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Half of the oddest prime.
This is more of a thing with words. The "oddest" prime in a way is 2 because it is the only even prime, so half of that is 1.
This can be solved by using definitions of the hyperbolic functions. The derivatives are \(\sinh'(\theta)=\cosh(\theta)\) and \(\sinh''(\theta)=\sinh(\theta)\). Then using the definitions in terms of \(e^\theta\), we have
\[\begin{align}&{\sinh(2\theta)\over\sinh'(\theta)\sinh''(\theta)} ={\sinh(2\theta)\over\cosh(\theta)\sinh(\theta)} ={{1\over2}\left(e^{2\theta}-e^{-2\theta}\right)\over {1\over2}\left(e^\theta+e^{-\theta}\right) {1\over2}\left(e^\theta-e^{-\theta}\right)}\\& =2{e^{2\theta}-e^{-2\theta}\over(e^\theta+e^{-\theta})(e^\theta-e^{-\theta})} =2{e^{2\theta}-e^{-2\theta}\over e^{2\theta}-e^{-2\theta}}=2 \end{align}\]A basic trigonometric identity tells us \(\cos(2\alpha)=\pm\sqrt{1-\sin^2(2\alpha)}=\pm\sqrt{1-0.6^2}=\pm0.8\). Since \(2\alpha\) is in quadrant 2, cosine is negative so we would have \(\cos(2\alpha)=-0.8\). Then we can find \(\tan(\alpha)\) with the following half angle identity, taking the positive answer since \(\alpha\) is in quadrant 1 so \(\tan(\alpha)\) will be positive.
\[\tan(\alpha)=\sqrt{1-\cos(2\alpha)\over1+\cos(2\alpha)} =\sqrt{1+0.8\over1-0.8}=\sqrt{1.8\over0.2}=\sqrt{9}=3\]How many double Mersenne primes are there conjectured to be?
Double Mersenne primes are numbers \(2^{2^p-1}-1\) for some prime \(p\). If the exponent \(2^p-1\) is not a Mersenne prime, then \(2^{2^p-1}-1\) is not prime. These double Mersenne numbers are prime for \(p=2,3,5,7\) and it is not known whether any more exist. These 4 are conjectured to be the only ones by estimating the probability of a number being prime. A number \(n\) has about a \(1/\ln(n)\) chance of being prime based on the prime number theorem. Then double Mersenne numbers grow so quickly that the chance they are prime approaches 0 faster than a geometric series, so the summation of the probabilities than these numbers are prime is tiny which is a heuristic to predict that no more double Mersenne primes exist.
The largest prime \(p\) such that neither \(p^2-1\) nor \(p^2+1\) are divisible by \(10\).
We notice that \(p=2\) would work. All other primes are odd so consider the 5 cases of congruency modulo 10.
| \(\mod10\) | \(p^2-1\) | \(p^2+1\) |
|---|---|---|
| 1 | 0 | 2 |
| 3 | 8 | 0 |
| 5 | 4 | 6 |
| 7 | 8 | 0 |
| 9 | 0 | 2 |
From this, we must have \(p\equiv5\mod10\) otherwise either \(p^2-1\) or \(p^2+1\) would be divisible by 10. The only such prime for which this is possible is \(p=5\).
This can be trivially evaluated to get 6. Another way to see it is that \(\sqrt{36}=6\) is a fixed point of \(\sqrt{30+x}\) so no matter how many times this function is applied (3 in this case), the result stays fixed at 6.
First rewrite \(f(y)\) in a way that is easier to integrate by using the geometric sum formula.
\[f(y)=\sum_{i=0}^\infty\left({y\over2}\right)^i={1\over1-{y\over2}} ={1\over{2-y\over2}}={2\over2-y}\]Note that this formula is valid only when \(f(y)\) converges, which is when \(\left|{y\over2}\right|<1\Rightarrow|y|<2\) so the integration bounds are valid. Next we can evaluate the integral by substituting \(z=2-y\).
\[S=\int_{y=-1}^{y=1}{2dy\over2-y}=\int_{z=3}^{z=1}{-2dz\over z} =2\int_{z=1}^{z=3}{dz\over z}=2\ln|z|\Big|_1^3=2\ln(3)\]Finally, \(x=e^S=(e^{\ln(3)})^2=3^2=9\).
