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Find the largest \(n\in\mathbb{Z}\) such that \(\exists\ a,b,c\in\mathbb{Z}^+\) that satisfy \(a^n+b^n=c^n\).
This is a statement of Fermat's last theorem which states that in integers there are no nontrivial solutions for \(n>2\). So the answer is \(n=2\).
This can be done with tangent identities for sum and difference of angles. \[\begin{align}&\tan\left({\pi\over12}\right) =\tan\left({\pi\over3}-{\pi\over4}\right) ={\tan(\pi/3)-\tan(\pi/4)\over1+\tan(\pi/3)\tan(\pi/4)} ={\sqrt{3}-1\over1+\sqrt{3}}\cdot{1-\sqrt{3}\over1-\sqrt{3}}=2-\sqrt{3} \end{align}\] \[\begin{align}&\tan\left({5\pi\over12}\right) =\tan\left({\pi\over3}+{\pi\over12}\right) ={\tan(\pi/3)+\tan(\pi/12)\over1-\tan(\pi/3)\tan(\pi/12)} ={\sqrt{3}+2-\sqrt{3}\over1-\sqrt{3}\left(2-\sqrt{3}\right)}\\ &={2\over4-2\sqrt{3}}={1\over2-\sqrt{3}}\cdot{2+\sqrt{3}\over2+\sqrt{3}} ={2+\sqrt{3}\over4-3}=2+\sqrt{3}\\\end{align}\] So we add these to results and the answer is \((2-\sqrt{3})+(2+\sqrt{3})=4\).
\[x={f(6)\over f(4)}\] where \(f(n)\) is the number of \(1\)'s in binary numbers with up to \(n\) digits.
We can consider binary numbers up to \(n\) digits as binary strings of length \(n\). For each position, there are \(2^{n-1}\) numbers with that digit as a 1 so \(f(n)=n\cdot2^{n-1}\). Therefore \[{f(6)\over f(4)}={6\cdot2^5\over4\cdot2^3}=6\]
Find the sum of the roots of \(x^3-7x^2-10x+16\).
Using Vieta's formula, we can do this with the 2 leading coefficients, and get \(-(-7)/1=7\). Alternatively, by factoring it, the rational root theorem can help us find \(x=-2\) is a factor. From there, divide out \(x+2\) and solve the remaining quadratic so the full factorization is \((x+2)(x-1)(x-8)\) and the sum of roots is \(-2+1+8=7\).
This rectangle has an isosceles right triangle. The non right angles of it are \(45^\circ\) so the red angle is also \(45^\circ\). We solve \(5x=45\Rightarrow x=9\).
Gwen spent $390. She used the same number of $5s and $1s, double that of $10s, and 3 more $20s than $10s. How many $10s did she use?
First we can write equations for this using \(x_1,x_5,x_{10},x_{20}\) as variables for how many of those dollar bills she used. The linear system is \[x_1+5x_5+10x_{10}+20x_{20}=390,\ x_1=x_5,\ x_{10}=2x_1,\ x_{20}=3+x_{10}\] This is simple enough to avoid solving with a matrix. Use the 2 equations in the middle to get \(26x_1+20x_{20}=390\) by replacing \(x_5\) and \(x_{10}\) with \(x_1\). Now we can replace \(x_{20}=3+x_{10}=3+2x_1\) so we have \(66x_1+60=390\Rightarrow x_1=5\). From this, we have \(x_5=5,x_{10}=10,x_{20}=13\). This is the unique solution to the system and the number we are looking for is \(x_{10}=10\).
What is the expected value of a roll of two fair 10-sided dice?
Assuming the numbers are \(1,2,3,4,5,6,7,8,9,10\), each with equal probability, the average of these is \(5.5\). Expected value of 2 independent rolls adds together so \(5.5+5.5=11\).
Find the larger eigenvalue \[\begin{bmatrix}7&5\\4&8\\\end{bmatrix}\]
\[\begin{vmatrix}7-\lambda&5\\4&8-\lambda\\\end{vmatrix} =(7-\lambda)(8-\lambda)-20=\lambda^2-15\lambda+36=(\lambda-3)(\lambda-12)\] Therefore the eigenvalues are \(\lambda=3,12\) and the larger is \(12\).
This is a modular arithmetic problem. It can be solved with the usual methods, including modular exponentiation by squaring, but here is a way to do it that involves easy numbers. Below, all modular congruences are modulo \(23\).
First compute a little bit by squaring. \(10^1\equiv10,10^2\equiv8,10^4\equiv-5, 10^8\equiv2\). Now the last 2 multiply to \(-10\). So we have \(10^{12}\equiv-10\). Then we can use this: \[10^{100}\equiv10^{96}\cdot10^4\equiv(10^{12})^8\cdot10^4\equiv (-10)^8\cdot10^4\equiv10^8\cdot10^4\equiv10^{12}\equiv-10\equiv13\]
Find the sum of all primes \(p\) where \(7p+4\) is a perfect square.
