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Some wack geometry problem with circles. I manually drew an SVG that looks like what they give you. For the solution, I had to learn about the assumptions to make from someone else because it is not obvious.
There are some problems with this. You are supposed to make some assumptions that are supposedly "obvious" but I would argue they are not. Those are that the outer box is a square, and the 2 circles at the corners have radius equal to half of the side length of the square. Anyway, let's label some things.
Now we can write some equations. For the right triangle, we have \(r^2+(2r-5)^2=(r+5)^2\). With some algebra, this simplifies to \(2r(2r-15)=0\) so \(r=15/2\) because \(r=0\) is physically impossible. Also observe that \(\cos(\theta)=\frac{2r-5}{r+5}=4/5\). We will use this in the next part. Now we can draw a different triangle, and I will use up a bunch of space for another SVG.
Now it is time for some law of cosines.
\[\begin{align} &\left(x+\frac{15}{2}\right)^2=(x+5)^2+\left(\frac{25}{2}\right)^2 -2(x+5)\left(\frac{25}{2}\right)\cos(\theta)\\ &\cancel{x^2}+15x+\frac{225}{4} =\cancel{x^2}+10x+25+\frac{625}{4}-25(x+5)\frac{4}{5}\\ &15x=10x+25+\frac{400}{4}-20x-100 \Rightarrow 25x=25\\ \end{align}\]So there you have it. The answer is \(x=1\).
We'll evaluate the inner summation first. By partial fractions, we solve:
\[\begin{align} \frac{A}{2n-1}+\frac{B}{2n+1}&=\frac{1}{(2n-1)(2n+1)}\\ A(2n+1)+B(2n-1)&=1\\ (2A+2B)n+(A-B)&=1\\ \Rightarrow A=\frac{1}{2},B=-\frac{1}{2} \end{align}\]Now we rewrite the inner summation:
\[\begin{align} &\sum_{n=1}^\infty\frac{1}{(2n-1)(2n+1)}=\sum_{n=1}^\infty\left( \frac{1}{4n-2}-\frac{1}{4n+2}\right)\\ &=\left(\frac{1}{2}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{10}\right) +\left(\frac{1}{10}-\frac{1}{14}\right)+\left(\frac{1}{14}-\frac{1}{18}\right) +\ldots\\ &=\lim_{n\to\infty}\left(\frac{1}{2}-\frac{1}{4n+2}\right)=\frac{1}{2} \end{align}\]Then we take the \(-1\)th power of this to get the answer: \(2\).
If we are concerned with only \(x\in\mathbb{R}\), then the problem is simple. The left side \(4^x+5^x+6^x\) is a differentiable and strictly increasing function (because the derivative is strictly positive everywhere), so there can only be one solution if it exists. Using a little "big brain" reasoning, we try \(x=3\) and find that it works. This is the only solution for \(x\in\mathbb{R}\). If there were 2 solutions, then the mean value theorem would require the left side has a 0 derivative somewhere, which cannot happen.
But the problem never specified the domain. What if \(x\in\mathbb{C}\)? The problem can get a whole lot harder. The function is holomorphic so we can apply Newton's method to find solutions of \(4^x+5^x+6^x-405=0\). I wrote the following Python code to utilize floating point operations on complex numbers for implementing Newton's method:
import sys
from cmath import log,exp
#f = lambda x: 4**x + 5**x + 6**x - 405
#fp = lambda x: (4**x)*log(4) + (5**x)*log(5) + (6**x)*log(6)
f = lambda x: exp(x*log(4)) + exp(x*log(5)) + exp(x*log(6)) - 405
fp = lambda x: exp(x*log(4))*log(4) + exp(x*log(5))*log(5) + exp(x*log(6))*log(6)
x = float(sys.argv[1])
y = float(sys.argv[2])
z0 = x + y*1j
print(z0)
z = z0
i = 0
while True:
i += 1
zz = z - f(z)/fp(z)
print(f'iter {i} {zz} {abs(zz-z)/abs(z)}')
if abs(zz-z)/abs(z) < 2**-50:
break
z = zz
print(z,f(z))
By playing around with different values for the initial point, I was able to find the following estimates of some other zeroes:
Since \(b^{x+iy}=e^{\ln(b)(x+iy)}=e^{x\ln(b)}e^{iy\ln(b)}\), the complex coordinate of the start point only affects angles of each exponent term, not magnitude. Based on my experimentation, it looks like all these solutions may have a real part a bit bigger than 3, and varying imaginary parts. The Newton's method iteration seemed to fail when trying near any other part of the complex plane. This result is entirely experimental and not a proof of any sort. It seems plausible that we may be able to choose almost any \(y\) value that distributes the angles around just right so we can rescale by adjusting \(x\) so we get a solution, but it is hard to say. Newton's method suggests there are many more zeroes if we extend solutions to the complex plane.
