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Find \(-ijk\) where \(i,j,k\) are quaternions.
Quaternions are an extension of complex numbers which is not commutative, but it is associative. Multiplication is defined where \(1\) is the identity element and the imaginary basis products are \[i^2=j^2=k^2=-1,\ ij=-ji=k,\ jk=-kj=i,\ ki=-ik=j\] So by using this \[-ijk=-(ij)k=-(k)k=-k^2=-(-1)=1\]
Find the remainder whehn \(x^6+1\) is divided by \(x^2-x+1\).
The most clear way to solve this would be to use polynomial division.
| \(x^4\) | \(+x^3\) | \(+0x^2\) | \(-x\) | \(-1\) | |||||
| \(x^2\) | \(-x\) | \(+1\) | \(x^6\) | \(+0x^5\) | \(+0x^4\) | \(+0x^3\) | \(+0x^2\) | \(+0x\) | \(+1\) |
| \(-\) | \((x^6\) | \(-x^5\) | \(+x^4)\) | ||||||
| \(x^5\) | \(-x^4\) | \(+0x^3\) | |||||||
| \(-\) | \((x^5\) | \(-x^4\) | \(+x^3)\) | ||||||
| \(-x^3\) | \(+0x^2\) | \(+0x\) | |||||||
| \(-\) | \((-x^3\) | \(+x^2\) | \(-x)\) | ||||||
| \(-x^2\) | \(+x\) | \(+1\) | |||||||
| \(-\) | \((-x^2\) | \(+x\) | \(-1)\) | ||||||
| \(2\) | |||||||||
This shows that the remainder is \(2\) and \[{x^6+1\over x^2-x+1}=x^4+x^3-x-1+{2\over x^2-x+1}\] One other possible way to see it for this problem is noticing that \(x^2-x+1\) is a factor of \(x^3+1\). Then \((x^3+1)(x^3-1)=x^6-1\) and adding the remainder of \(2\) shows that \(x^6+1=(x^6-1)+2\) where \(x^2-x+1\) is a factor of \(x^6-1\). Or it can also be found by seeing how to factor \[x^6-1=(x^3+1)(x^3-1)=(x+1)(x^2-x+1)(x-1)(x^2+x+1)\]
The 2nd summation is a typical geometric sum. There is a way to modify the first so we can evaluate it with geometric sums. Ignore \(k=0\) since it is 0 and consider the grid \[\begin{array}{|c|c|c|c|c|}\hline 2^1&&&&\\\hline 2^2&2^2&&&\\\hline 2^3&2^3&2^3&&\\\hline \vdots&&&\ddots&\\\hline 2^9&2^9&\ldots&2^9&2^9\\\hline \end{array}\] If we expand \(k2^k=2^k+2^k+\ldots+2^k\) then the summation sums row by row in this grid. We can change it to sum column by column instead. \[\sum_{k=0}^9k2^k=\sum_{k=1}^9k2^k=\sum_{i=1}^9\left[\sum_{j=i}^92^j\right]\] Next we replace the inner summation with 2 summations and proceed to solve it \[\begin{align}&=\sum_{k=1}^9\left[\sum_{j=0}^92^j-\sum_{j=0}^{i-1}2^j\right] =\sum_{k=1}^9\left[(2^{10}-1)-(2^i-1)\right] =\sum_{k=1}^9\left[2^{10}-2^i\right]\\&=9\cdot2^{10}-\sum_{k=1}^92^i =9\cdot2^{10}-(2^{10}-2)=8\cdot2^{10}+2=2^{13}+2 \end{align}\] Then we just subtract \[(2^{13}+2)-\sum_{j=0}^{12}2^j=(2^{13}+2)-(2^{13}-1)=3\]
Find the smaller solution \[\sqrt{85-x}+\sqrt{x}=11\] We can square each side to eliminate square roots, but at the end we will have to check for extraneous solutions. \[(85-x)+x+2\sqrt{(85-x)x}=11^2=121\Rightarrow\sqrt{85x-x^2}=(121-85)/2=18\] Then square again to eliminate the new square root. \[85x-x^2=18^2=324\Rightarrow x^2-85x+324=0\Rightarrow(x-4)(x-81)=0\] So the solutions are \(x=4,x=81\). Checking shows that both are solutions and we take the smaller one which is \(x=4\).