We need to find the perimeter of the shaded area, marked in light blue. The amount the white square is angled is not specified (and turns out to not matter) so we can mark that angle as \(\theta\) where \(0<\theta<\pi/2\). In the diagram below, we form 2 right triangles inside the white square. Both of these triangles have a hypotenuse of length 1 so their leg lengths are \(\sin(\theta)\) and \(\cos(\theta)\). It should be clear that the red part has a length of 8, so we must find the length of the lightblue segments labeld \(a,b\) and the blue segments of the smaller right triangles. By using the measurements of the right triangles inside the white square, we find \(a=1-\cos(\theta)\) and \(b=1-\sin(\theta)\).
Next, we focus on the smaller right triangles (not drawn to scale below). The orange leg lengths can be determined by using the measurements of the right triangles inside the white squares.
Now with some right triangle trigonometry, focus on the triangle shown on the left which corresponds to \(\triangle A\) in the previous diagram. We have \[\tan(\theta)={c_1\over1-\sin(\theta)}\Rightarrow c_1=(1-\sin(\theta))\tan(\theta)\] and \[h_1^2=(1-\sin(\theta))^2+c_1^2 =(1-\sin(\theta))^2(1+\tan^2(\theta))=(1-\sin(\theta))^2\sec^2(\theta)\] so \(h_1=(1-\sin(\theta))\sec(\theta)\). Then we add them to get \[h_1+c_1=(1-\sin(\theta))(\sec(\theta)+\tan(\theta)) =(1-\sin(\theta)){1+\sin(\theta)\over\cos(\theta)} ={1-\sin^2(\theta)\over\cos(\theta)}=\cos(\theta)\]
Next, focus on the other triangle, \(\triangle B\). We have \[\tan(\theta)={1-\cos(\theta)\over c_2}\Rightarrow c_2\tan(\theta)=1-\cos(\theta)\Rightarrow c_2=(1-\cos(\theta))\cot(\theta)\] and \[h_2^2=(1-\cos(\theta))^2(1+\cot^2(\theta)) =(1-\cos(\theta))^2\csc^2(\theta)\] so \(h_2=(1-\cos(\theta))\csc(\theta)\). Then \[h_2+c_2=(1-\cos(\theta))(\csc(\theta)+\cot(\theta)) =(1-\cos(\theta)){1+\cos(\theta)\over\sin(\theta)} ={1-\cos^2(\theta)\over\sin(\theta)}=\sin(\theta)\]
Now to complete the work for the nontrivial part of the perimeter, we add \(a+b\) with the lengths of the blue sides of the smaller right triangles. \[(1-\cos(\theta))+(1-\sin(\theta))+\cos(\theta)+\sin(\theta)=2\] Finally add this 2 to the 8 we found earlier and the perimeter is 10. From this work, we can also see that the length of the nontrivial parts of the perimeter is independent of \(\theta\) (as long as it is in the range that makes sense for this problem).
An alternative solution is drawing the following isosceles triangles. They are isosceles because 2 of the side lengths are the same. The blue segments are each 1 so those triangles with the blue and orange edges are isosceles. In each of those triangles, the angles next to the orange edge are of equal measure. Using complementary angles, the angles on the other side of each orange edge have equal measure so the triangles formed with the orange, red, and green edges are isosceles. From this, we can replace 2 segments of the perimeter with the green and red edges at the far right. This tells us that the perimeter we are looking for is the perimeter of the 2x2 outer square added to the 2 blue edges of the white/rotated square, so the perimeter is 10.
