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Find the number of proper normal subgroups of \(A_5\), the alternative group on five elements.
For \(n\geq5\), the alternating group \(A_n\) is a simple group, whose only normal subgroups are \(\{e\}\) and \(A_n\). Only \(\{e\}\) would be a proper subgroup so the answer is 1.
If \(N\) is a normal subgroup of \(G\) (\(N \trianglelefteq G\)), then all \(n\in N\) and \(g\in G\) satisfy \(gng^{-1}\in N\). Proving that \(A_5\) is simple seems a bit challenging. Some relevant theorems are that \(|N|\) is the sum of sizes of some conjugacy classes of \(G\) and \(|N|\) divides \(|G|\).
We will be adding terms so we can try an arctangent addition formula: \[\arctan(a)+\arctan(b)=\arctan\left({a+b\over1-ab}\right)\] This formula can be derived by substituting \(\alpha=\arctan(a)\) and \(\beta=\arctan(b)\) into the sum formula: \[\tan(\alpha+\beta)={\tan(\alpha)+\tan(\beta)\over1-\tan(\alpha)\tan(\beta)}\] Now we evaluate a few partial sums to look for a pattern. The first term is \(\arctan(1)\). Then \[\arctan(1)+\arctan\left({1\over(1)^2+(1)+1}\right)=\arctan(1) +\arctan\left({1\over3}\right)\] \[=\arctan\left(1+{1\over3}\over1-{1\over3}\right) =\arctan\left({4/3\over2/3}\right)=\arctan(2)\] One more term: \[\arctan(2)+\arctan\left({1\over(2)^2+(2)+1}\right)=\arctan(2) +\arctan\left({1\over7}\right)\] \[=\arctan\left({2+{1\over7}\over1-{2\over7}}\right) =\arctan\left({15/7\over5/7}\right)=\arctan(3)\] So the pattern appears to be that the sum from \(n=0\) to \(N\) is \(\arctan(N+1)\). We already have a base case and can prove by induction. \[\arctan(N)+\arctan\left({1\over N^2+N+1}\right) =\arctan\left({N+{1\over N^2+N+1}\over1-{N\over N^2+N+1}}\right)\] \[=\arctan\left({N(N^2+N+1)+1\over N^2+N+1-N}\right) =\arctan\left({N^3+N^2+N+1\over N^2+1}\right)\] That numerator can be factored by grouping \[N^3+N^2+N+1=N^2(N+1)+(N+1)=(N^2+1)(N+1)\] So everything simplifies to \(\arctan(N+1)\) completing the proof. The summation is taken as \(N\to\infty\) so we get \(\pi/2\). Therefore \(x=2\).
The equations for \(r,s\) are identical in structure so we just solve one to find possible values for each. Multiply each side by \(e^r\). \[e^{2r}+16=10e^r \Rightarrow e^{2r}-10e^r+16=0 \Rightarrow (e^r-2)(e^r-8)=0\] Solving it like a quadratic equation, we find \(e^r=2\) or \(e^r=8\) so \(r=\ln(2)\) or \(r=\ln(8)=3\ln(2)\). There is a mistake because we should get an answer of \(3\). Either they meant \(r>s\) or \(x=s/r\). With either correction, we find \(x=3\) after picking \(r,s\) from these 2 possible values.
First we let \(s\) be the side length of the square and draw a right triangle.
Since the square sits in the middle of the 2 circles, we have a right triangle with longer leg \(5-s/2\) and shorter leg \(5-s\). The hypotenuse is a radius so it is \(5\). Now use the pythagorean theorem. \[(5-s/2)^2+(5-s)^2=5^2 \Rightarrow 25-5s+s^2/4+25-10s+s^2=25\] \[\Rightarrow (5/4)s^2-15s+25=0 \Rightarrow 5s^2-60s+100=0\] \[\Rightarrow s^2-12s+20=0 \Rightarrow (s-2)(s-10)=0\] If \(s=10\) then the square just fills the whole space between the circle centers and the right triangle we drew turns into \(5^2+0^2=5^2\). So we must have \(s=2\) for the side length and an area of \(x=4\).
Find the digit \(O\).
Going through the steps of multiplication, we first see that \[O\times O\equiv O\ \text{mod}\ 10\] This restricts possibilities to \(O=1,5,6\). First we try \(O=1\).