How many times would you expect to have to flip a coin if you flip until two tails are flipped in a row?
Let \(E\) be the expected number of flips. Consider the following tree.
If we start with heads, then we are back to the start waiting for 2 consecutive tails. This has probability \(1/2\) and expected flips \(E+1\). If we flip tails and then heads, it is a similar situation of returning to the start. The probability is \(1/4\) and expected flips is \(E+2\). Finally, flipping 2 tails has probability \(1/4\) and expected flips \(2\). Combining all this information gives us the following equation for \(E\) \[E={1\over2}\cdot(E+1)+{1\over4}\cdot(E+2)+{1\over4}\cdot2\] Multiply each side by \(4\) \[4E=2(E+1)+(E+2)+2=3E+6\] Then subtract to get \(E=6\). Therefore 6 flips are expected.
This problem can be generalized to \(k\) tails instead of \(2\). It can be shown by similar reasoning that the solution is \(2^{k+1}-2\).
Find the slope of the line through \((12,87)\) and \((-12,-81)\)
Using the 2 point slope formula \[{y_2-y_1\over x_2-x_1}={-81-87\over-12-12}={-168\over-24}=7\]
\(247,623,212,35x\) is divisible by \(9\). \(x\) is a single digit.
When writing a base 10 number with digits \(d_nd_{n-1}\ldots d_1d_0\) as a sum, we can see the following reduction modulo 9. \[\sum_{i=0}^nd_i\cdot10^i\equiv\sum_{i=0}^nd_i1^i=\sum_{i=0}^nd_i\] This shows a number is divisible by 9 when its digit sum is divisible by 9. We have the following sum \[2+4+7+6+2+3+2+1+2+3+5+x=37+x\] which must be divisible by 9 for some \(0\leq x<10\). The solution is \(x=8\).
What is the tenths digit of \(\left(17+\sqrt{280}\right)^2\)
I did not find an elegant way to solve this. When expanding this thing, the term we have to care about is \(34\sqrt{280}\) since it creates the fractional part, the integer part does not affect the answer. If we multiply it by 10, then we are looking for the ones digit. Then \[340\sqrt{280}=\sqrt{340^2\cdot280}=\sqrt{32368000}\] and we could find by integer square root \[5689^2<32368000<5690^2\] So we take the ones digit from the floor which is \(9\). This can also be solved by calculator.
How many ways can one choose 2 pieces of fruit from a choice of nectarine, tangerine, mango, apple, pear?
The fruits available are irrelevant. The part that matters is there are 5 of them. This is 5 choose 2 in combinatorics. \[{5\choose2}={5!\over2!(5-2)!}={5!\over2!3!}={5\cdot4\over2}=10\]
This problem does not appear to have an elegant solution and is probably easiest solved by division directly which shows the remainder is \(11\).
\[{1\over\sum_{n=1}^\infty{H_n\over n2^n\pi^2}}\] where \(H_n\) are the harmonic numbers.
The \(\pi^2\) can be taken out so the essential part of this problem is evaluating \[\sum_{n=1}^\infty{H_n\over n2^n},\quad H_n=\sum_{k=1}^n{1\over k}\] Solving this problem will require a few parts. The main ideas for this solution were posted on the Daily Epsilon twitter.
