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Every known perfect number except \(6\) leaves remainder \(x\) when divided by \(9\).
All known perfect numbers are in the form of \(2^{p-1}(2^p-1)\) where \(2^p-1\) is a Mersenne prime. The divisors of such numbers are \(1,2,4,\ldots,2^{p-1}\) and each of these multiplied by \(2^p-1\), so they all sum to \((2^p-1)(1+2^p-1)=2^p(2^p-1)\), which means the number is perfect. It is known that all even perfect numbers are of this form, but that is much harder to prove. It is unknown whether any odd perfect numbers exist.
This form is \(6\) when \(p=2\). All other cases are for some odd \(p\geq3\). Let \(p=2k+1\) for some \(k>0\). Then
\[2^{p-1}(2^p-1)=2^{2k}(2^{2k+1}-1)=2^{4k+1}-2^{2k}\]To find its congruence modulo \(9\), let's examine \(2^0,2^1,2^2,\ldots\) modulo \(9\).
\[1,2,4,8,7,5,1,2,4,8,7,5,\ldots\]So the powers of \(2\) modulo \(9\) repeat in a cycle of length \(6\). Next we have to consider a few cases of \(k\). If \(k\equiv0\pmod{3}\) then \(4k+1\equiv1\pmod{6}\) and \(2k\equiv0\pmod{6}\) so \(2^{4k+1}-2^{2k}\equiv2^1-2^0\equiv1\pmod{9}\). If \(k\equiv1\pmod{3}\) then \(4k+1\equiv5\pmod{6}\) and \(2k\equiv2\pmod{6}\) so \(2^{4k+1}-2^{2k}\equiv2^5-2^2\equiv1\pmod{9}\). Finally if \(k\equiv2\pmod{3}\) then \(4k+1\equiv3\pmod{6}\) and \(2k\equiv4\pmod{6}\) so \(2^{4k+1}-2^{2k}\equiv2^3-2^4\equiv1\pmod{9}\). Therefore, \(x=1\).
This can be interpreted as \((((((3^{-1})^{-1})^{-1})^{-1})^{-1})^{-1}=3\).
First, we need to know the integral which can be done using integration by parts with \(u=x,v=e^x\).
\[\int xe^x=xe^x-\int e^xdx=xe^x-e^x+C=(x-1)e^x+C\]Then we evaluate the answer.
\[3\ln\left(\int_1^2xe^xdx\right)=3\ln\left((2-1)e^2-(1-1)e^1\right)=3\cdot2=6\]If we rewrite all the \(2\)s as \(1+1\), then we get:
\[x=1+\frac{6}{1+1+\frac{6}{1+1+\frac{6}{1+1+\ldots}}}=1+\frac{6}{1+x} \Rightarrow x(1+x)=(1+x)+6 \Rightarrow x^2=7\]Since \(x^2\) is the answer we are looking for, the answer is \(7\). We could also show that it doesn't make sense for \(x=-\sqrt{7}\) and conclude that \(x=\sqrt{7}\).
Let \(x\) be equal to this whole thing. Then we have \(x=6+\sqrt{x} \Rightarrow x=(x-6)^2 \Rightarrow x^2-13x+36=0 \Rightarrow (x-9)(x-4)=0\). So either \(x=4\) or \(x=9\) but of course \(x=9\) because \(x=4\) does not make sense here.
This is Chinese for "one plus two plus three plus four" so we have \(1+2+3+4=10\).
Assuming \(x\) and \(y\) are both positive real numbers, substitute the first equation into the second to get \(y^{y^{y^{11}}}=11\). Then from here, we can form arbitrarily long power towers like this: \(y^{y^{\unicode{x22F0}^{y^{11}}}}=11\). Then by taking log of both sides, we would find \(11=\log_y11 \Rightarrow y^{11}=11\) so \(x=11\).
This might not quite be the most satisfying/rigorous solution. Another way to see this is that \(y^{y^x}\) is monotonic so finding \(y=\sqrt[11]{11}\) as a solution means it is the only one so we can conclude \(x=11\).
Find the number of positive integers less than \(50000\) than are divisible by \(4,7,11,13\).
To solve this, we need the lowest common multiple. Each of these numbers are relatively prime to all the others so it's just the product of them \(4\cdot7\cdot11\cdot13=4004\). Next we would compute \(\lfloor49999/4004\rfloor=12\). The 12 such integers are \(4004,8008,\ldots,48048\).