If \(O=1\) then there are no carries for the first row, and \(TGO=GO\) so \(T=0\). But then on the second row, \(T\times O=T\equiv O\) is not possible so \(O\neq1\).
If \(O=6\), then \(O\times O\equiv O\) gives us a carry of 3 for the next step. \(O\times G+3=6G+3\equiv G\). But then \(5G\equiv-3\) is impossible modulo 10 so \(O\neq6\).
So the solution is \(O=5\) and we can work out the full puzzle. We carry 2 from the first \(O\times O\equiv O\) step. Then for \(O\times G+2\equiv G \Rightarrow4G\equiv-2\), we can find \(G=2\) and \(G=7\) as possibilities. Then since \(5\times G+2=TG\), we can match these to a value of \(T\), leaving us with either \((T,G)=(1,2)\) or \((T,G)=(3,7)\).
Moving on to the second row, \(T\times O\equiv O\) is compatible with either possible value of \(T\). Next, \(T\times G\equiv G\) but it cannot carry because we have no more digits to the left. It turns out \((T,G)=(1,2)\) works here but not \((T,G)=(3,7)\). So we have \(O=5,G=2,T=1\) and can fully work out the puzzle to find \(F=3,R=7\).
\[\begin{array}{ccc} &2&5\\ \times&1&5\\ \hline 1&2&5\\ 2&5&\\ \hline 3&7&5\\ \end{array}\]It turns out a problem like this is very difficult to solve in general. We can find some trivial solutions if we try \(x=1\) (finding \(y=7850\)) or \(y=1\) (finding \(x=-7848\)). Otherwise, we can search.
| \(x\) | Explanation or work |
|---|---|
| \(x=2:y^2-2^y\) | We can't make \(y\) large enough to get \(y^2\) near \(7849\)
before \(2^y\) grows way too fast. \(2^2-2^2=0\) \(3^2-2^3=1\) \(4^2-2^4=0\) \(5^2-2^5=-7\) \(6^2-2^6=-28\) |
| \(x=3:y^3-3^y\) | Similarly, \(3^y\) grows too quickly. \(2^3-3^2=-1\) \(3^3-3^3=0\) \(4^3-3^4=-17\) \(5^3-3^5=-118\) |
| \(x=4:y^4-4^y\) | Once again, the quick growth of \(4^y\) dominates. \(2^4-4^2=0\) \(3^4-4^3=17\) \(4^4-4^4=0\) \(5^4-4^5=-399\) |
| \(x=5:y^5-5^y\) | The numbers get a little bigger, but they still get negative fast. \(2^5-5^2=7\) \(3^5-5^3=118\) \(4^5-5^4=399\) \(5^5-5^5=0\) \(6^5-5^6=-7849\) |
| \(x=6:y^6-6^y\) | Above we saw the constant we were looking for. Here we find it again. \(2^6-6^2=28\) \(3^6-6^3=513\) \(4^6-6^4=2800\) \(5^6-6^5=7849\) \(6^6-6^6=0\) \(7^6-6^7=-162287\) |
So a solution we found is \(x=6\) and \(y=5\). Showing rigorously that this is the only solution is challenging. Below are some observations.
>>> [y**6-6**y for y in range(2,8)] [28, 513, 2800, 7849, 0, -162287] >>> [y**7-7**y for y in range(2,9)] [79, 1844, 13983, 61318, 162287, 0, -3667649] >>> [y**8-8**y for y in range(2,10)] [192, 6049, 61440, 357857, 1417472, 3667649, 0, -91171007] >>> [y**9-9**y for y in range(2,11)] [431, 18954, 255583, 1894076, 9546255, 35570638, 91171007, 0, -2486784401]
We may be able to show we have searched far enough if past a certain choice of \(x\), we can show it increases, then decreases to 0, and keeps decreasing. We would only have to check near \(2^x-x^2\) and \((x-1)^x-x^{x-1}\). Furthermore, we may be able to show that past a certain point, \(3^x-x^3>7849\) and \((x-1)^x-x^{x-1}>7849\) so we do not have to check everything in the middle.
Find the smallest natural number \(x\) such that \(10!\times x\) is a square number.