For the first part, we will establish a few identities with polylogarithm functions. The polylogarithm function (of order \(s\)) is the following power series \[\text{Li}_s(z)=\sum_{n=1}^\infty{z^n\over n^s}\] which is convergent when \(|z|<1\). It can be extended by analytic continuation, but we do not need that for this problem. For this problem, we will need orders 1 and 2 and only to consider \(z\in(0,1)\). First we show an expression for \(\text{Li}_1(x)\) by starting with the derivative. \[{d\over dx}\text{Li}_1(x)={d\over dx}\sum_{n=1}^\infty{x^n\over n} =\sum_{n=1}^\infty{x^{n-1}}=1+x+x^2+\ldots={1\over1-x}\] Then by integration we find \[\text{Li}_1(x)=\int{dx\over1-x}=-\int{d(1-x)\over1-x}=-\ln(1-x)+c\] And \(c=0\) because \(\text{Li}_1(0)=0\). Next we do something similar for \(\text{Li}_2(x)\). \[{d\over dx}\text{Li}_2(x)={d\over dx}\sum_{n=1}^\infty{x^n\over n^2} =\sum_{n=1}^\infty{x^{n-1}\over n}\] Then multiply each side by \(x\) \[x{d\over dx}\text{Li}_2(x)=\sum_{n=1}^\infty{x^n\over n}=\text{Li}_1(x) =-\ln(1-x)\] So from this we find \[\text{Li}_2(x)=-\int{\ln(1-x)dx\over x}\] Later we will also need this which comes from substituting \(1-x\) in place of \(x\) \[\text{Li}_2(1-x)=-\int{\ln(1-(1-x))d(1-x)\over1-x}=\int{\ln(x)dx\over1-x}\] Note that these identities are for \(x\in(0,1)\). At one point we will use the limit as \(x\to0^+\).
For the next part, we will show how the essential summation from this problem arises when trying to find an expression for \(\ln^2(1-x)\). The insight for this part is unclear, but the idea was posted on the Daily Epsilon twitter. \[\ln^2(1-x)=\left(-\ln(1-x)\right)^2=\left(\text{Li}_1(x)\right)^2 =\left(\sum_{n=1}^\infty{x^n\over n}\right)^2\] Think of this squared summation like adding the terms in the following table which we will rearrange. \[\begin{array}{c|ccccc} &x&x^2/2&x^3/3&x^4/4&\ldots\\\hline x&x^2&x^3/2&x^4/3&x^5/4&\ldots\\ x^2/2&x^3/2&x^4/4&x^5/6&x^6/8&\\ x^3/3&x^4/3&x^5/6&x^6/9&x^7/12&\\ x^4/4&x^5/4&x^6/8&x^7/12&x^8/16&\\ \vdots&\vdots&&&&\ddots\\ \end{array}\] This can be summed by diagonals with powers \(x^2,x^3,x^4,\ldots\). The diagonal with \(x^{n+1}\) has denimonators \({1\over1\cdot n},{1\over2\cdot(n-1)},{1\over3\cdot(n-2)}\ldots\). \[\sum_{n=1}^\infty\left(x^{n+1}\sum_{k=1}^n{1\over k\cdot(n+1-k)}\right)\] Here we can use the identity \(1/a+1/b=(a+b)/ab\) with \(a=k\) and \(b=n+1-k\). \[\sum_{n=1}^\infty\left({x^{n+1}\over n+1} \sum_{k=1}^n{n+1\over k\cdot(n+1-k)}\right)=\sum_{n=1}^\infty\left( {x^{n+1}\over n+1}\sum_{k=1}^n\left({1\over k}+{1\over n+1-k}\right)\right)\] For the inner summation both \(1/k\) and \(1/(n+1-k)\) run through \(1+1/2+1/3+\ldots+1/n\) so this is just \(2H_n\). \[2\sum_{n=1}^\infty{x^{n+1}H_n\over n+1}=2\sum_{n=2}^\infty{x^nH_{n-1}\over n} =2\sum_{n=2}^\infty{x^n(H_n-1/n)\over n}=2\sum_{n=1}^\infty{x^n(H_n-1/n)\over n} =2\sum_{n=1}^\infty{x^nH_n\over n}-2\text{Li}_2(x)\] So what we have shown is \[\sum_{n=1}^\infty{x^nH_n\over n}={1\over2}\ln^2(1-x)+\text{Li}_2(x)\] Then with \(x=1/2\), we have an expression for that essential summation we need. \[\sum_{n=1}^\infty{H_n\over n2^n}={1\over2}\ln^2\left({1\over2}\right) +\text{Li}_2\left({1\over2}\right) ={1\over2}\ln^2(2)+\text{Li}_2\left({1\over2}\right)\]