We have \(7p+4=q^2\Rightarrow7p=q^2-4=(q+2)(q-2)\) for some integer \(q\). This means \(7|(q+2)(q-2)\) and \(p|(q+2)(q-2)\). Since \(7\) is prime, \(7|(q+2)\) or \(7|(q-2)\). Since \(p\) is prime, we can make this a little stronger and say \(7=q+2\) or \(7=q-2\) since after dividing \(7\), what remains must be prime. If \(7=q+2\), then \(p=q-2=3\). If \(7=q-2\) then \(p=q+2=11\). Therefore, the 2 solutions are \(p=3,11\) and the sum is \(14\).
The only comment here is that the angles should also be specified as degrees. The 2 non right angles of a right triangle sum to \(90^\circ\) so we have \(4x+2x=90\Rightarrow x=15\).
Find the smallest \(n\in\mathbb{Z}\) you can write as \(x^y=y^x\) where \(x\neq y\) and \(x,y\in\mathbb{Z}\).
By trying a few numbers, we can find \(2^4=4^2\) so we have \(n=16\) as a possibility. From here, the only possible integer powers (other than 1) which are smaller are \(2^3=8\) and \(3^2=9\), but these do not form a solution.
How many wallpaper groups are there?
Wallpaper groups describe symmetries of 2d repeating patterns, classified by translations, rotations, reflections, glide reflections, and combinations of them. There are 17 such wallpaper groups.
How many degrees does one exterior angle of a regular 20-sided polygon have?
Exterior angles of a polygon sum to \(360^\circ\). Since there are 20 sides, each exterior angle is \(360^\circ/20=18^\circ\).
The area of the square is \(16\) so the side length is \(s=\sqrt{16}=4\). There are 5 segments making the perimeter, each of equal length, so the perimeter is \(5\cdot4=20\).
How many edges does the complete graph with 7 vertices have?
Assuming a standard undirected graph, there are \({7\choose6}={7\cdot6\over2}=21\) possible ways to choose 2 vertices to connect with an edge, so there are 21 edges in the complete graph.
Find the positive root of \(x^2-2x-440\).
This can be factored by whatever means to \((x+20)(x-22)=0\). The roots are \(x=-20,22\) so the positive root is 22.
Seven consecutive odd numbers sum to 119. \(x\) is the largest.
We can express this as a sum (where \(y\) is an odd integer) \[y+(y+2)+(y+4)+\ldots+(y+12)=7y+2(1+2+3+4+5+6)=7y+42=119\] So solving this we find \(y=11\). The largest in the sequence is \(x=y+12=23\).
If we extend the 156° angle clockwise a little, we get a 180° angle on the left side of the line going through the parallel lines. The added part to the angle is also \(x^\circ\) by the geometry theorem for alternate interier angles. So we have \(156+x=180\Rightarrow x=24\).
First solve for \(y\) using integration by parts \(u=\ln(t),dv=-t^4dt\) \[\begin{align}&y=\int_0^1udv=uv\Big|_0^1-\int_0^1vdu =\ln(t){-t^5\over5}\Big|_0^1-\int_0^1{-t^5\over5}{dt\over t}\\& ={-t^5\ln(t)\over5}\Big|_0^1+\int_0^1{t^4\over5}dt ={-t^5\ln(t)\over5}\Big|_0^1+{t^5\over25}\Big|_0^1 =\left[{t^5\over25}-{t^5\ln(t)\over5}\right]_0^1\end{align}\] At \(t=1\), we get \(1/25\) since \(\ln(1)=0\). For \(t=0\), we have to evaluate a limit since this is actually an improper integral because \(\ln(0)\) is undefined. \[\lim_{t\to0^+}\left({t^5\over25}-{t^5\ln(t)\over5}\right) =\lim_{t\to0^+}{t^5\over25}-\lim_{t\to0^+}{t^5\ln(t)\over5} =\lim_{t\to0^+}{-\ln(t)\over5t^{-5}}\] We can apply LHopital's rule here because both \(-\ln(t)\to+\infty\) and \(5t^{-5}\to+\infty\) as \(t\to0^+\). \[=\lim_{t\to0^+}{-1/t\over-25t^{-6}}=\lim_{t\to0^+}{t^{-1}\over25t^{-6}} =\lim_{t\to0^+}{t^5\over25}=0\] So we have \(y=1/25\) and the problem asks for \(x=1/y=25\).
Find the number of triples \((a,b,c)\) such that \(a,b,c\in\mathbb{Z}\), \(a+b>0\), \(b+c>0\), \(c+a>0\), and \((a+b)(b+c)(c+a)=140\).