This is an integral with \(e^{-t^2}\) like the one that shows up in the normal distribution. We can evaluate this by converting to polar coordinates from squaring the integral and combining them into a double integral because of independent variables.
\[\left(\int_0^\infty e^{-4t^2\over625}dt\right)^2 =\left(\int_0^\infty e^{-4v^2\over625}dv\right) \left(\int_0^\infty e^{-4u^2\over625}du\right) =\left(\int_0^\infty\int_0^\infty e^{-4(u^2+v^2)\over625}dvdu\right)\]Next, we convert the area \([0,\infty]\times[0,\infty]\) in rectangular coordinates to \([0,\pi/2]\times[0,\infty]\) in polar coordinates.
\[\begin{align}& =\int_0^{\pi/2}\int_0^\infty e^{-4r^2\over625}rdrd\theta ={\pi\over2}\int_0^\infty e^{-4r^2\over625}rdr\\& ={\pi\over2}{625\over2\cdot4} \int_\infty^0e^{-4r^2\over625}d\left({-4r^2\over625}\right) ={625\pi\over16}e^{-4r^2\over625}\Big|_\infty^0={625\pi\over16} \end{align}\]Now we take the square root and round, finding \({25\over4}\sqrt{\pi}\approx11\).
We have 2 planes and a sphere. From the equations for the 2 planes, we can replace \(y\) and \(z\) with expressions in terms of \(x\). Subtract 3 times the first plane from the second and we find \(x+2z=28\). From 2 times the first plane, \[2y=72-2x-2z=72-2x-(28-x)=44-x\] Next multiply the sphere equation by 4 and we can substitute these \[\begin{align} &(2x)^2+(2y)^2+(2z)^2=1856 \\& (2x)^2+(44-x)^2+(28-x)^2=1856 \\& 4x^2+44^2-88x+x^2+28^2-56x+x^2=1856\\& 6x^2-144x=-864 \\& x^2-24x+144=0\\& (x-12)^2=0\end{align}\] so we find \(x=12\). Using the 2 planes, we can also find \(y=16,z=8\) and these satisfy the sphere equation as well. The 2 planes intersect at a line, which intersects the sphere at exactly 1 point.
One way to see the solution is to factor \(312=24\cdot13\) and \(264=24\cdot11\) and notice that \(11+13=24\). But to solve this with algebra, first factor the equations to find: \[\begin{align}&x(x+y)=24\cdot13\\&y(x+y)=24\cdot11\end{align}\] Next, notice that each of the quantities \(x,y,x+y\) are nonzero. Then if we divide the 2 equations we get \({x\over y}={13\over11}\). So substitute \(y={11\over13}x\) into the first equation and we have: \[x\left(x+{11\over13}x\right)={24\over13}x^2=24\cdot13 \Rightarrow x^2=13^2 \Rightarrow x=\pm13\] This tells us the answer is \(x=13\). The 2 solutions to this system are \(x=13,y=11\) and \(x=-13,y=-11\).
Another useful observation is that by adding the 2 equations, we get \[x^2+2xy+y^2=576 \Rightarrow (x+y)^2=\pm24\] which leads to the same solutions by using the equations written as factored above.
What is the expected value of the total pips from a throw of four fair-sided dice?
Each dice throw is independent so let \(X\) be a random variable for 1 dice throw. Then \[E(X)={1\over6}(1+2+3+4+5+6)={7\over2}\] because each roll has the same probability. Since there are 4 dice, we are looking for \(4\cdot E(X)=14\).
Due to the nature of this problem, I will always use \(\cdot\) to explicitly show multiplication, otherwise digits are concatenated. First square each side to find \[abcd=(10\cdot a+d)^2=100\cdot a^2+20\cdot a\cdot d+d^2\] Since \(0<d\leq9\), we have \[100\cdot a^2+20\cdot a\cdot d+d^2 \leq 100\cdot a^2+180\cdot a+81\] and we can show \[100\cdot a^2+180\cdot a+81 <a000=1000\cdot a\] Subtracting, we have the quadratic inequality \[100\cdot a^2-820\cdot a+81<0\] which can be shown true by finding the zeroes which are \({1\over10},{81\over10}\). This shows that \((ad)^2\) cannot be big enough which \(a<9\). This gives us \(a=9\). Then we need \(d^2\equiv d\mod10\) so the ones digits match. The only possible \(d\) that satisfy this modular equation are \(1,5,6\). By testing each, we find that \(96^2=9216\) so the answer is \(x=a+d=9+6=15\).
