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Let \(H_n\) be the generalized Harmonic numbers. \(x=H_{1\over2}+2\ln(2)\)
The harmonic numbers are defined as \[H_n=1+{1\over2}+{1\over3}+\ldots+{1\over n}=\sum_{k=1}^n{1\over k}\] The generalized harmonic numbers can be defined from turning this summation into an integral as follows \[\begin{align}&\sum_{k=1}^n{1\over k} =\sum_{k=1}^n\left(\int_0^1x^{k-1}dx\right) =\int_0^1\left(\sum_{k=1}^nx^{k-1}\right)dx\\ &=\int_0^1\left(1+x+x^2+\ldots+x^{n-1}\right)dx =\int_0^1{1-x^n\over1-x}dx\end{align}\] So we can compute this generalized harmonic number as \[H_{1\over2}=\int_0^1{1-\sqrt{x}\over1-x}dx =\int_0^1{1-\sqrt{x}\over(1-\sqrt{x})(1+\sqrt{x})}dx =\int_0^1{dx\over1+\sqrt{x}}\] Here, substitute \(u=1+\sqrt{x}\) which gives us \(du={dx\over2\sqrt{x}} \Rightarrow dx=2(u-1)du\). The bounds change to \(u(0)=1,u(1)=2\). \[\int_1^2{2(u-1)du\over u}=\int_1^2\left(2-{2\over u}\right)du =\left[2u-2\ln|u|\right]_1^2=2-2\ln(2)\] Then finally \(x=H_{1\over2}+2\ln(2)=(2-2\ln(2))+2\ln(2)=2\)
Consider the side-angle-side area formula for the "Area=3" triangle. It would be \({1\over2}ab\sin(\theta)\) where \(a,b\) are \(3,\sqrt{5}\) and \(\theta\) is its angle next to the point in the middle of the figure. Now the angle right across it, part of the \(x\) triangle, is \(180^\circ-\theta\). We can apply the same side-angle-side area formula and only the angle is different because the adjacent sides are the same \(3,\sqrt{5}\). Therefore, \(x={1\over2}ab\sin(180^\circ-\theta)\). It turns out that \(\sin(180^\circ-\theta)=\sin(\theta)\) so it is the same as the "Area=3" triangle. Therefore, \(x=3\)
What is the maximum area among all rhombuses of side length 2?
Intuitively, we would think this is the square, the special case where all the angles are right angles, which would have area \(2\times2=4\). If we consider this more rigorously, a parallelogram has area \(ab\sin(\theta)\) where the side lengths are \(a,b\). The rhombus is the special case where \(a=b\). So the area of the rhombus in this problem is \(a^2\sin(\theta)=4\sin(\theta)\) and is maximized when \(\theta=\pi/2\).
What percent of integers is divisible by 2 and 5 but not 4?
Since we are dealing with cardinality of integers, this may be better phrased in terms of natural density which is the ratio of integers \(1,2,3,\ldots,n\) satisfying a property as \(n\to\infty\). The ratio of integers divisible by 2 and 5 is \(1/\text{lcm}(2,5)=1/10\) since \(2\mid n\) and \(5\mid n\) implies \(\text{lcm}(2,5)\mid n\). Then we subtract integers divisible by 10 but not 4. Any \(10k\) is divisible by 4 when \(k\) is even and not divisible by 4 when \(k\) is odd. Therefore it is half of these integers. So the ratio we are looking for is \({1\over10}\times{1\over2}={1\over20}\) which is 5%.
How many ways can Emmy, Maryam, and Federico line up?
This is permutations of 3 objects taken from a set of 3, so \(3!=6\).
The largest zero of \(x^4-5x^3-19x^2+29x+42\).
The zeroes can be found by the rational root test by testing \(\pm1,\pm2,\pm3,\pm6,\pm7,\pm14,\pm21,\pm42\). By trial and error, we can find all the zeroes with this method are \(-1,2,-3,7\) so the largest is \(7\).