The sum of \(x-1\) angles of a convex polygon with \(x\) sides is \(1920\).
This problem was not quite clear, but the explanation about \(x-1\) sides is included in the solution below.
With \(x\) sides, the sum of interior angles is \(180(x-2)\). The sum of \(x-1\) of the sides must be somewhere between this and \(180(x-3)\) since the measure of each angle is between 0 and 180 degrees. So next we look for \(x\) such that 1920 is between these. That happens when \(x=13\) since \(180\cdot10=1800<1920<1980=180\cdot11\). Therefore, the only possible value for \(x\) is \(13\).
Difference between min and max \(y\) values of \(r=\frac{56}{9}(1-\sin(\theta))\).
First, we can multiply each side by \(\sin(\theta)\) so on the left we have \(r\sin(\theta)=y=\frac{56}{9}(\sin(\theta)-\sin^2(\theta))\). Now we have the \(y\) coordinate as a function of \(\theta\). We can differentiate next to find where the change of \(y\) is 0, that is, the tangent line is flat. This will tell us where the maximum and minimum points are.
\[\frac{dy}{d\theta}=\frac{56}{9}(\cos(\theta)-2\sin(\theta)\cos(\theta)) =\frac{56}{9}\cos(\theta)(1-2\sin(\theta))=0\]The solutions to this are when \(\cos(\theta)=0\) or \(\sin(\theta)=1/2\). This gives us \(\theta=\pi/2,3\pi/2,\pi/6,5\pi/6\) as solutions. When testing these in the equation for \(y\), the values we get are \(\frac{56}{9}\frac{1}{4}\) for \(\theta=\pi/6,5\pi/6\), \(0\) for \(\theta=\pi/2\), and \(-2\frac{56}{9}\) for \(\theta=3\pi/2\). That \(0\) actually corresponds to a point where the graph is not differentiable with respect to \(x\). So the difference between the max and min \(y\) is:
\[\frac{56}{9}\frac{1}{4}-\left(-2\frac{56}{9}\right)=\frac{56}{9}\frac{9}{4} =14\]The 2 side lengths next to that \(2\pi/3\) angle are known so we can use the law of cosines to find the length of the remaining side:
\[\begin{align} &x^2=(5\sqrt{3})^2+(5\sqrt{3})^2-2(5\sqrt{3})(5\sqrt{3})\cos(2\pi/3)\\ &x^2=75+75-150(-1/2)=225 \Rightarrow x=15\\ \end{align}\]Find the area of the square.
Let \(s\) be the side length of the square and \(t\) be the length of the altitude drawn from the top left corner. Then by similar right triangles, we have (ratio of hypotenuse to longer leg)
\[\frac{s}{16/\sqrt{17}}=\frac{1}{t} \Rightarrow st=16/\sqrt{17}\]Next, with the right triangle with the explicitly labeled right angle, we can use the Pythagorean theorem and replace \(t\) with \(s\) using the above equation
\[\left(\frac{16}{\sqrt{17}}\right)^2+t^2=s^2 \Rightarrow \frac{256}{17}+\left(\frac{16}{s\sqrt{17}}\right)^2=s^2 \Rightarrow 256s^2+256=17s^4\]Then we substitute \(u=s^2\) and by solving for \(u\), we find the area of the square. The quadratic equation is factorable as \(17u^2-256u-256=(17u+16)(u-16)\) so we have \(u=16\) or \(u=-16/17\). Only the positive answer makes sense here so the area of the square is 16.
This is a simple dot product. The answer is \(1\cdot3+2\cdot-1+4\cdot4=17\).
First let's label 2 additional points.
Let \(c=m\angle ACD=m\angle DCB,a=m\angle A,b=m\angle B,y=\overline{AD}\). Using \(\triangle ABC\), we have:
\[\frac{\sin(a)}{36}=\frac{\sin(b)}{24} \Rightarrow \frac{\sin(a)}{\sin(b)}=\frac{3}{2}\]Using \(\triangle ACD\), we have
\[\frac{\sin(c)}{y}=\frac{\sin(a)}{z}\]Using \(\triangle BCD\), we have
\[\frac{\sin(c)}{x}=\frac{\sin(b)}{z}\]Using these 2 equalities, we can find
\[z=\frac{x\sin(b)}{\sin(c)}=\frac{y\sin(a)}{\sin(c)} \Rightarrow \frac{x}{y}=\frac{\sin(a)}{\sin(b)}=\frac{3}{2}\]So now, knowing \(x+y=30\), we can substitute \(y=2x/3\) and find that \(x=18\).