First rewrite this with exponent base 2 \[2^{3x}+2^{2x-16}=2^{2x+8}+2^x\] Then multiply each side by \(2^{-x}\) which subtracts \(x\) from each exponent \[2^{2x}+2^{x-16}=2^{x+16}+1\] then if we multiply each side by \(2^{16}\), it might look a little nicer \[2^{2x+16}+2^x=2^{x+32}+2^{16}\] Now with some not as analytical thinking, what if the terms corresponded to each other in pairs? If we make \(2^x=2^{16}\), then \(x=16\) works for the other pair of terms. This can be shown to be the only solution but that is kind of hard to do rigorously. By plotting the log of each side, it shows that they have 1 intersection point.
Here we have used squaring to compute \(2^8\mod19\) for the 2s factored from 1000. Then we leave the 5s since the power 125 is close to 128, a power of 2. Next, square 7 times to compute \(9^{128}\mod19\). We need to subtract 3 from that exponent which can be done with the modular inverse of 9 modulo 19 which is 17 (or -2). This allows us to evaluate the answer without a whole lot of tedious work. The most tedious part here is computing \(9^{128}\mod19\).
\(y\) is a two-digit decimal number that is not divisible by 10. \(x\) is the sum of the digits of \(99y\).
Let \(y=10a+b\) for the digits \(0<a,b\leq9\). Then we can compute \(100y-y\) by subtraction with borrowing. \[\begin{array}{c|c|c|c|c} &&&9&10\\&&b-1&\cancel{10}&\\&a&\cancel{b}&0&0\\ -&&&a&b\\\hline&a&b-1&9-a&10-b\end{array}\] Now we just add the 4 digits of the result to get \(a+(b-1)+(9-a)+(10-b)=18\).
For any function \(f:\mathbb{R}^+\to\mathbb{R}^+\) such that \(\forall x,y:x^2(f(x)+f(y))=(x+y)f(f(x)y)\) find \(f(0.05)\).
First substitute \(x=y\) to see what happens. \[2x^2f(x)=2xf(f(x)x) \Rightarrow xf(x)=f(xf(x))\] This shows that all possible values of \(xf(x)\) are fixed points of \(f\). Notice how we have \(x+y\) on one side and \(f(x)+f(y)\) on the other side. We can remove one usage of \(f\) if we make \(x\) and \(y\) fixed points. Make the substitution \(x=tf(t)\) and \(y=sf(s)\) for some \(s,t\). This allows us to divide the \(x+y\) and \(f(x)+f(y)\) parts. \[\begin{align}&t^2f(t)^2(f(tf(t))+f(sf(s)))=(tf(t)+sf(s))f(f(tf(t))sf(s))\\ &t^2f(t)^2(tf(t)+sf(s))=(tf(t)+sf(s))f(tf(t)sf(s))\\&t^2f(t)^2=f(tf(t)sf(s)) \end{align}\] If instead we substitute \(x=sf(s)\) and \(y=tf(t)\) we get a similar expression. \[s^2f(s)^2=f(tf(t)sf(s))\] Now if we substitute \(s=1\) in both, the right sides are equal so we find \(f(1)^2=t^2f(t)^2 \Rightarrow f(1)=tf(t)\) which we can also write as \(f(t)={f(1)\over t}\). All that is left is to find what \(f(1)\) is. \[f(1)=tf(t)=f(tf(t))=f(f(1))={f(1)\over f(1)}=1\] Therefore \(f(t)={1\over t}\) is the only solution and \(f(0.05)=20\).