Consider the factorization of \(10!\). \[10!=2^8\times3^4\times5^2\times7\] Square numbers have even exponents for all their prime factors, so we only need to multiply to make the exponents even. Multiplying by \(7\) is sufficient here since it is the only prime with an odd exponent, so \(x=7\).
Let \(r\) be the radius of the smaller circle and let \(R\) be the radius of the larger circle. We can apply the pythagorean theorem on the right triangle. \[3^2+r^2=(R+r)^2\Rightarrow3^2=R^2+2rR=R(R+2r)\] We have 1 equation and 2 variables, so we can't exactly solve for the variables. But notice that one side of the rectangle is \(R\) and the other is \(R+2r\). So what we have is exactly the rectangle area, which is \(x=9\).
For the parabola \(y=ax^2+bx+c\) that passes through \((0,12),(-2,4),(2,4)\) find \(a+b+c\).
In general, we can make 3 equations and solve a linear system. But here we have symmetry to do this. Clearly \((0,12)\) is the vertex because going 2 steps left or right decreases \(y\) from 12 to 4. The parabola is \(y=a(x-0)^2+c\) for value we need to find. At \(x=0\), we find \(c=12\). At \(x=2\), we find \(4=a(-2)^2+12\Rightarrow4=4a+12\Rightarrow a=-2\). We find the equation \(y=-2x^2+12\) with \(a=-2,b=0,c=12\). We can check that all 3 points are on it and \(x=a+b+c=10\).
Let \(y=4x/(x-9)\) and then we have a quadratic equation. \[y^2+88=26y\Rightarrow y^2-26y+88=0\Rightarrow(y-22)(y-4)=0\] Now try solving for \(x\) with each of these values. \[{4x\over x-9}=4\Rightarrow4x=4(x-9)\Rightarrow0=-36\Rightarrow\bot\] That gives us no solution, but the other one \[{4x\over x-9}=22\Rightarrow4x=22(x-9)=22x-198\Rightarrow18x=198 \Rightarrow x=11\]
Use the chinese remainder theorem. \(33=3\cdot11\) so we compute this expression module 3 and 11. For mod 3, we also reduce the exponent mod \(\phi(3)=2\), and see that 2025 is divisible by 3. \[2025^{2025}\equiv0^1\equiv0\] For mod 11, 2025 is congruent to 1 and we reduce the exponent mod 10 \[2025^{2025}\equiv1^5\equiv1\] So now we know the solution is \(x\) where \(x\equiv0\ \text{mod}\ 3\) and \(x\equiv1\ \text{mod}\ 11\). There are few enough numbers to look at a cycle and find \(x=12\).
A class averaged \(6\) out of \(20\) on a quiz, so the teacher gave half credit on all missed questions. What is the adjusted average?
The average person missed \(14\) questions so they gain \(7\) on average, which brings the average up by \(7\) to \(6+7=13\).
Substitute \(s=45/r\) to get an equation with a single variable. \[r^2-\left({45\over r}\right)^2=56\] Multiply \(r^2\) to get a quadratic in \(r^2\) \[r^4-45^2=56r^2\] The numbers are large but this can be factored \[(r^2+25)(r^2-81)=0\] In real numbers, we only can get 0 from \(r=\pm9\). The constraint is \(r>0\) so we pick \(r=9\). Then \(s=5\). We can verify that this solves the equations and then find \(x=r+s=9+5=14\).
Get all the linear factors in the same form to make it a litte easier to see what we do next. \[x=-(t-3)(t-2)(t-1)(t)(t+1)(t+2)\] Now pick pairs of terms whose constant part adds to \(-1\), so we will end up with \(t^2-t\) and can substitute \(t^2-t=5\) from the quadratic we are given. \[(t-3)(t+2)=t^2-t-6=5-6=-1\] \[(t-2)(t+1)=t^2-t-2=5-2=3\] \[(t-1)(t)=t^2-t=5\] Then \(x=-(-1)(3)(5)=15\)
Separate some things and see that we have powers of \(x\) that can be combined. \[x^x=2^{2x}\cdot2^{32}=4^x\cdot2^{32} \Rightarrow\left({x\over4}\right)^x=2^{32}\] Let \(y=x/4\). Then \[y^{4y}=2^{32}\Rightarrow y^y=2^8=256\] This is easier to look for a solution in and we find \(y=4\) so \(x=16\).