To make this problem with many constraints simpler, we can observe the symmetry. The sum of any pair of the 3 numbers is positive and the product of those 3 sums is 140. Now without loss of generality, consider a solution \((a,b,c)\) with \(a\leq b\leq c\). Once we know all solutions sorted, the other solutions are found as permutations of any of the sorted solrutions.
So with \(a\leq b\leq c\), we also have \(a+b\leq a+c\leq b+c\) which means \(140=(a+b)(b+c)(c+a)\geq(a+b)^3\) so we have bounds \(0<a+b\leq\lfloor\sqrt[3]{140}\rfloor=5\). We also know \((a+b)|140\) so the possibilities are \(a+b=1,2,4,5\). From here, we work through each case.
If \(a+b=5\), then \((b+c)(c+a)=28\). But also \(5\leq c+a\leq b+c\) so this is impossible by considering divisors of 28.
If \(a+b=4\), then \((b+c)(c+a)=35\). The only way to factor 35 like this so \(4\leq c+a\leq b+c\) is with \(c+a=5,b+c=7\). Solving for the 3 variables, we find \((a,b,c)=(1,3,4)\), which has 6 permutations so this solution counts for 6 triples.
If \(a+b=2\), then \((b+c)(c+a)=70\). Either \(b+c\) or \(c+a\) is even, but not both. If \(b+c\) is even, then subtracting \(a+b\) which is even, we find that \(c-a\) is even and so is \(a+c\). This would mean \(4|70\) which is a contradiction. We get a similar result by assuming \(c+a\) is even, so there are no solutions from this case.
If \(a+b=1\), then \((b+c)(c+a)=140\). Either \(b+c\) is odd or even. If it is even, then \(a+2b+c\) is odd and so is \(a+c\). If it is odd, then \(a+2b+c\) is even and \(a+c\) is also even. This means one of them is even and the other is odd. Since \(4|140\), then we must have \(4|(b+c)\) or \(4|(c+a)\). Now we use this to select from all possible cases found by considering divisors of 140.
| \(c+a\) | \(b+c\) | \(2a+b+c\) | \(2a\) | \(a\) | \(b\) | \(c\) | Permutations |
|---|---|---|---|---|---|---|---|
| \(1\) | \(140\) | \(2\) | \(-138\) | \(-69\) | \(70\) | \(70\) | \(3\) |
| \(2\) | \(70\) | ||||||
| \(4\) | \(35\) | \(5\) | \(-30\) | \(-15\) | \(16\) | \(19\) | \(6\) |
| \(5\) | \(28\) | \(6\) | \(-22\) | \(-11\) | \(12\) | \(16\) | \(6\) |
| \(7\) | \(20\) | \(8\) | \(-12\) | \(-6\) | \(7\) | \(13\) | \(6\) |
| \(10\) | \(14\) |
In this table are the steps for computing the needed values in the cases where we have both 1 odd and 1 even for \(b+c\) and \(c+a\). The red rows are the 2 cases that do not satisfy that condition. Here, we find solutions \((-69,70,70)\) with 3 permutations, \((-15,16,19)\) with 6 permutations, \((-11,12,16)\) with 6 permutations, and \((-6,7,13)\) with 6 permutations. Adding the permutations for all 5 solutions found gives us a total of \(6+3+6+6+6=27\) solutions.
A parabola with a vertical line of symmetry passes through \((0,-1),(1,7),(2,17)\) and \((3,x)\).
The parabola has an equation \(y=ax^2+bx+c\). We could use the 3 given points to solve this with linear algbra but some of the points given are convenient for solving this without the matrix method. Using \(x=0\), we find \(c=-1\). Then point \((1,7)\) tells us that \(a+b-1=7\Rightarrow a+b=8\). Finally, \((2,17)\) tells us that \(4a+2b-1=17\Rightarrow2a+b=9\). Now from this system of 2 equations we can find \(a=1,b=7\). So the equation is \(y=x^2+7x-1\) and substituting \(x=3\) tells us the missing point is \((3,29)\) so the answer is \(29\). We reused \(x\) for our parabola equation which may be a bit of a mistake but be aware that throughout this solution, \(x\) has nothing to do with the \(x\) stated in the problem.
Find the half-life of a radioactive sample that started with 1056 grams and had 33 grams remaining after 150 days.
The amount of sample remaining after \(n\) half lives is \(1056\cdot2^{-n}\). The number of half lives passed will be 150 days divided by the half life in days, so \(150/x\). Therefore, we obtain the equation \[1056\cdot2^{-150/x}=33\] and solving it \[\Rightarrow2^{-150/x}=33/1056=1/32\Rightarrow-150/x=-5\Rightarrow x=30\]