+Find the number of weeks in 231840 minutes. +
+ ++By factoring, we find \(2^5\cdot3^2\cdot5\cdot7\cdot23\). We divide out +\(60\cdot24=2^5\cdot3^2\cdot5\) for minutes in a day and \(7\) for days in a +week. What remains in the factor \(23\) +
++The 3 equations have a nice factorization if 1 is added to each side. We get +\[\begin{align}&(x+1)(y+1)=525=3\cdot5\cdot5\cdot7\\ +&(y+1)(z+1)=147=3\cdot7\cdot7\\&(z+1)(w+1)=105=3\cdot5\cdot7\end{align}\] +From the first 2 of these, \(y+1\) divides both \(525\) and \(147\) so it +divides their GCD which is \(21\). Similarly, \(z+1\) divides \(21\). Since +\(y>0\), the 3 cases are \(y+1=3,7,21\). If \(y+1=3\), then \(z+1=49\) which +does not divide \(21\) so it can't be that. If \(y+1=7\), then \(x+1=75\) so we +find \(37\) is a factor of \(x\) but it is not a factor of \(8!\) so this case +does not work either. Finally, if \(y+1=21\), then \(z+1=7,x+1=25,w+1=15\). So +by multiplying the variables \(x=24,y=20,z=6,w=14\), we can confirm \(xyzw=8!\). +Therefore the solution is \(x=24\). +
++Alec lost \(20\%\) of his money. He then gained \(x\%\) of what he had +remaining, just breaking even. +
+ ++By losing \(20\%\), Alec has \(0.8M\) where \(M\) is the amount he started with. +From here, we solve the equation \(0.8M(1+x/100)=M\) to find what \(\%\) he has +to gain to get back to \(M\). +\[0.8M(1+x/100)=M\Rightarrow1+x/100=1.25\Rightarrow x=25\] +
++Find the smallest positive integer \(x\) such that the sum of \(x\) and the next +\(50\) consecutive numbers is a square number. +
+ ++The sum is +\[x+(x+1)+(x+2)+\ldots+(x+50)=51x+{50\cdot51\over2}=51(x+25)\] +The factors are \(51=3\cdot17\) so \(x+25\) must have these same factors to make +this quantity a square. Therefore, \(x+25=51\Rightarrow x=26\). +
++It turns out there is actually no point in 3D space that simultaneously +satisfies all 3 of these equations. We can find a solution for \(x\) which may +correspond to imaginary solutions. Start with the 2nd equation, multiply \(x\), +then substitute the 3rd +\[x^2+xy+xz=72x\Rightarrow yz+xy+xz=72x\] +Now square both sides of the 2nd equation and use some substitutions +\[\begin{align}&x^2+y^2+z^2+2xy+2xz+2yz=72^2\\ +&1296+2(xy+xz+yz)=72^2\\ +&36^2+2(72x)=72^2\\ +&144x=3\cdot36^2\Rightarrow x=27\\\end{align}\] +Now if we substitute this into the equations, we get a new system in 2 variables +\[\begin{align}&y^2+z^2=567\\&y+z=45\\&yz=729\\\end{align}\] +But by using the first 2 equations +\[y^2+(45-y)^2=567\Rightarrow y^2-45y+729=0\Rightarrow +y={45\pm9\sqrt{-11}\over2}\] +Then we can use this to show that there are 2 solutions in complex 3D space that +satisfy all 3 equations simultaneously +\[\begin{align}&x=24,y={45+9\sqrt{-11}\over2},z={45-9\sqrt{-11}\over2}\\ +&x=24,y={45-9\sqrt{-11}\over2},z={45+9\sqrt{-11}\over2}\end{align}\] +
+