Another idea for solving this is suppose we can factor it into 2 quadratic polynomials with integer constants. Then let those constants be a factorization of 42 and see what happens. It turns out multiple choices work as long as each factor is the product of 2 of the zeroes. Let's first try \(6\times7=42\). We would be looking for \(a,b\) such that the polynomial factors to \[(x^2+ax+6)(x^2+bx+7)\] This expands to \[x^4+x^3(a+b)+x^2(13+ab)+x(7a+6b)+42\] Now using the \(x^3\) and \(x\) terms, we can solve the linear system \[a+b=-5,\quad7a+6b=29\] Which has solution \[a=59,\quad b=-64\] Then we have to check with the \(x^2\) term to make sure this works, but it does not since \(13+ab=-3763\). So we try another choice, such as \(3\times14=42\). \[(x^2+ax+3)(x^2+bx+14)=x^4+x^3(a+b)+x^2(17+ab)+x(14a+3b)+42\] This time the linear system is \[a+b=-5,\:14a+3b=29\quad\Rightarrow\quad a=4,\:b=-9\] And we find \(17+ab=-19\) so this is a valid factorization. Now we can find the zeroes by factoring quadratic polynomials \[x^2+4x+3=(x+1)(x+3),\quad x^2-9x+14=(x-2)(x-7)\]
How many binary numbers \(<100000_2\) have exactly three \(1\)'s?
This is counted from 5 bit numbers so we can view this as a combinatorics problem where the 3 positions for the ones are chosen. The answer is \({5\choose3}=10\)
Find the order of the Galois group for \(f(x)=x^4-8x+12\).
Solving this problem requires techniques from Galois theory. Some of them are described here.
The discriminant method can be used for part of it. Since \(f(x)\) is a depressed quartic (missing the \(x^3\) term), the discriminant formula is a bit simpler. For \(f(x)\), the discriminant is \(-27(-8)^4+256(12)^3=331776\) (the formula is on the Wikipedia page). Since \(331776=576^2=24^4\), it is a perfect square which shows that the Galois group of \(f(x)\) is a subgroup of the alternating group \(A_4\) which has order 12.
The next method that can be used is factoring \(f(x)\) over integer fields modulo a prime that does not divide the discriminant. The following factorizations were posted on the Daily Epsilon twitter. \[f(x)\equiv(x+4)(x^3+x^2+x+3)\mod5\] \[f(x)\equiv(x^2+8x+4)(x^2+11x+3)\mod19\] This shows that \(f(x)\) has cycle types \((1,3)\) and \((2,2)\). Therefore, both 3 and 4 divide the order of the Galois group. Combining both results, the order must be 12.
How many Archimedean solids are there?
Archimedean solids are one of 13 convex polyhedra such that all their faces are one of at least 2 types of regular polygons and all vertices are symmetric. They are listed on Wikipedia. Two of them are chiral so sometimes they may be counted as 15 solids.
\(7x\) is the largest two-digit number divisible by \(7\).
This means that \(7x\leq99\). Since \(7x\) is divisible by \(7\), \(x\) is an integer and equal to \(\lfloor99/7\rfloor=14\).
Find thet smallest composite number, \(n\), such that there is only one group of order \(n\).
There are many theorems in group theory about properties of groups depending on the prime factorization of the group order. If the group order can be factored into distinct primes \(p\times q\) with \(p<q\), there is only 1 group when \(p\nmid q-1\). If \(p=2\) then \(q-1\) is even so this will not be satisfied. When \(p=3\) and \(q=5\), then \(q-1=4\) and \(3\nmid4\). So there is only 1 group of order 15 and this is the smallest such composite order group.
The triangles are
Water flows at 2m/s through a pipe of radius 3cm into a pipe of radius 1cm, where it flows at a rate of \(x\) m/s.
In order to maintain the same flow rate, the water speed times the pipe area needs to stay the same. When the radius changes from 3cm to 1cm, the area becomes \(1/9\) of the original, so the speed must be multiplied by \(9\) to maintain the flow rate. The new speed is \(x=2\cdot9=18\).
How many rooted trees have \(6\) vertices?