How many times do \(y=\sin(x)\) and \(y=\frac{x}{10\pi}\) intersect?
Both functions are odd so we can restrict our search to \(x\geq0\) and do some counting. On this interval, we need \(\frac{x}{10\pi}\in[0,1]\) which restricts our search to \([0,10\pi]\). Intervals where \(\sin(x)\geq0\) are \([0,\pi],[2\pi,3\pi],[4\pi,5\pi],[6\pi,7\pi],[8\pi,9\pi]\).
Next, we will solve the problem on one interval in a way that applies to all 5 intervals. Let \(0\leq u<v<1\) and define \(f(x)\) as the line such that \(f(0)=u,f(\pi)=v\). We will now prove that \(f(x)=\sin(x)\) has exactly 2 solutions on \([0,\pi]\). Note: the line is \(f(x)=\frac{v-u}{\pi}x+u\).
Now let's consider the function \(d(x)=\sin(x)-f(x)\). The derivative is \(d'(x)=\cos(x)-\frac{v-u}{\pi}\) and the 2nd derivative is \(d''(x)=-\sin(x)\).
Now where is \(d'(x)=0\)? That is when \(\cos(x)=\frac{v-u}{\pi}\) which happens exactly once since \(\cos(x)\) is strictly decreasing on \([0,\pi]\) from \(1\) to \(-1\) and \(\frac{v-u}{\pi}\) is in \((-1,1)\). Let \(c\) be the unique point on \((0,\pi)\) where \(d'(x)=0\). So now since \(d(x)\) is concave down on \([0,\pi]\) by the 2nd derivative test, we can see that it must be strictly increasing on \([0,c]\) and strictly decreasing on \([c,\pi]\).
Next, we can find that \(d(0)\leq0,d(\pi/2)>0,d(\pi)<0\) from the way we defined it and chose \(u,v\). By the intermediate value theorem, we must have points \(c_1\in[0,\pi/2),c_2\in(\pi/2,\pi)\) such that \(d(c_1)=d(c_2)=0\). This means that by the mean value theorem, there exists a point in \((c_1,c_2)\) where \(d'(x)=0\). This point must be the \(c\) we identified earlier and it must be such that \(d(c)>0\).
So now we have that on \([0,c]\), \(d(x)\) is strictly increasing and and changes from negative to positive (or 0 to positive when \(u=0\)). Then on \([c,\pi]\), \(d(x)\) is strictly decreasing and changes from positive to negative. Therefore, on \([0,\pi]\), \(d(x)=0\) at exactly 2 points, thus \(\sin(x)=f(x)\) has 2 solutions.
Now we apply this to the 5 interval identified earlier. On each interval, \(\sin(x)\) is exactly the same as it is on \([0,\pi]\) and the segment of \(y=\frac{x}{10\pi}\) on that interval corresponds to some choice of \(u,v\) in our subproof. Therefore, we have 5 intervals and 2 intersections per interval so 10 intersections. Then we go back to our reasoning about odd functions and realize this must be doubled, but then we subtract 1 for the intersection \((0,0)\) which would be counted twice. Therefore, there are exactly 19 intersections.
Note: the reasoning's precise meaning depends a little on the specifiecs about whether to use closed or open intervals in theorems like the mean value theorem. I tried to handle them to work with the possibility of \(u=0\) in the subproof but it may need to be adjusted in a few places.
Below is a plot which can be used to solve the problem graphically.
To evaluate this limit, first replace the exponent base with \(x=e^{\ln(x)}\). Then the limit is
\[\lim_{x\to1}x^{\frac{3x}{\sin(x-1)}} =\lim_{x\to1}e^{\frac{3x\ln(x)}{\sin(x-1)}} =e^{\lim_{x\to1}\frac{3x\ln(x)}{\sin(x-1)}}\]So now we have a new limit to evaluate which is in the indeterminate form \(\frac{0}{0}\) so we can use LHopital's rule.
\[\lim_{x\to1}\frac{3x\ln(x)}{\sin(x-1)} =\lim_{x\to1}\frac{3\ln(x)+3}{\cos(x-1)} =\frac{3\cdot0+3}{\cos(1-1)}=3\]So now the answer is \(\text{Round}(e^3)\) which is 20.