This one is a bit difficult and takes quite a bit of work but once you see how to do it, it is mostly familiar partial fractions and telescoping series. First let's focus on the inner summation, factoring out \(12/n\) \[{12\over n}\sum_{m=1}^\infty{1\over m(m+n+2)}\] then using partial fractions to decompose \[\begin{align}&{A\over m}+{B\over m+n+2}={1\over m(m+n+2)}\\ &A(m+n+2)+Bm=1\\&(A+B)m+A(n+2)=1\\ &A={1\over n+2},B={-1\over n+2}\end{align}\] Then apply this decomposition to the summation of \(m\) \[\begin{align}&{12\over n}\sum_{m=1}^\infty \left[{1/(n+2)\over m}-{1/(n+2)\over m+n+2}\right] ={12\over n(n+2)}\sum_{m=1}^\infty\left[{1\over m}-{1\over m+n+2}\right]\\ &{12\over n(n+2)}\left[ \begin{array} {1\over1}&+&{1\over2}&+&\ldots&+&{1\over n+2}& +&{1\over n+3}&+&{1\over n+4}&+&\ldots\\ &&&&&&&-&{1\over n+3}&-&{1\over n+4}&-&\ldots\\ \end{array}\right]\\& ={12\over n(n+2)}\sum_{m=1}^{n+2}{1\over m}\end{align}\] We still have to sum this over \(n=1\) to \(\infty\). First we can apply partial fraction decomposition to \({12\over n(n+2)}\). It is just like the partial fraction decomposition we already did but replace 12 with 1, remove the \(n\), and change each remaining \(m\) to \(n\). Then combining this with summing over \(n\), we have \[\sum_{n=1}^\infty{12\over n(n+2)}\sum_{m=1}^{n+2}{1\over m} =6\sum_{n=1}^\infty\left[{1\over n}-{1\over n+2}\right] \sum_{m=1}^{n+2}{1\over m} =6\sum_{n=1}^\infty\left[{1\over n}-{1\over n+2}\right]S(n)\] where \(S(n)=\sum_{m=1}^{n+2}{1\over m}\). Now we use telescoping series reasoning on the \(1/n-1/(n+2)\) part \[\begin{align}&6\left[\begin{array} {1\over1}S(1)&+&{1\over2}S(2)&+&{1\over3}S(3)&+&{1\over4}S(4)&+&\ldots\\ &&&-&{1\over3}S(1)&-&{1\over4}S(2)&-&\ldots\\ \end{array}\right]\\ &=6\left[S(1)+{1\over2}S(2)+\sum_{r=3}^\infty{1\over r} \left(S(r)-S(r-2)\right)\right]\\ &=6\left[{11\over6}+{1\over2}\cdot{25\over12} +\sum_{r=3}^\infty{1\over r} \left({1\over r+1}+{1\over r+2}\right)\right]\\ &=11+{25\over4}+6\sum_{r=3}^\infty{1\over r(r+1)} +6\sum_{r=3}^\infty{1\over r(r+2)}\end{align}\] We are almost to the solution. Next we apply partial fractions to these 2 summations which is just like what we did before. Then they are going to be familiar telescoping series. \[\sum_{r=3}^\infty{1\over r(r+1)} =\sum_{r=3}^\infty\left({1\over r}-{1\over r+1}\right) ={1\over3}\] \[\sum_{r=3}^\infty{1\over r(r+2)} ={1\over2}\sum_{r=3}^\infty\left({1\over r}-{1\over r+2}\right) ={1\over2}\left({1\over3}+{1\over4}\right)={7\over24}\] Then we use these to finalize our answer. \[11+{25\over4}+6{1\over3}+6{7\over24}={69\over4}+6{5\over8} ={69\over4}+{15\over4}={84\over4}=21\]
Find the radius of the circle \[{x^2\over4}-x+{y^2\over4}-2y-116=0\]
\[x^2-4x+y^2-8y-464=0\] \[(x-2)^2-2^2+(y-4)^2-4^2-464=0\] \[(x-2)^2+(y-4)^2=484=22^2\] Therefore the radius is 22 and the circle is centered at (2,4).