Now we focus on the summation on the exponent.
\[\begin{align} &\sum_{n=1}^\infty\frac{n}{2^n}=\sum_{n=1}^\infty\sum_{m=1}^n\frac{1}{2^n} \end{align}\]Now we show the terms in a table to describe how to rewrite the summation.
\[\begin{array}{c|ccccc} &m=1&m=2&m=3&\ldots\\\hline n=1&1/2\\ n=2&1/4&1/4\\ n=3&1/8&1/8&1/8\\ \vdots&\vdots&&&\ddots \end{array}\]The double summation above has the outer summation go over the rows. We will rewrite it with an outer summation over the columns which gives us a way to use geometric series.
\[\begin{align} &\sum_{n=1}^\infty\sum_{m=1}^n\frac{1}{2^n} =\sum_{m=1}^\infty\sum_{n=m}^\infty\frac{1}{2^n} =\sum_{m=1}^\infty\sum_{n=0}^\infty\frac{1}{2^{n+m}} =\sum_{m=1}^\infty\frac{1}{2^m}\sum_{n=0}^\infty\frac{1}{2^n}\\ &=\sum_{m=1}^\infty\frac{1}{2^m}\frac{1}{1-\frac{1}{2}} =2\sum_{m=1}^\infty\frac{1}{2^m} =2\left(\frac{1}{1-\frac{1}{2}}-1\right)=2\\ \end{align}\]So plug in this 2 for the exponent and the answer is \(2^2=4\).
Someone pointed out to me that the series summation can be solved with calculus instead of a clever double summation trick. We have this geometric series when \(|x|<1\).
\[\frac{1}{1-x}=\sum_{n=0}^\infty x^n=1+x+x^2+x^3+\ldots\]Then we can take the derivative of both sides:
\[\frac{1}{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}=1+2x+3x^2+\ldots\]So if we multiply this summation by \(x\) to get the exponent to be \(n\) instead of \(n-1\), we have:
\[\frac{x}{(1-x)^2}=\sum_{n=1}^\infty nx^n\]Now just plug in \(x=1/2\) so it matches the series we came up with before and we will find that the summation is \(2\).
The "big brain" reasoning shows us that \(x=5\) is a solution. Let's see if the derivative of the left side changes sign:
\[\frac{4}{\ln(5)(2x-5)}-\frac{1}{\ln(4)(x-1)} =\frac{4\ln(4)(x-1)-\ln(5)(2x-5)}{\ln(5)\ln(4)(2x-5)(x-1)}\]The function is only defined for \(x>5/2\) so the bottom is always positive, which means we need to analyze the sign of the top:
\[(4\ln(4)-2\ln(5))x+(5\ln(5)-4\ln(4))\]The slope is \(4\ln(4)-2\ln(5)\approx2.326\) and the zero is \(x\approx-1.076\). Therefore, this function is always increasing on its entire domain, so there can be no other solution.
What about a more satisfying solution? A friend suggested ignoring the logs and thinking of it as \(2y-z=1\). If \(y=z=1\) then it would be solved. It turns out that the single choice of \(x=5\) makes \(y=z=1\). This still isn't the most satisfying solution though. WolframAlpha does not show a clever way to solve it, instead showing the solution as a result of plots.
So is there a more satisfying way to solve this that doesn't exactly end up being a form of "guess and check"?
In base 10, how many distinct digits can be the last digit of a perfect square?
Consider the numbers modulo 10. If \(n=10k+d\) for some \(k\) and \(0\leq d<10\), then \(n^2=100k^2+20kd+d^2\equiv d^2\:(\text{mod}\:10)\). So the squares of single digits determine what the last digits of a perfect square can be. The squares of the single digits are 0,1,4,9,16,25,36,64,81. Taking these modulo 10, the unique digits are 0,1,4,5,6,9 so there are 6.
How many frieze patterns are there?
A frieze patern repeats in one direction in 2 dimensions. There are 7 types classified by their symmetries, as stated on Wikipedia.