This problem can be solved with dynamic programming methods. Let \(T(n)\) be the number of rooted trees with \(n\) vertices. Then \(T(1)=1\). For \(n\) vertices, we have a root and subtrees totaling \(n-1\) vertices. The way to form these subtrees can be counted by partitioning \(n-1\). The recurrence relation is \[T(n)=\sum_{t_1,t_2,\ldots,t_k\in P(n-1)}T(t_1)T(t_2)\ldots T(t_k)\] where \(P(n-1)\) is all the partitions of \(n-1\). For each number in the partition, choose one of the possible rooted trees with that many vertices and that will form one of the subtrees for the \(n\) vertex rooted tree. The partitions are
Find the smallest number of congruent squares needed to fill a region that is \(\sqrt{363}\times\sqrt{12}\).
In order for the squares to fill this region, its side lengths must be a rational ratio. \[{\sqrt{363}\over\sqrt{12}}={\sqrt{3\cdot11\cdot11}\over\sqrt{2\cdot2\cdot3}} ={11\over2}\] So if \(n\) squares fill the \(\sqrt{12}\) side length, then \(11n/2\) squares fill the \(\sqrt{363}\) side length. The smallest \(n\) that makes both integers is \(2\). So 22 congruent squares are needed at minimum.
Four people write their name on a slip of paper. Each is randomly given one of the slips. What is the ratio of the odds that someone doesn't have their name to the odds that everyone has their name?
There are \(4!=24\) ways to assign the papers. Only 1 of these permutations gives everyone their name. All 23 other permutations have someone without their name. Therefore the ratio is 23.
What is the largest number divisible by all numbers less than its square root?
We can see this is possible with \(1,2,3,4\). Their lowest common multiple is \(12\) so \(12\) and \(24\) are divisible by all of them. Since \(4<\sqrt{24}<5\), \(24\) satisfies the required property. Now the challenging part is to show it is the largest.
Suppose we have some integer \(N\) such that \(n^2<N\leq(n+1)^2\) for some integer \(n\geq5\) and require \(N\) is divisible by all of \(1,2,\ldots,n\). Since \(n\mid N\), either \(N=n(n+1)\) or \(N=n(n+2)\). We must also have \((n-1)\mid N\). Since \(\gcd(n,n-1)=1\), we also have \(n(n-1)\mid N\). For the first case (\(N=n(n+1)\)), \[n(n-1)\mid n(n+1)\Rightarrow(n-1)\mid(n+1)\] which is impossible because \({n+1\over2}<n-1<n+1\) when \(n\geq5\). For the second case (\(N=n(n+2)\)), \[n(n-1)\mid n(n+2)\Rightarrow(n-1)\mid(n+2)\] which is similarly impossible when \(n\geq5\). If you were to check \(n=4\), this corresponds to the solution \(N=24\).
By similar triangles, we have the ratios \[{x+y\over30}={6\sqrt{11}\over h}\] We can find \(x+y\) with the pythagorean theorem \[30^2+\left(6\sqrt{11}\right)^2=(x+y)^2\Rightarrow x+y=36\] Then using this we find \(h=5\sqrt{11}\) by substituting. Now use another similar triangles ratio \[{x+y\over6\sqrt{11}}={6\sqrt{11}\over y}\Rightarrow y(x+y)=36\cdot11 \Rightarrow y=11\] Then \(x=36-y=25\)
The sum of two numbers is \(34\). Their product is \(120\). Find their difference.
This can be written as a quadratic equation. Let \(a,b\) be the numbers so we have \(a+b=34\) and \(ab=120\). Then \(a^2+ab=34a\Rightarrow a^2-34a+120=0\) which can be solved by factoring \((a-30)(a-4)=0\). Therefore, \(a=4,30\). Since the equations are symmetric, we expect this to correspond to \(b=30,4\) which is correct. The difference is \(30-4=26\).
Find the coefficient of \(x^6\) in the expansion of \((x+1)^8\).
This is a simple binomial coefficents problem and the answer is \({8\choose6}={8\cdot7\over2}=56/2=28\).
The sum of angles is \(360^\circ\) for a quadrilateral so \(x+94+72+163=360\). We can solve this for \(x=31\).