For the next part, we need to show that \[\lim_{x\to0^+}\ln(x)\ln(1-x)=0\] We can rewrite this in indeterminate form \(0/0\) and use LHopital's rule. \[\lim_{x\to0^+}{\ln(1-x)\over(\ln(x))^{-1}} =\lim_{x\to0^+}{-1/(1-x)\over-(\ln(x))^{-2}x^{-1}} =\lim_{x\to0^+}{x\ln^2(x)\over1-x} =\lim_{x\to0^+}{1\over1-x}\lim_{x\to0^+}x\ln^2(x) =\lim_{x\to0^+}x\ln^2(x)\] This is true if this limit exists, which will be shown next. Use the indeterminate form \(\infty/\infty\) and Lhopital's rule again. \[\lim_{x\to0^+}x\ln^2(x)=\lim_{x\to0^+}{\ln^2(x)\over x^{-1}} =\lim_{x\to0^+}{2\ln(x)x^{-1}\over-x^{-2}}=-2\lim_{x\to0^+}x\ln(x)\] \[=-2\lim_{x\to0^+}{\ln(x)\over x^{-1}}=-2\lim_{x\to0^+}{x^{-1}\over-x^{-2}} =-2\lim_{x\to0^+}(-x)=0\]
Now we combine the results. The insight for this is not clear and the idea is from a comment on the Daily Epsilon twitter. Start by creating the following integral and using the \(\text{Li}_2(x)\) results from the first part. \[\ln(x)\ln(1-x)=\int d\left(\ln(x)\ln(1-x)\right) =\int\left({\ln(1-x)\over x}-{\ln(x)\over1-x}\right) =-\text{Li}_2(x)-\text{Li}_2(1-x)+c\] By using the limit result from part 3, we solve for \(c\) with \[0=-\text{Li}_2(0)-\text{Li}_2(1)+c\Rightarrow c=\text{Li}_2(1)= \sum_{n=1}^\infty{1\over n^2}={\pi^2\over6}\] Where we use the well known \(\pi^2/6\) identity (Basel problem). Now use \(x=1/2\) to find \[\ln^2\left({1\over2}\right)=-2\text{Li}_2\left({1\over2}\right)+{\pi^2\over6} \Rightarrow{1\over2}\ln^2(2)={\pi^2\over12}-\text{Li}\left({1\over2}\right)\] Then we substitute this into the result from part 2 and find \[\sum_{n=1}^\infty{H_n\over n2^n}={\pi^2\over12}\] Now we put this into the given expression for the problem and the solution is \[{1\over{1\over\pi^2}\sum_{n=1}^\infty{H_n\over n2^n}} ={1\over{1\over\pi^2}{\pi^2\over12}}={1\over1/12}=12\]
If \(x-8<0\) then the equation because \(x-(x-8)=18\Rightarrow8=18\) so no solution. If \(x-8\geq0\) then we find the solution \[x+(x-8)=18\Rightarrow2x-8=18\Rightarrow2x=26\Rightarrow x=13\] This can be checked to ensure it is not an extraneous solution.
There are various methods to solve this problem. One is to split the object into triangles and add their areas. Triangle areas can be computed by using 2 of their sides as vectors. The determinant of 2 vectors \((a,b),(c,d)\) is \(ad-bc\) and its absolute value is the area of the parallelogram determined by it. Half of this parallelogram is a triangle with area \(|ad-bc|/2\). Below is the diagram with points labeled.