I probably need to understand more group theory to give a better explanation about this stuff. The Wikipedia page shows examples in a table, including Conway's names for them describing the symmetries with feet.
First use logs to find \(y\ln(2)=\ln(9)=2\ln(3) \Rightarrow y=\frac{2\ln(3)}{\ln(2)}\) and \(z\ln(3)=\ln(16)=4\ln(2) \Rightarrow z=\frac{4\ln(2)}{\ln(3)}\). Then
\[x=yz=\left(\frac{2\ln(3)}{\ln(2)}\right)\left(\frac{4\ln(2)}{\ln(3)}\right) =8\]What is the order of magnitude difference between a thousand and a trillion?
Normally when talking about order of magnitude, we mean in base 10. We can about order of magnitude mathematically with base 10 logarithms. So a thousand is \(10^3\) and a trillion is \(10^{12}\). The base 10 logarithms of these are 3 and 12 respectively so the difference is 9.
The only assumption we have to make is that this thing is in a square. It may not be 100% obvious because the bottom edge is not labeled as equal length like the others with side length 8, and the top right corner is not labeled as a right angle. Now let's draw a right triangle and label some things.
There are 3 right triangles. Let's call them "left" (with the green sides), "middle" (with the right sides), and "right" (with the bottom right right angle). For the right one, we have side lengths 4 and 8 (from the right side of the square) so the remaining side length is \(\sqrt{4^2+8^2}=\sqrt{80}\). Now for the middle triangle, we have hypotenuse \(\sqrt{80}\) and a side of length 4 so the remaining side is \(\sqrt{80-4^2}=8\). Now we know that \(x=8+y\) so we must solve for \(y\). Let's focus on different triangles next:
The 2 blue sides are length 4 and they share the long green side. This means they are similar right triangles so the 2 shorter green sides are the same length. Now we can write out an equation for the right triangle formed with \(x\), the top edge (8), and the left side \(8-y\).
\[\begin{align} &x^2=8^2+(8-y)^2\Rightarrow (8+y)^2=8^2+(8^2-16y+y^2)\\ &\Rightarrow(\cancel{8^2}+16y+\cancel{y^2})=\cancel{8^2}+8^2-16y+\cancel{y^2} \Rightarrow 32y=64\\ \end{align}\]So there we get \(y=2\) which means \(x=8+y=10\).
Find the area bounded by \(y=\frac{4}{9}x+3,x=3\) and the axes.
On the \(x\) axis, we have \(x=0\) and \(x=3\) as bounds, so we can integrate on that interval. We have \(y=0\) on the \(y\) axis, and a function that is positive on our entire interval \(x\in[0,3]\). One way to solve this is by integration.
\[\begin{align} &\int_0^3\left(\frac{4}{9}x+3\right)dx=\left[\frac{2}{9}x^2+3x\right]_0^3 =\frac{2}{9}(3)^2+3(3)=2+9=11 \end{align}\]Alternatively, we can use geometry. The plot looks like this:
So we have a trapezoid with a base length of \(3\), left height of \(3\), and right height of \(\frac{4}{3}+3=\frac{13}{3}\). Using the trapezoid area formula:
\[\frac{1}{2}(3)\left(3+\frac{13}{3}\right)=\frac{3}{2}\frac{22}{3}=11\]Find the product of eigenvalues of \(\begin{bmatrix}1&2&3\\2&1&3\\3&2&1\\\end{bmatrix}\)
One way is to compute the eigenvalues directly. The characteristic polynomial is:
\[\begin{align} &\begin{vmatrix}1-\lambda&2&3\\2&1-\lambda&3\\3&2&1-\lambda\\\end{vmatrix} =(1-\lambda)\begin{vmatrix}1-\lambda&3\\2&1-\lambda\\\end{vmatrix} -2\begin{vmatrix}2&3\\3&1-\lambda\\\end{vmatrix} +3\begin{vmatrix}2&1-\lambda\\3&2\\\end{vmatrix}\\ &=(1-\lambda)[(1-\lambda)^2-6]-2[2(1-\lambda)-9]+3[4-3(1-\lambda)]\\ &=(1-\lambda)^3-6(1-\lambda)-4(1-\lambda)+18+12-9(1-\lambda)\\ &=(1-\lambda)^3-19(1-\lambda)+30=-\lambda^3+3\lambda^2+16\lambda+12 \end{align}\]We could factor this in a logical way, or we can try something dumb I came up with. Let's group the cubic+linear and quadratic+constant terms:
\[-\lambda(\lambda^2-16)+3(\lambda^2+4)\]Now let's adjust the constants in the parenthesis to each be -4 so we have a common factor of \(\lambda^2-4\).