For a continuous real function \(f\), \(\forall x:f(2x)=11f(x)\) and \(\int_0^1f(x)dx=1\). Find \(\int_1^2f(x)dx\).
Let's first integrate both sides of \(f(2x)=11f(x)\) on \([0,1]\) with the substitution \(u=2x\).
\[\int_0^1f(2x)dx=\int_0^111f(x)dx \Rightarrow \frac{1}{2}\int_0^2f(u)du=11\int_0^1f(x)dx\]Now multiply both sides by 2 and split up the integral on \([0,2]\).
\[\int_0^1f(u)du+\int_1^2f(u)du=22\int_0^1f(x)dx \Rightarrow \int_1^2f(u)du=21\int_0^1f(x)dx\]Since that integral on the right is 1, the answer is 21.
Square the part with the square roots so there isn't a square root inside a square root.
\[\left(\frac{\sqrt{2}+\sqrt{10}}{\sqrt{3+\sqrt{5}}}\right)^2 =\frac{12+4\sqrt{5}}{3+\sqrt{5}}=\frac{4(3+\sqrt{5})}{3+\sqrt{5}}=4\]So that whole part with the square roots simplifies to \(2\) (since it must be positive, \(-2\) does not makes sense). The answer is \(11\cdot2=22\).
Find the \(x\)-intercept of the line through \((5,3)\) and \((11,2)\).
The slope is \(m=\frac{2-3}{11-5}=\frac{-1}{6}=-\frac{1}{6}\). Using the point \((5,3)\), we can get the line in point-slope form \(y-3=m(x-5)\Rightarrow y=3-\frac{1}{6}(x-5)\). The \(x\)-intercept is when \(y=0\) so \(3-\frac{1}{6}(x-5)=0\Rightarrow x=23\).
Find the number of trailing zeroes in \(100!\).
This is the same as how many times we can divide \(10\), which reduces to counting multiplicity of factors \(2,5\). For 2, we have it counted once for each even number (50 times), then again for each multiple of 4 (25 times), and so on. The total is \(50+25+12+6+3+1=97\). For 5, similarly we get \(20+4=24\). So 10 is a factor with multiplicity 24 since we are limited by the multiplicity of 5. Therefore, \(100!\) has 24 trailing zeroes.
Find the area of a square with perimeter \(20\).
The side length is \(20/4=5\) so the area is \(5^2=25\).
First, using the difference of cubes formula gives us \(t=27^3-25^3=(27-25)(25^2+25\cdot27+27^2)\). Now let \(u=26\) and this is \(2((u-1)^2+(u-1)(u+1)+(u+1)^2)=2(3u^2+1)=6u^2+2\). Now we use this to solve for \(x=\sqrt{\frac{t-2}{6}}=\sqrt{\frac{6u^2}{6}}=\sqrt{u^2}=u=26\).
\(a+b+c=3\) and \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{3}\). Find \(a^3+b^3+c^3\).
First we find the cube of each side of the first equation.
\[(a+b+c)^3=a^3+b^3+c^3+3\left(a^2b+a^2c+ab^2+b^2c+ac^2+bc^2+2abc\right)=27\]Next, multiply each side of the 2nd equation by \(3abc\).
\[3(bc+ac+ab)=abc\]Now multiply this equation by \(a+b+c\) on the left and \(3\) on the right since they are equal from the first equation.
\[3(bc+ac+ab)(a+b+c)=3abc \Rightarrow abc=3abc+a^2b+a^2c+ab^2+b^2c+ac^2+bc^2\]Now just subtract \(abc\) from each side and rearrange a little to get this.
\[a^2b+a^2c+ab^2+b^2c+ac^2+bc^2+2abc=0\]This is exactly what was in the parenthesis earlier, and since it is equal to \(0\), that part goes away leaving \(a^3+b^3+c^3=27\).
This is in base 3 which is \(1\cdot3^3+0\cdot3^2+0\cdot3^1+1\cdot3^0=28\).