\[\begin{align}&=5^2+(4+3)(4-3)-(2^2-1)^2=5^2-(2^2-1)^2+(4+3)(4-3) =5^2-(4-1)^2+(4+3)(4-3)\\&=5^2-3^2+(4+3)(4-3)=(5+3)(5-3)+(4+3)(4-3) =8\cdot2+7\cdot1=16+7=23\end{align}\]
Notice how the exponent bases are 2 and 4, which suggests the bottom is like a quadratic polynomial. First factor \(2^{-2}\) from the numerator and then use \(u=2^n\) to observe that it is just like a rational function. \[\left({1\over4}\sum_{n=1}^\infty {2^n\over8\cdot4^n-6\cdot2^n+1}\right)^{-1} \qquad{u\over8u^2-6u+1}\] Then we can factor \(8u^2-6u+1=(2u-1)(4u-1)\) which prompts us to do the following partial fraction decomposition. \[\begin{align}&{A\over2u-1}+{B\over4u-1}={u\over8u^2-6u+1}\\ &A(4u-1)+B(2u-1)=u\\&(4A+2B)u-A-B=u\\&4A+2B=1,\quad-A-B=0\\ &A={1\over2},\quad B={-1\over2}\\&{2^n\over8\cdot4^n-6\cdot2^n+1}= {1/2\over2\cdot2^n-1}-{1/2\over4\cdot2^n-1}\end{align}\] Now we use this to write the summation as \[\left({1\over4}\sum_{n=1}^\infty {2^n\over8\cdot4^n-6\cdot2^n+1}\right)^{-1} =8\left(\sum_{n=1}^\infty{1\over2\cdot2^n-1} -\sum_{n=1}^\infty{1\over4\cdot2^n-1}\right)^{-1}\] The 2 summations look similar. Changing \(4\cdot2^n=2\cdot2^{n+1}\) gives us \[8\left(\sum_{n=1}^\infty{1\over2\cdot2^n-1} -\sum_{n=1}^\infty{1\over2\cdot2^{n+1}-1}\right)^{-1}\] We can substitute \(m=n+1\) into the summation to get \[8\left(\sum_{n=1}^\infty{1\over2\cdot2^n-1} -\sum_{m=2}^\infty{1\over2\cdot2^m-1}\right)^{-1}\] Now the summations are the same except for the indexes. Notice how subtracting leaves us with only the \(n=1\) term which is \(1/3\). The answer is \(8(1/3)^{-1}=8\cdot3=24\).
There are a few ways to solve this problem. First I will present the first way I tried which gives a tough system of equations to solve. Draw the 2 right triangles shown below.
The side length of the gray square is \(\sqrt{75}=5\sqrt{3}\). The diagonal of the big square is \(t\sqrt{2}\). For the red right triangle, we have the equation \[r^2+(2r+5\sqrt{3})^2=(t\sqrt{2})^2\] and for the green one \[t^2+(t-\sqrt{r^2+75})^2=(2r\sqrt{2})^2\] This system is difficult to solve, but plotting it in a graphing calculator shows there is a single solution. Next we will see easier methods for finding the answer.
Here, we draw in orange a square around the big square, which is like the one used in a proof of the Pythagorean theorem. Using the bottom orange edge and the vertical line shared by the 3 smaller squares, we find \[a+b=2b+r=2r+5\sqrt{3}\] where \(r=\sqrt{x}\). Next we draw the red right triangle which also has legs of lengths \(a,b\) like the right triangles on the outside with the orange legs. So we can see that \[b={r+5\sqrt{3}\over2}\] Then \[a=r+b={3r+5\sqrt{3}\over2}\] Now we use similar triangles. The yellow shaded triangle is similar to one of the right triangles with legs \(a,b\) so \[\begin{align} &{a\over b}={5\sqrt{3}\over r}={3r+5\sqrt{3}\over r+5\sqrt{3}}\\ &5r\sqrt{3}+75=3r^2+5r\sqrt{3}\\&3r^2=75\Rightarrow r=5\Rightarrow x=r^2=25 \end{align}\]
An alternative solution uses the sum of angles for tangent. See the 2 angles in the diagram below.