\[-\lambda(\lambda^2-4-12)+3(\lambda^2-4+8) =-\lambda(\lambda^2-4)+12\lambda+3(\lambda^2-4)+24\]Ok now we factor out \(\lambda^2-4\) and simplify the \(12\lambda+24\).
\[(-\lambda+3)(\lambda^2-4)+12(\lambda+2)\]Look at that, \(\lambda+2\) is now a factor of both parts. So let's factor it even more.
\[(-\lambda+3)(\lambda+2)(\lambda-2)+12(\lambda+2) =(\lambda+2)[(-\lambda+3)(\lambda-2)+12]\]Now that part in brackets can be expanded to another quadratic polynomial and those are easier to factor.
\[(\lambda+2)[-\lambda^2+5\lambda+6]=-(\lambda+2)(\lambda+1)(\lambda-6)\]So there, the eigenvalues are \(-1,-2,6\) and the product is \(12\). That was a bit painful to factor. Instead, since we only need to know the product and not the individual eigenvalues, we can use a theorem from linear algebra which is that the determinant is equal to the product of the eigenvalues. So we just need to find the determinant:
\[\begin{align} &\begin{vmatrix}1&2&3\\2&1&3\\3&2&1\\\end{vmatrix} =1\begin{vmatrix}1&3\\2&1\\\end{vmatrix} -2\begin{vmatrix}2&3\\3&1\\\end{vmatrix} +3\begin{vmatrix}2&1\\3&2\\\end{vmatrix}\\ &=(1-6)-2(2-9)+3(4-3)=-5+14+3=12\\ \end{align}\]Find the median of \(\{15,25,0,7,19,4,11,21\}\).
Today is an easy one. We first sort the list in order: \(\{0,4,7,11,15,19,21,25\}\). Then we pick the item in the middle, but since this list is even length, we use the average of the 2 items in the middle. That is, \(\frac{11+15}{2}=13\).
The first way I came up with to solve this is really stupid because the algebra becomes disgusting. We use Heron's formula. The area for the full triangle is:
\[A=\sqrt{s(s-7)(s-9)(s-x)},s=\frac{7+9+x}{2}=8+\frac{x}{2}\]This can be simplified somewhat with algebra. I won't go into all the details because I don't feel like typing all the disgusting algebra.
\[4A=\sqrt{(256-x^2)(x^2-4)}\]This area is equal to the areas of the 2 smaller triangles added together. Let's find formulas for those areas similarly. Again, I don't feel like writing out all the work, but these are the formulas I obtained:
\[16A_1=\sqrt{(676-x^2)(x^2-100)},16A_2=\sqrt{(484-x^2)(x^2-36)}\]So now we know that \(A=A_1+A_2 \Rightarrow A^2=A_1^2+A_2^2+2A_1A_2\). Since we are squaring something with square roots, we may introduce extraneous solutions. I'm going to multiply each side by 256 to make it a little easier to plug in the formulas above, so we have \(256A^2=256A_1^2+256A_2^2+512A_1A_2\) which can also be written as \(16(4A)^2=(16A_1)^2+(16A_2)^2+2(16A_1)(16A_2)\). Now we plug in the Heron's formula results. What happens is we will end up with a square root and some polynomial terms. I will omit all the tedious algebra, but after separating the polynomial stuff and the square root on different side, you should have this:
\[-14x^4+2864x^2+68640=2\sqrt{(676-x^2)(x^2-100)(484-x^2)(x^2-36)}\]Now, square both sides so we can write it as a single polynomial. This gets very disgusting.
\[192x^8-75008x^6+4326400x^4+587862016x^2=0\]Now let's substitute \(u=x^2\) since all the \(x\) degrees are even. We can divide 64 out of each coefficient too.
\[u(3u^3-1172u^2+67600u+9185344)=0\]It's a lot of work by hand, but using a tool like WolframAlpha, this can be factored. With that assistance, the factorization I found is:
\[u(u-196)(3u^2-584u-46864)=0\]So from this, we have a few solutions. \(x^2=0\Rightarrow x=0\) doesn't work because it is physically impossible. \(x^2=\frac{292-8\sqrt{3529}}{3}\) gives imaginary solutions so that doesn't work. For \(x^2=\frac{292+8\sqrt{3529}}{3}\), we get a solution \(x\approx15.992\) which is not physically possible, but that takes a bit more work to show so I will omit that as well. So that leaves \(x^2=196\Rightarrow x=\pm14\) so the only solution that does work is \(x=14\).