First take a factor of \(-16\) and then use long division on \(x^4/(x^2-9)\).
| \(x^2\) | \(+0x\) | \(+9\) | |||||
| \(x^2\) | \(+0x\) | \(-9\) | \(x^4\) | \(+0x^3\) | \(+0x^2\) | \(+0x\) | \(+0\) |
| \(-\) | \((x^4\) | \(+0x^3\) | \(-9x^2)\) | ||||
| \(9x^2\) | \(+0x\) | \(+0\) | |||||
| \(-\) | \((9x^2\) | \(+0x\) | \(-81)\) | ||||
| \(81\) | |||||||
So the quotient is \(x^2+9\) and remainder \(81\). \[\int_0^2{16x^4dx\over9-x^2}=-16\int_0^2{x^4dx\over x^2-9} =-16\int_0^2\left(x^2+9+{81\over x^2-9}\right)dx\] \[=\left[{-16x^3\over3}\right]_0^2+\left[-144x\right]_0^2 -1296\int_0^2{dx\over x^2-9}\] \[={-128\over3}-288-1296\int_0^2{dx\over(x-3)(x+3)}\] Apply partial fractions \[\int_0^2{dx\over(x-3)(x+3)} =\int_0^2\left({1/6\over x-3}-{1/6\over x+3}\right)dx\] \[={1\over6}\left[\ln|x-3|-\ln|x+3|\right]_0^2 ={1\over6}(\ln(1)-\ln(5)-\ln(3)+\ln(3))=-{\ln(5)\over6}\] Then use this in what we had before to finish the problem: \[{-128\over3}-288+1296{\ln(5)\over6}=216\ln(5)-{992\over3} \approx16.972\approx17\]
Evaluate the integral for \(y\). This requires a bit of a symmetry trick. First factor away the \(3\)s, then divide both the numerator and denominator by \(t^4\). \[\int_0^\infty{t^4+t^2\over(3t^4-3t^2+3)^2}dt ={1\over9}\int_0^\infty{1+t^{-2}\over(t^2-1+t^{-2})^2}dt\] Next we substitute \(u=t-t^{-1}\). This changes the bounds to \((-\infty,+\infty)\). Notice that \(du=(1+t^{-2})dt\) matches the numerator and the denominator is almost \(u^2=t^2-2+t^{-2}\). \[={1\over9}\int_{-\infty}^{+\infty}{du\over(u^2+1)^2}\] Now use the substitution \(u=\tan(\theta)\) with \(du=\sec^2(\theta)d\theta\) and the identity \(\tan^2(\theta)+1=\sec^2(\theta)\). This changes the bounds to \((-\pi/2,+\pi/2)\) \[={1\over9}\int_{-\pi/2}^{+\pi/2}{\sec^2(\theta)d\theta \over(\tan^2(\theta)+1)^2}={1\over9}\int_{-\pi/2}^{+\pi/2}{\sec^2(\theta)d\theta \over\sec^4(\theta)}\] \[={1\over9}\int_{-\pi/2}^{+\pi/2}{d\theta\over\sec^2(\theta)} ={1\over9}\int_{-\pi/2}^{+\pi/2}{\cos^2(\theta)d\theta}\] Use the identity \(\cos^2(\theta)=(1+\cos(2\theta))/2\) to finish. \[={1\over9}\int_{-\pi/2}^{+\pi/2}\left(1+\cos(2\theta)\over2\right)d\theta ={1\over9}\left({1\over2}\int_{-\pi/2}^{+\pi/2}d\theta +{1\over2}\int_{-\pi/2}^{+\pi/2}\cos(2\theta)d\theta\right)\] \[={1\over18}\left[\left[\theta\right]_{-\pi/2}^{+\pi/2} +\left[{1\over2}\sin(2\theta)\right]_{-\pi/2}^{+\pi/2}\right] ={1\over18}(\pi+0)={\pi\over18}\] Then \(x=18\).