The area of the "outer" quadrilateral from points \(A,D,E,F\) can be determined from adding the following 2 triange areas: \[{1\over2}\begin{vmatrix}F-A\\D-A\\\end{vmatrix} ={1\over2}\begin{vmatrix}1&6\\6&1\\\end{vmatrix}={35\over2}\] \[{1\over2}\begin{vmatrix}F-E\\D-E\\\end{vmatrix} ={1\over2}\begin{vmatrix}-4&-1\\1&-6\\\end{vmatrix}={25\over2}\] So the "outer" quadrilateral has area \(30\). For the "inner" quadrilateral \(A,B,C,D\), we use the following 2 triangles: \[{1\over2}\begin{vmatrix}B-A\\D-A\\\end{vmatrix} ={1\over2}\begin{vmatrix}2&4\\6&1\\\end{vmatrix}={22\over2}\] \[{1\over2}\begin{vmatrix}B-C\\D-C\\\end{vmatrix} ={1\over2}\begin{vmatrix}-2&-1\\2&-4\\\end{vmatrix}={10\over2}\] So the "inner" quadrilateral has area \(16\). By subtracting, the area of the shaded part is \(30-16=14\).
Another method is to use Pick's theorem which shows a way to calculate the area of simple polygons with integer coordinates by counting integer coordinate points. There are 11 interier points and 8 border points so the area is \[i+{b\over2}-1=11+{8\over2}-1=14\]
Find the number of ways that a red and blue die can be rolled so that their product is a multiple of 6.
This could be done by writing the multiplication table and counting, but can be counting a bit more intelligently. One condition is that either is 6 so by inclusion-exclusion, it is 6 for the red die, 6 for the blue die, and subtract 1 for when both are 6 so there are 11 ways. Without a 6, there are the following pairs \((2,3),(3,2),(3,4),(4,3)\). So there is a total of 15. Below is the table with highlighted (yellow) multiples of 6.
| 1 | 2 | 3 | 4 | 5 | 6 | |
| 1 | 1 | 2 | 3 | 4 | 5 | 6 |
| 2 | 2 | 4 | 6 | 8 | 10 | 12 |
| 3 | 3 | 6 | 9 | 12 | 15 | 18 |
| 4 | 4 | 8 | 12 | 16 | 20 | 24 |
| 5 | 5 | 10 | 15 | 20 | 25 | 30 |
| 6 | 6 | 12 | 18 | 24 | 30 | 36 |
First consider a domino along the thin edge. There are 2 possibilities shown below.
In the first possibility, placing a domino adjacent on the left of the red tile leaves a single space that cannot be filled with a \(2×1\) domino. So we use the second possibility. By applying similar reasoning to the other 4 edges, we must have dominoes in the following positions.
This leaves 4 \(2×2\) spaces. Each of these has 2 possibilities: 2 vertical dominoes or 2 horizontal dominoes. The total number of possible tilings is \(2^4=16\).
The messy way to solve this is by squaring to remove the square roots. First let \[x=\sqrt{784-440\sqrt{3}}+\sqrt{325-100\sqrt{3}}\] Then \[\begin{align}&x^2=\left(784-440\sqrt{3}\right)+\left(325-100\sqrt{3}\right) +2\sqrt{\left(784-440\sqrt{3}\right)\left(325-100\sqrt{3}\right)}\\ &=1109-540\sqrt{3}+2\sqrt{386800-221400\sqrt{3}} =1109-540\sqrt{3}+20\sqrt{3868-2214\sqrt{3}}\\\end{align}\] Now to get rid of the nested root, we rearrange to \[x^2-1109+540\sqrt{3}=20\sqrt{3868-2214\sqrt{3}}\] Then square each side \[x^4+x^2\left(-2218+1080\sqrt{3}\right)+\left(2104681-1197720\sqrt{3}\right) =400\left(3868-2214\sqrt{3}\right)\] \[x^4+x^2\left(-2218+1080\sqrt{3}\right)+\left(557481-312120\sqrt{3}\right)=0\] Now using the quadratic formula