The Heron's method way was really stupid because if you are experienced with algebra, you'll realize that if anything you plug in at the start is mildly disgusting, its disgustingness will be amplified by multiplying and squaring several mildly disgusting formulas together.
It turns out that since the edge of length 4 is a median (because it bisects the edge of length \(x\)), we could use Apollonius's theorem, which was new to me. Instead, I will present a solution using the law of cosines which is similar to a proof of Apollonius's theorem. Let \(\alpha\) be the angle between the sides of lengths \(x\) and \(7\). Then we can use the law of cosines on that angle for both triangles it is contained in.
\[9^2=x^2+7^2-2(x)(7)\cos(\alpha) \Rightarrow \cos(\alpha)=\frac{x^2-32}{14x}\] \[4^2=\left(\frac{x}{2}\right)^2+7^2-2(7)\left(\frac{x}{2}\right)\cos(\alpha) \Rightarrow \cos(\alpha)=\frac{x^2+132}{28x}\]So we have 2 things equal to \(\cos(\alpha)\) so let's solve for \(x\).
\[\frac{x^2-32}{14x}=\frac{x^2+132}{28x} \Rightarrow 2x^2-64=x^2+132 \Rightarrow x^2-196=0\]So \(x=\pm14\) and of course \(x=14\) is the only possible solution.
This is a problem using properties of angles. Let's first label unknown angles so the equations are not confusing.
Now we know \(a=48\). Since \(a\degree+30\degree+c\degree=180\degree\) from the left triangle, we get \(c=102\). By supplementary angles, \(c\degree+d\degree=180\degree\) so \(d=78\). Now using the middle triangle, \(a\degree+d\degree+e\degree=180\degree\), so \(e=54\). Using supplementary angles again, wew have \(e\degree+2b\degree=180\degree\) so \(b=63\). And finally from the triangle at the right, we use \(c\degree+b\degree+x\degree=180\degree\) to find \(x=15\).
Even if you did this incorrectly with left associativity, you would still get \(2^{2^2}=4^2=16\).
Find the base 10 integer with Zeckendorf representation \(100101\).
The Zeckendorf representation is a sum of Fibonacci numbers. It comes from Zeckendorf's theorem which states every positive integer has a unique representation as the sum of nonadjacent Fibonacci numbers. For this number, using the positions of the ones, it is \(F_6+F_3+F_1=13+3+1=17\).
Another geometry problem. The only slightly not obvious things to note here are that \(x\) is the area of the shaded region, \(3\) is the side length of the shaded region, and the outer shape is a square. First we shall label 2 right triangles and their points.
We would like to show \(\triangle ABE \sim \triangle EDC\) (the 2 triangles are similar). By using complementary angles, we can show that \(\angle BAE = \angle DEC\) and \(\angle BEA = \angle DCE\). Since all 3 angles have matching measures, they are similar. Next, we know \(\overline{CD}=2\overline{BE}\) because \(\overline{CD}\) is the same length of a side of the square and \(\overline{BE}\) is half of the side length of the square. Therefore, by triangle similarity, \(\overline{CE}=2\overline{AE}=2\times3=6\). So then we can determine that the area of the shaded rectangle is \(3\times6=18\).
This is another geometry problem. We can solve it with the intersecting chords theorem, sometimes called power of a point. Using that theorem, we have:
\[2x(2x-10)=3x(2x-20) \Rightarrow 4x^2-20x=6x^2-60x \Rightarrow 2x^2-40x=0 \Rightarrow x(x-20)=0\]So \(x=0\) or \(x=20\). Since only one of these is physically possible, the answer is \(x=20\).
How many factors does \(576\) have?
First factor it. \(576=2\cdot288=2^2\cdot144=2^3\cdot72=2^4\cdot36= 2^5\cdot18=2^6\cdot9=2^6\cdot3^2\). Therefore, every factor of \(576\) is a number \(n=2^a\cdot3^b\) where \(a=0,1,2,3,4,5,6\) and \(b=0,1,2\). There are 7 choices for \(a\) and 3 choices for \(b\) so it has \(21\) factors.