The base 2 number is \(2^6+2^4+2^1+2^0=64+16+2+1=83\). Since there is a digit \(7\), we should find \(x\geq8\). Solving the base \(x\) equation: \[4x+7=83\Rightarrow4x=76\Rightarrow x=19\]
Find the derivative of \(f(x)=(x^3-9x^2+27x-27)(x^2-12)\) at \(x=4\).
First recognize that the cubic polynomial is just \((x-3)^3\). Then we can use the product rule to avoid lots of algebra. \[{d\over dx}f(x)={d\over dx}((x-3)^3(x^2-12))=3(x-3)^2(x^2-12)+(x-3)^3(2x)\] Then at \(x=4\) \[3(4-3)^2(4^2-12)+(4-3)^3(2\cdot4)=3\cdot1^2\cdot4+1^3\cdot8=12+8=20\]
If the length and width of a rectangle are increased by 10%, by what percent is the area increased?
If the area before is \(LW\), then the area after is \((1.1L)(1.1W)=1.21LW\). Therefore, the increase is \(21\%\).
Find the number of regular polygons with integer degree interior angles.
If there are \(n\) sides, the interior angles sum to \(180^\circ(n-2)\). Each interior angle is \[{180^\circ(n-2)\over n}={180^\circ n-360^\circ\over n} =180^\circ-{360^\circ\over n}\] So \(n\) must divide \(360\). We can count factors of \(360=2^3\cdot3^2\cdot5\) to be \((3+1)(2+1)(1+1)=24\). But \(n=1\) and \(n=2\) are excluded since a regular polygon must have at least \(3\) sides. So there are \(22\) possible.
Start with the first equation, substituting \(y=x-21\). \[2xy=117+x+(x-21)=2x+96\Rightarrow xy-x=48\Rightarrow x(y-1)=48\] Then substitute \(y-1=x-22\) and we get a quadratic equation. \[x(x-22)=48\Rightarrow x^2-22x-48=0\Rightarrow(x+2)(x-24)=0\] So \(x=-2\) (and \(y=-23\)) would not satisfy the constraints. But \(x=24\) and \(y=3\) does satisfy the constraints.
Find the sum of the squares of all complex solutions to \((z-4)^4=(z-2)^4\).
Take the square root of each side and we have plus/minus cases. \[(z-4)^2=\pm(z-2)^2\] For the plus case, expand, see that \(z^2\) cancels, so we get 1 solution. \[z^2-8z+16=z^2-4z+4\Rightarrow-4z+12=0\Rightarrow z=3\] For the minus case, we get a quadratic equation \[z^2-8z+16=-z^2+4z-4\Rightarrow2z^2-12z+20=0\Rightarrow z^2-6z+10=0\] The solutions are \[z={6\pm2i\over2}=3\pm i\] And their squares are \[z^2=9+i^2\pm3i=8\pm 3i\] So now if we add squares of all 3 solutions we have \[3^2+(8+3i)+(8-3i)=9+8+8=25\]
The root contains a perfect square \((x+2)^2\). \[x^2+4x-724=2(x+2)=2x+4\Rightarrow x^2+2x-728=0\] \[\Rightarrow(x+28)(x-26)=0\] We find \(x=-28\) is extraneous and does not satisfy the constraint. The solution is \(x=26\) which satisfies the constraints.
Find the sum of the proper divisors of \(28\).
This is a perfect number, which are numbers whose sum of proper divisors equals itself. We can verify it by factoring and summing its divisors. \(28=2^2\cdot7\) so the divisors are \(1+2+4+7+14=28\). If \(2^p-1\) is a mersenne prime then \(2^{p-1}(2^p-1)\) is a perfect number, and all even perfect numbers have this property.
Split up the fraction and notice how the terms are square roots of what we see in the equation with \(x\), so square both sides. \[7y+{1\over2y}=6\Rightarrow49y^2+{1\over4y^2}+7=36\] Then we can find \[x=49y^2+{1\over4y^2}=36-7=29\]
Find the sum of the digits plus the number of digits of \(8^8\times5^{18}\).
We can simplify this a bit and find a big power of \(10\). \[8^8\times5^{18}=2^{24}\cdot5^{18}=2^6\cdot(2\cdot5)^{18}=64\cdot10^{18}\] So the number is \(64\) followed by 18 \(0\)s. The digit sum is \(6+4\) and the number of digits is \(18+2\), so the solution is \(30\).