is not right. You want it in a grid.

But you have not said what shape the grid is. I suppose it is supposed to be a 3 by 3 grid but this has not been stated.

So I might think about it again......

Since area of DAB, ABC, and CDA are 1, 6, and 2, respectively, we have

\(a+b=1\),

\(b+c=6\), and

\(a+d=2\).

We need to find the area of BCD, that is \(c+d\).

Add equations two and three to get

\(a+b+c+d=8\).

Subtract equation one to get

\(c+d=8-1=7\)

\(T=8,\ E=5,\ and\ N=0; \) so

\(TEN=850\).

\(F=2,\ O=9,\ R=7,\ and\ Y=6\); so

\(FORTY=29786\).

\(S=3,\ I=1,\ and\ X=4;\) so

\(SIXTY=31486\). Check to make sure \(850+850+29786=31486\).

One major key  to this puzzle is the possible carry over from the ones digits and the tens digits: that gave me the clue that one of the two letters N and E must be 0 and the other 5, and E cannot be 0 since then the carry over from the ones digits would make \(1+T=T\), which cannot be. The second clue came from the first two digits, from the left, of \(FORTY\) and \(SIXTY. \) I reasoned that O must be a very large digit, either 8 or 9, since the carry over from the hundreds digits could be at most 2. But 2 and 8 make 10 and I had already assigned 0; so I decided O must be 9 and I = 1, and S = F+1. The carry over from the hundreds digit being 2 forces \(1+T+T+R=2X.\) So T and R must be large digits and X a rather small digit. The process of trial and error did the rest of the work. Thanks for the puzzle-I had quite a lot of fun.

Solve the puzzle FORTY + TEN + TEN = SIXTY.  (Every letter stands for a different digit.)

FORTY + TEN + TEN = SIXTY
29786 + 850 + 850 = 31486

As EP  pointed out, it will fit diagonally...

Let the bottom left vertex of the square   be located at  (0,0)

And the two  vertexes  of the short side of the rectangle  can be  located at  (0, 10/√2)  and  ( 10/√2, 0)

And let the other two  vertexes  be located at   (40 - 10/√2, 40)  and ( 40, 40 - 10/√2)

And  the distance  between     (10/√2, 0 )   and  ( 40 , 40 - 10/√2)  =

√ [ (40 - 10/√2)^2   + (40 - 10/√2)^2 ]  =   √2 * (40 - 10/√2)  = 40√2 - 10   ≈  46.56  

So.....the rectangle will fit

Very nice, chenxander   !!!!!

Scroll down for the answer.

Since the three products are the same, it is impossible for one product to have a factor that another does not. 

Therefore, 7 and 5 don't work because they are coprime (don't share any factors) with every other number on the list.

Of course, 0 doesn't work because then one or two products will be zero while the other is a positive number.

The remaining numbers, 1, 2, 3, 4, 6, 8, and 9 , are the possible digits. 

Notice that every number only has factors of 2 and 3. So, separate the numbers into two groups depending on their factors:

factors of 2: (2,4,6,8)

factors of three: (3,6,9)

There are a total of seven factors of 2 and four factors of 3.

This is somewhat hard to explain, but this means that you must put a factor of 2 as G because otherwise one product will always have one more factor of two than another. Since G belongs to only one product, putting a factor of two as G allows the remaining six factors of 2 to be split equally between AxBxC and DxExF.

If you aren't able to completely follow along below, it's fine, as I am just guessing-and-checking. Just scroll down some more.

Try setting 2 as G. Since 8 has three factors and 2 combined with 4 has three factors, set 8 as A and D,F as 4 and 6 respectively.

You will find that this doesn't work. Instead of putting a 2 on the side, put a six as F. Then, set B as 9 and G as 3. This will balance out the factors of three evenly. 

Finally, put one as C and two as G.

My final configuration is (there are many, put they all have 2 as G):

8      3

9  2  4

1      6

Answer: The value of G is 2.

The strategy here is to find the boundary of the graph and then figure out what the actual graph is. First, plot some points for |x| + |y| = 5 on the Cartersian plane; (5,0), (4,1), (3,2), and combinations of their opposites work as well (ex. -4,1).

The graph of |x| + |y| = 5 becomes a diamond centered at the origin with perpendicular diagonals of length 10 on the x and y-axis. 

If you think about it, |x| + |y| < 5 must be all of the area within the diamond.

At this point, you can count the number of lattice points by drawing a diagram. Be careful, as the inequality sign is less than, not less than or equal to, so do not count the lattice points on the diamond itself.

By counting with symmetry, you get

6*4 + 4*4 + 1 = 41. 

The 6*4 is for the points between the axes, the 4*4 is the points on the axes (excluding the origin), and the 1 is for the origin.

 good luck

The answer for number 3 is not 11

See the graph here :  https://www.desmos.com/calculator/9riuxf41mb

We have 16 lattice points above and below the x axis and 9 points on the x axis....so....the total number of lattice points  =

16 + 16  + 9   =

41

23 mod N =A
44 mod N =A
N = 44 - 23 = 21 - The largest N
23 mod 21 =2
44 mod 21 =2

(1,1), (3,5), and (-1,4)

There will  be three possibilities

We can find each by adding  the coordinates of any two of the points  and subtracting the  coordinates of the remaining point

So  we  have

(1, 1) + ( 3,5)  - ( -1, 4)  =   ( 5, 2)

(1, 1) + ( -1, 4)  - (3,5)  =  (-3, 0)

(3,5) + ( -1, 4) - (1,1)  =  ( 1, 8)

Here's a pic :   https://www.desmos.com/calculator/jvrousklbs