This problem is going to involve a few steps, each requiring its own SVG that isn't a copy of the above + annotations (so a lot of tedious work because I've been writing these SVGs manually). First let's focus on the square and solve for \(r\).
The side length of the square is \(4r\) from the 2 circles of radius \(r\) so the square we have drawn in the top right corner has side length \(2r\). Then by using the radii of the circles, we can find the blue right triangle. By the Pythagorean theorem, we get (\(r=0\) is impossible):
\[r^2+(2r-11)^2=(11+r)^2 \Rightarrow 2r(2r-33)=0 \Rightarrow r=\frac{33}{2}\]Next we focus on the right triangle on the top.
To solve for \(s\), we use the Pythagorean theorem.
\[(4r)^2+(r+s)^2=(3r+s)^2 \Rightarrow 8r^2=4rs \Rightarrow s=2r\]Now we know this right triangle has side lengths \(3r,4r,5r\). By triangle similarity, we can find the ratio of the bottom left triangle to this triangle (using the sides on the right) and use that ratio to solve for \(x\). These right triangles are similar because the angles are all the same, because of the theorem about a line intersecting 2 parallel lines.
\[\frac{4r}{3r}=\frac{4}{3} \Rightarrow \frac{x}{r}=\frac{4}{3} \Rightarrow x=22\]Both parts of the fraction have consecutive powers so let's factor to be able to write the power sum part as \(1+x+x^2+x^3\).
\[\frac{x+x^2+x^3+x^4}{x^2+x+1+x^{-1}} =\frac{x(1+x+x^2+x^3)}{\frac{1}{x}(x^3+x^2+x+1)} =x^2\:\text{when}\:1+x+x^2+x^3\neq0\]Since the sum \(1+x+x^2+x^3\) is all positive numbers, it is nonzero so the answer is \(x^2=23\).
Find \(x=\frac{\text{area}(ABC)}{\text{area}(PQR)}\)
This is doable with triangle median theorems. Each median divides the triangle into 2 equal halfs and all 3 medians divide the triangle into 6 equal areas. First let's label the areas so we can work with them.
Using the median with point \(B\), we have
\[D+F+H=E+G+J\]Since \(PQ||AC\), we have similar triangles and can find \(D=E\) so we can simplify the above to
\[F+H=G+J\]Now by similar triangles, \(D+F+H=4D\) so we can substitute \(F+H=3D\)
\[3D=G+J\]Now let \(X\) be the area of \(\triangle ABC\). We have \(D+F+H=X/2\) so \(D=E=X/8\). From the equal areas theorem, \(J=X/3\) since it contains 2 of the 6 areas created by the medians. Now we plug those in
\[3X/8=G+X/3 \Rightarrow G=X/24\]So we find that if \(\text{area}(ABC)=X\) then \(\text{area}(PQR)=G=X/24\). Therefore, \(x=24\).
Anoher way we can write this is \(49\uparrow\uparrow5\mod47\) using Knuth's up arrow notation. 2 up arrows is for power towers. Repeated multiplication creates a cycle in modular arithmeic, so we will have to look at these power cycles to solve this problem.
First we look at powers of \(49\) modulo \(47\). Since \(49\equiv2\mod47\), we can look at powers of \(2\) modulo \(47\). Since \(\gcd(2,47)=1\), we can find some power \(p\) so \(2^{p}\equiv1\mod47\). From number theory, \(p|\phi(47)\) so since \(\phi(47)=46\), \(p\) is one of \(1,2,23,46\). We could use \(p=46\) since that will work, but the math gets a little easier if we can use something smaller. It turns out that \(2\) is not a primitive root modulo \(47\) and that the smallest such \(p\) that works is \(23\). So we have \(49\uparrow\uparrow5\equiv2^{49\uparrow\uparrow4}\mod47\). If we can express \(49\uparrow\uparrow4=23u+v\), then we would know the answer is \(2^v\mod47\). This has reduced the power tower size by 1 and the modulus we are working with from 47 to 23. We have to repeat this a few times so the work is summarized in the table below.