with \(y=x^2\), we find \[\begin{align}y&={\left(2218-1080\sqrt{3}\right)\pm\sqrt{ \left(2218-1080\sqrt{3}\right)^2-4\left(557481-312120\sqrt{3}\right)}\over2} \\&={\left(2218-1080\sqrt{3}\right)\pm\sqrt{6188800-3542400\sqrt{3}}\over2}\\ \end{align}\] If we expect an integer answer, we should find that the square root part is the square of some integer plus/minus \(1080\sqrt{3}\). By factoring we can see \[3542400\sqrt{3}=2\cdot1640\cdot1080\sqrt{3}\] This suggests we should be able to have this as the \(2ab\) part of \((a+b)^2=a^2+b^2+2ab\). It turns out this factoring works and \[\begin{align}y&={\left(2218-1080\sqrt{3}\right)\pm\sqrt{ \left(1080\sqrt{3}-1640\right)^2}\over2}\\ \end{align}\] The first solution \((+)\) gives us \[y={2218-1080\sqrt{3}+1080\sqrt{3}-1640\over2}={2218-1640\over2}={578\over2} =289\Rightarrow x^2=289\Rightarrow x=\pm17\] So clearly \(x=17\) is the solution and \(x=-17\) is extraneous. For the other solution to \(y\) \[y={3858-2160\sqrt{3}\over2}=1929-1080\sqrt{3}\approx58.385\] This is extraneous since our first squaring shows \[y=x^2\geq1109-540\sqrt{3}\approx173.693\]
Another way is to consider that the values inside the square roots look like the result of \(\left(a-b\sqrt{3}\right)^2=a^2+3b^2+2ab\sqrt{3}\). So if both of them are like this, we would expect to find the same multiple of \(\sqrt{3}\) for each. The \(\sqrt{3}\) parts should match the \(2ab\sqrt{3}\) \[440\sqrt{3}=2\cdot220\sqrt{3},\quad100\sqrt{3}=2\cdot50\sqrt{3}\] So we need a \(\sqrt{3}\) multiple that divides \(\gcd(220,50)=10\). So we can try \(10\sqrt{3}\). For the first root, \[2\cdot220\sqrt{3}=2\cdot22\cdot10\sqrt{3}\Rightarrow \left(22-10\sqrt{3}\right)^2=484-440\sqrt{3}+100\cdot3=784-440\sqrt{3}\] And for the second root, \[2\cdot50\cdot\sqrt{3}=2\cdot5\cdot10\sqrt{3}\Rightarrow \left(10\sqrt{3}-5\right)^2=100\cdot3-100\sqrt{3}+25=325-100\sqrt{3}\] Therefore \[\begin{align}&\sqrt{784-440\sqrt{3}}+\sqrt{325-100\sqrt{3}} =\sqrt{\left(22-10\sqrt{3}\right)^2}+\sqrt{\left(10\sqrt{3}-5\right)^2}\\ &=22-10\sqrt{3}+10\sqrt{3}-5=22-5=17\\\end{align}\]
These are supplementary angles so \(4x+6x=180\Rightarrow10x=180\Rightarrow x=18\)
Find the larger solution to \(4x^2-36x=x^2+20x+19\)
\[3x^2-56x-19=0\] By factoring \[(3x+1)(x-19)=0\Rightarrow x=-1/3,19\] By quadratic formula \[\begin{align}&x={56\pm\sqrt{56^2+4\cdot3\cdot19}\over6} ={56\pm\sqrt{56^2+2\cdot2\cdot56+2^2}\over6}\\ &={56\pm\sqrt{(56+2)^2}\over6}={56\pm58\over6}=-1/3,19\end{align}\]
How many times does 2 appear in the natural numbers under 100?
In single digits, it is just the number 2. In double digits, there are 8 numbers with a tens digit other than 2 and a ones digit of 2. If the tens digit is 2, then 2 appears 10 times as the tens digit and once as the ones digit. In total, there are 20 appearances.
Find the (rounded) slope of the line tangent to \(y=3^x\) at \(x=2.77\).
The derivative is \(y'=3^x\ln(3)\). The slope is \(3^{2.77}\ln(3)\approx23\) by calculator.
This is a 4 vertex complete graph. Any permutation of the vertices is a valid path so the number of Hamiltonian paths is \(4!=24\).
