+The torque applied is the mass times the distance from the center. For each of +these blocks, we use the midpoint. The values to balance form the equation +\[7\cdot2.5+5\cdot1.5=1\cdot2.5+x\cdot4.5\Rightarrow17.5+7.5=2.5+4.5x +\Rightarrow22.5=4.5x\Rightarrow x=5\] +
++Find the sum of the roots of \(x^8-7x^7+8\) +
+ ++Using Vieta's formula, is is \(-a_7/a_8=-(-7)/1=7\). +
++\(y\) is a positive integer and \(x=1+y+y^2+\ldots+y^7\) is a power of a prime. +
+ ++By finite geometric sum, \(x={y^8-1\over y-1}\) when \(y>1\). This can be +further factored: +\[\begin{align}&x={y^8-1\over y-1}={(y^4+1)(y^4-1)\over y-1} +={(y^4+1)(y^2+1)(y^2-1)\over y-1}\\& +={(y^4+1)(y^2+1)(y+1)(y-1)\over y-1}=(y^4+1)(y^2+1)(y+1)\end{align}\] +From here, \(y+1\) is a power of a prime. Suppose \(y+1=p^a\) for some prime +\(p\) and integer \(a>0\). Then +\[y=p^a-1\Rightarrow y^2=p^{2a}-2p^a+1\Rightarrow y^2+1=p^{2a}-2p^a+2\] +This quantity must be equal to some other power of the same prime so suppose for +some integer \(b>0\) +\[p^{2a}-2p^a+2=p^b\Rightarrow 2=p^b-p^{2a}+2p^a\] +We see that 2 is a multiple of \(p\) therefore our prime must be \(p=2\). This +means \(y+1=2^a\). Then +\[y^2=(2^a-1)^2=2^{2a}-2\cdot2^a+1\Rightarrow y^2+1=2^{2a}-2\cdot2^a+2=2^b\] +The only time this is true is when \(a=1\). We can see this by factoring it to +\[y^2+1=2\cdot(2^{2a-1}-2^a+1)\] +which is twice an odd number whenever \(a>1\). Therefore, the only +possibility is \(y+1=2\Rightarrow y=1\). In this case, \(x=8=2^3\) which is the +solution. +
++Find the number of factors of 36. +
+ ++Factor it to \(36=2^2\cdot3^2\). For each prime, there are 3 power choices to +form a factor (0,1,2). Therefore the number of factors is \(3\times3=9\). +
++\[x{42-30+27-35\over45}={40\over45}\Rightarrow 4x=40\Rightarrow x=10\] +
++We should focus on the first equation to find what \(\theta\) is. This might +make us think of the 3-4-5 right triangle which is related. That can be used to +find \(\sin(\theta)=3/5,\cos(\theta)=4/5\). These values can be used to find +\(x\) in the second equation +\[5\sin\theta+5\cos\theta+3\cot\theta=5(3/5)+5(4/5)+3{4/5\over3/5}=3+4+4=11\] +
++Multiply by \(2/2\) so the denominator becomes \(2^x\) +\[2\sum_{x=0}^\infty{2x+1\over2^x}=2\left[2\sum_{x=0}^\infty{x\over2^x} ++\sum_{x=0}^\infty2^{-x}\right]\] +The simpler sum is +\[\sum_{x=0}^\infty2^-x={1\over1-1/2}={1\over1/2}=2\] +The other sum can be rewritten as a double summation by changing the order of +summation thinking about the following table +\[\begin{array}{cccc}1/2&&&\\1/4&1/4&&\\1/8&1/8&1/8&\\\vdots&&&\ddots\\ +\end{array}\] +Each term corresponds to a row so now if we sum column by column +\[\begin{align}&\sum_{x=1}^\infty{x\over2^x}=\sum_{x=1}^\infty +\sum_{y=x}^\infty{2^{-y}}=\sum_{x=1}^\infty\left[\sum_{y=0}^\infty2^{-y} +-\sum_{y=0}^{x-1}2^{-y}\right]=\sum_{x=1}^\infty\left[ +{1\over1-1/2}-{1-2^{-x}\over1-1/2}\right]\\& +=\sum_{x=1}^\infty(2-2+2\cdot2^{-x})=2\sum_{x=1}^\infty2^{-x} +=2\left[{1\over1-1/2}-2^0\right]=2\end{align}\] +So it turns out both of these summations are 2 and if we substitute them into +the expression we found earlier, the answer is \(6\). +
++Find the largest solution of \((25-2y)^{(y^2-25)}=1\). +
+ ++If \(|25-2y|\neq1\) then we need \(y^2-25=0\Rightarrow y=\pm5\). +
+ ++If \(|25-2y|=1\) then \(y=12,13\). The only requirement is the exponent is +positive to avoid the division by zero problem, which is true for these values. +Therefore, the largest solution is \(y=13\). +
++Two natural numbers differ by 4. The sum of their squares is 106. Find their +sum. +
+ ++Without loss of generality, let \(a,b\) be the 2 numbers with \(a>b\). Then +\(a-b=4\) and \(a^2+b^2=106\). By substitution +\[(b+4)^2+b^2=106\Rightarrow2b^2+8b+16=106\Rightarrow b^2+4b-45=0 +\Rightarrow(b+9)(b-5)=0\] +The solutions are \(b=5,-9\) so the only natural number solution is \(b=5\). +Then \(a=b+4=9\) so the sum is \(a+b=14\). +
++The easiest way to solve this is probably by first finding \(\tan(\pi/12)\) +which can be done with the angle difference identity +\[\tan(\pi/3-\pi/4)={\tan(\pi/3)-\tan(\pi/4)\over1+\tan(\pi/3)\tan(\pi/4)} +={\sqrt{3}-1\over1+\sqrt{3}}\] +Then from this we can easily find +\[\cot(\pi/12)={1+\sqrt{3}\over\sqrt{3}-1}\] +These can be substituted in with some gross algebra +\[\begin{align}&{\cot^3(\pi/12)-\tan^3(\pi/12)\over2\sqrt{3}} +={\left({1+\sqrt{3}\over\sqrt{3}-1}\right)^3 +-\left({\sqrt{3}-1\over1+\sqrt{3}}\right)^3\over2\sqrt{3}} +={{6\sqrt{3}+10\over6\sqrt{3}-10}-{6\sqrt{3}-10\over6\sqrt{3}+10} +\over2\sqrt{3}}\\& +={\left(6\sqrt{3}+10\right)^2-\left(6\sqrt{3}-10\right)^2 +\over2\sqrt{3}\left(6\sqrt{3}-10\right)\left(6\sqrt{3}+10\right)} +={36\cdot3+100+120\sqrt{3}-36\cdot3-100+120\sqrt{3} +\over2\sqrt{3}\left(36\cdot3-100\right)} +={240\sqrt{3}\over16\sqrt{3}}=15\end{align}\] +
+