| step | compute this | base reduction | power cycle |
|---|---|---|---|
| 1 | \(49\uparrow\uparrow5\mod47\) | \(2^{49\uparrow\uparrow4}\mod47\) | \(2^{23}\equiv1\mod47\) |
| 2 | \(49\uparrow\uparrow4\mod23\) | \(3^{49\uparrow\uparrow3}\mod23\) | \(3^{11}\equiv1\mod23\) |
| 3 | \(49\uparrow\uparrow3\mod11\) | \(5^{49\uparrow\uparrow2}\mod11\) | \(5^5\equiv1\mod11\) |
| 4 | \(49^{49}\mod5\) | \(4^{49}\mod5\) | \(4^2\equiv1\mod5\) |
| 5 | \(49\mod2\) | \(1\) |
Next, we have to work backwards. We know that \(49\equiv1\mod2\). Using this, we have \(49^{49}\equiv4^{49}\equiv4^1\equiv4\mod5\). Next, \(49\uparrow\uparrow3\equiv5^{49^{49}}\equiv5^4\equiv9\mod11\). We repeat this and the work is summarized in the following table.
| step | compute this | work |
|---|---|---|
| 1 | \(49\mod2\) | \(1\) |
| 2 | \(49^{49}\mod5\) | \(\equiv4^{49}\equiv4^{2\cdot24+1}\equiv4^1\equiv4\mod5\) |
| 3 | \(49\uparrow\uparrow3\mod11\) | \(\equiv5^{49\uparrow\uparrow2}\equiv5^4\equiv9\mod11\) |
| 4 | \(49\uparrow\uparrow4\mod23\) | \(\equiv3^{49\uparrow\uparrow3}\equiv3^9=18\mod23\) |
| 5 | \(49\uparrow\uparrow5\mod47\) | \(\equiv2^{49\uparrow\uparrow4}\equiv2^{18}\equiv25\mod47\) |
Therefore, the answer is 25.
We can split this into odds and evens and use summation properties.
\[(2+4+\ldots+52)-(1+3+\ldots+51)=2\sum_{i=1}^{26}i-\sum_{i=1}^{26}(2i-1) =26(27)-26^2=26\]We can also make the following observation:
\[(-1+2)+(-3+4)+\ldots+(-51+52)=1+1+\ldots+1=26\]There can only be a possible value for \(x\) if \(3r+3s+3t\) is linearly dependent on the first 2 equations so let's see if we can find constants \(a,b\) to add multiples of them to get \(3r+3s+3t\).
\[a(3r+6s+t)+b(5r+11+t)=3r+3s+3t \Rightarrow 3a+5b=3,6a+11b=3,a+b=3\]We have 2 variables and 3 equations. By solving the first 2, we find \(a=6,b=-3\) and this also satisfies the 3rd. So this means \(x=21a+33b=6\cdot21-3\cdot33=126-99=27\).
Alternatively, we can write this as a matrix and do row elimination.
\[\begin{align} &\begin{bmatrix} 3&6&1&\big|&21\\ 5&11&1&\big|&33\\ 3&3&3&\big|&x\\ \end{bmatrix}\sim \begin{bmatrix} 1&1&1&\big|&x/3\\ 3&6&1&\big|&21\\ 5&11&1&\big|&33\\ \end{bmatrix}\sim \begin{bmatrix} 1&1&1&\big|&x/3\\ 0&3&-2&\big|&21-x\\ 0&6&-4&\big|&33-5x/3\\ \end{bmatrix}\\ \sim& \begin{bmatrix} 1&1&1&\big|&x/3\\ 0&3&-2&\big|&21-x\\ 0&0&0&\big|&(33-5x/3)-2(21-x)\\ \end{bmatrix} \end{align}\]For this matrix to have a solution, we must have \(0=(33-5x/3)-2(21-x)=x/3-9\Rightarrow x=27\).
Multiply each side by 10
\[6x-25=2x+1+90\]Group similar terms and solve.
\[4x=116 \Rightarrow x=29\]The line passing through \((-1,3)\) and \((-3,-2)\) is perpendicular to the line passing through \((-5,3)\) and \((x,-11)\).
The slope of the first line is \(\frac{-2-3}{-3-(-1)}=\frac{-5}{-2}=\frac{5}{2}\) so the slope of a perpendicular line is \(-1/\frac{5}{2}=\frac{-2}{5}\). Using point-slope form for the known point on the 2nd line, we get \(y-3=\frac{-2}{5}(x+5)\). Now we substitute \(y=-11\) and solve to find that \(x=30\).
We could also do this with vectors. A vector describing the first line is \((-2,-5)\) and a vector describing the second line is \((x+5,-14)\). The dot product must be zero for them to be perpendicular so we solve an equation: \(-2(x+5)+(-5)(-14)=0\Rightarrow x=30\).