We can find \[\tan(\alpha)={r\over5\sqrt{3}}, \qquad\tan(\beta)={r\over2r+5\sqrt{3}}\] Observe that \(\alpha+\beta=\pi/4\) so \(\tan(\alpha+\beta)=\tan(\pi/4)=1\) Using a trigonometric identity, \[\begin{align}&\tan(\alpha+\beta)={\tan(\alpha)+\tan(\beta) \over1-\tan(\alpha)\tan(\beta)}={{r\over5\sqrt{3}}+{r\over2r+5\sqrt{3}} \over1-{r\over5\sqrt{3}}{r\over2r+5\sqrt{3}}}=1\\ &1-{r\over5\sqrt{3}}{r\over2r+5\sqrt{3}} ={r\over5\sqrt{3}}+{r\over2r+5\sqrt{3}}\\ &5\sqrt{3}(2r+5\sqrt{3})-r^2=r(2r+5\sqrt{3})+5r\sqrt{3}\\ &10r\sqrt{3}+75-r^2=2r^2+10r\sqrt{3}\\ &3r^2=75\Rightarrow r=5\Rightarrow x=r^2=25\end{align}\]
First to simplify it, let \(u=t+1\) and notice how we can use this to complete the square inside the square root and simplify. \[{52\over\pi}\int_{u=1}^{u\to\infty}{du\over u\sqrt{u^2-1}}\] This looks like a good integral to use trigonometric substitution with. The identity \(\tan^2(\theta)+1=\sec^2(\theta)\) tells us what we should pick to simplify the square root part, which is \(u=\sec(\theta)\). This substitution changes the domain to \([0,\pi/2)\) because \(\sec(0)=1\) and \(\lim_{\theta\to\pi/2^-}\sec(\theta)=\infty\). \[\begin{align}&{52\over\pi}\int_{\theta=0}^{\theta\to\pi/2^-} {\sec(\theta)\tan(\theta)d\theta\over \sec(\theta)\sqrt{\sec^2(\theta)-1}} ={52\over\pi}\int_{\theta=0}^{\theta\to\pi/2^-} {\sec(\theta)\tan(\theta)d\theta\over\sec(\theta)\tan(\theta)}\\ &={52\over\pi}\int_{\theta=0}^{\theta=\pi/2}d\theta ={52\over\pi}{\pi\over2}=26\end{align}\] The integral found after substituting \(u\) is also a familiar integral for inverse trigonometric functions so it can be solved with that formula.
Here, we can use Stirling's approximation which is \[n!\sim\sqrt{2\pi n}\left(n\over e\right)^n\] which means the ratios of these 2 approaches 1 as \(n\to\infty\). \[\begin{align}&\lim_{n\to\infty}{10n\over(10n!)^{1/n}} =\lim_{n\to\infty}{10n\over10^{1/n}(n!)^{1/n}} =\lim_{n\to\infty}{10n\over10^{1/n}(\sqrt{2\pi n})^{1/n}({n\over e})}\\ &=\lim_{n\to\infty}{10e\over10^{1/n}(\sqrt{2\pi})^{1/n}(\sqrt{n})^{1/n}} \end{align}\] Since \(10\) and \(\sqrt{2\pi}\) are constants, \(10^{1/n}\to1\) and \((\sqrt{2\pi})^{1/n}\to1\) as \(n\to\infty\). We need to focus on the remaining part which is \[\lim_{n\to\infty}(\sqrt{n})^{1/n}=\lim_{n\to\infty}n^{1\over2n} =\lim_{n\to\infty}e^{\ln(n)\over2n}=e^{\lim_{n\to\infty}{\ln(n)\over2n}} =e^0=1\] So everything in the denominator is 1, which makes the limit equal to \(10e\). The closest integer is 27.
First use the geometric series summation to evaluate that the summation is equal to \(2^{501}-2\). Then by efficient modular exponentiation, we have \(2^{501}\mod223=30\) so subtract 2 and the answer is \(28\).
Find the sum of the solutions to \(x^2+3x+16\equiv0\mod32\).
We can solve this with Hensel's lifting lemma. The modulus is \(32=2^5\) which only has 1 prime, so we do not need to use the Chinese remainder theorem. First, we find solutions \(x^2+3x+16\equiv0\mod2\) and then lift the solutions \(\mod4,8,16,32\). Both \(x\equiv0\) and \(x\equiv1\) are solutions so we have to lift these 2 solutions with the lemma. Let \(f(x)=x^2+3x+16\) and \(f'(x)=2x+3\).