Compute the derivative \(f'(x)=7\cos(x)-24\sin(x)\) and then a maximum will occur at one of the solutions to \[7\cos(x)-24\sin(x)=0\Rightarrow7\cos(x)=24\sin(x) \Rightarrow\tan(x)={7\over24}\Rightarrow x=\arctan\left({7\over24}\right)\] This gives a unique solution on \((-\pi/2,\pi/2)\) and 2 solutions on \((-\pi,\pi)\). Using this angle, we can draw the following right triangle.
Let \(\theta\) be this angle in \((0,\pi/2)\). Then \(\theta+\pi\) is also a solution. Therefore we have 2 points to evaluate \(f(x)\) at. \[f(\theta)=7\left({7\over25}\right)+24\left({24\over25}\right) ={7^2+24^2\over25}={25^2\over25}=25\] \[\begin{align}&f(\theta+\pi)=7\sin(\theta+\pi)+24\cos(\theta+\pi) =7\sin(\theta)\cos(\pi)+7\cos(\theta)\sin(\pi) +24\cos(\theta)\cos(\pi)-24\sin(\theta)\sin(\pi)\\& =-7\sin(\theta)-24\cos(\theta)=-f(\theta)=-25 \end{align}\] This gives us a maximum of \(25\) and minimum of \(-25\).
Find the sum of the number of faces, vertices, and edges in a cube.
There are 6 faces, 8 vertices, and 12 edges. If we did not know one of these, we could find the remaining quartity with Euler's formula \[\text{faces}+\text{vertices}=\text{edges}+2\] The total is \(6+8+12=26\).
If we draw a hexagon through the centers of the 6 smaller circles, we will find that it is a regular hexagon with side length \(2r\) (where \(r=\sqrt{3\over\pi}\) is the radius of a small circle). A regular hexagon can be split nicely into 6 equilateral triangles so the distance from 1 point to the opposite point is twice the smaller radius, \(2r\). Then subtract both of the half radii of smaller circles and the remaining distance is \(r\) for the radius of the circle that fits in the center. From here, we can see the radius of the larger circle is \(3r\) so its area is \(9\times3=27\).
Find the order of the dihedral group of a regular 14-gon.
Dihedral groups describe rotating a n-gon so it fits back over its original position. This also includes flipping it over. The number of ways this can happen is twice the number of sides/vertices so \(2\times14=28\).
To count these elements, we have to find which last 2 digits stay the same when squaring the number. We look for \(z\in\{0,1,2,\ldots,99\}\) that solve \(z\equiv z^2\) mod 100 \(\Rightarrow z(z-1)\equiv0\). We can factor \(100=4\times25\) which are coprime and using the Chinese remainder theorem we must have \(z(z-1)\equiv0\) modulo both 4 and 25. For modulo 4, since \(z\) and \(z-1\) are coprime, 4 must divide one of them so the only solutions are \(z\equiv0,1\). Similarly modulo 25, we must have \(z\equiv0,1\). These can be combined into the following 4 solutions.
| mod 4 | mod 25 | mod 100 |
|---|---|---|
| 0 | 0 | 0 |
| 1 | 0 | 25 |
| 0 | 1 | 76 |
| 1 | 1 | 1 |
This shows that each of the internals \([100,199],[200,299],\ldots,[700,799]\) have 4 solutions, for a total of 28. Then we add the solution \(y=800\) for a total of \(|S|=29\).
The Nim game with piles of 9, 23, and \(x\) is a second player win.
Nim has been solved and it is shown that the first player has a winning strategy iff the nim sum of the piles is nonzero. So the second player has a winning strategy iff the nim sum of the piles is zero. Nim sum is another terminology for binary xor. So we need to solve \[9\oplus23\oplus x=0\Rightarrow x=9\oplus23\] In binary, \(9=1001_2\) and \(23=10111_2\) so \(9\oplus23=11110_2=30\).
What is the third Mersenne prime?
The first 3 primes are \(2,3,5\) so the first Mersenne numbers are \(2^2-1,2^3-1,2^5-1\) or \(3,7,31\). All 3 of these are prime so the third Mersenne number is 31.