By Hensel's lemma, if \(f(a_k)\equiv0\mod p^k\) for some prime \(p\) and integer \(k\), and \(f'(a_k)\not\equiv0\mod p\), then \[a_{k+1}\equiv a_k-f(a_k)(f'(a_1))^{-1}\mod p^{k+1}\] is a lifted solution, \(f(a_{k+1})\equiv0\mod p^{k+1}\). First we lift the \(x\equiv0\) solution. We have \(a_1=0\) and \(f'(a_1)=3\). The table below summarizes the steps which does involve a few tedious calculations.
| \(k\) | \(a_k\) | \(f(a_k)\) | \(3^{-1}\) | \(a_{k+1}\) |
|---|---|---|---|---|
| \(1\) | \(0\) | \(16\equiv0\mod4\) | Not needed | \(0-16\cdot3^{-1}\equiv0\mod4\) |
| \(2\) | \(0\) | \(16\equiv0\mod8\) | Not needed | \(0-16\cdot3^{-1}\equiv0\mod8\) |
| \(3\) | \(0\) | \(16\equiv0\mod16\) | Not needed | \(0-16\cdot3^{-1}\equiv0\mod16\) |
| \(4\) | \(0\) | \(16\equiv16\mod32\) | \(11\) | \(0-16\cdot11\equiv16\mod32\) |
Then we lift the \(x\equiv1\) solution. We have \(a_1=1\) and \(f'(a_1)=5\). Below is a similar table.
| \(k\) | \(a_k\) | \(f(a_k)\) | \(5^{-1}\) | \(a_{k+1}\) |
|---|---|---|---|---|
| \(1\) | \(1\) | \(20\equiv0\mod4\) | Not needed | \(1-20\cdot5^{-1}\equiv1\mod4\) |
| \(2\) | \(1\) | \(20\equiv4\mod8\) | \(5\mod8\) | \(1-20\cdot5^{-1}\equiv5\mod8\) |
| \(3\) | \(5\) | \(56\equiv8\mod16\) | \(13\mod16\) | \(5-56\cdot5^{-1}\equiv13\mod16\) |
| \(4\) | \(13\) | \(224\equiv0\mod32\) | Not needed | \(13-224\cdot5^{-1}\equiv13\mod32\) |
So the solutions add to \(16+13=29\).
The tricky part for this one is evaluating the summation. Consider the binomial expansion \[(1+x)^{30}=\sum_{j=0}^{30}{30\choose j}x^j\] When \(x=1\), then the terms for \(j=0,3,6,\ldots,30\) match what we are looking for. So what we need is choices of \(x\) so we can add/subtract to cancel the terms when \(j\) is not divisible by 3. It turns out that the cube roots of unity are a good choice. We would evaluate the sum \[(1+1)^{30}+(1+e^{2i\pi/3})^{30}+(1+e^{-2i\pi/3})^{30} =\sum_{j=0}^{30}{30\choose j}(1+e^{2i\pi j/3}+e^{-2i\pi j/3})\] On the right, the imaginary parts cancel so we have \[1+e^{2i\pi j/3}+e^{-2i\pi j/3}=1+2\cos\left({2\pi j\over3}\right)= \begin{cases}3&j\equiv0\mod3\\0&\text{otherwise}\\\end{cases}\] This is exactly the summation we are looking for. On the left, we have \[1+e^{2i\pi/3}=1+{-1\over2}+{\sqrt{3}\over2}i={1\over2}+{\sqrt{3}\over2}i =e^{\pi i/3}\] and \[1+e^{-2i\pi/3}=1+{-1\over2}+{-\sqrt{3}\over2}i ={1\over2}+{-\sqrt{3}\over2}i=e^{-\pi i/3}\] Then the left side is \[2^{30}+(e^{\pi i/3})^{30}+(e^{-\pi i/3})^{30} =2^{30}+e^{10\pi i}+e^{-10\pi i}=2^{30}+2\] Now putting this together, we have \[\log_2\left(\sum_{k=0}^{10}3{30\choose3k}-2\right)=\log_2(2^{30}+2-2) =\log_2(2^{30})=30\]