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My friends rolled up outside of my house at 8am on Monday, and said what amounted to, “Good, you’re not dead. Now get in loser, we’re going hiking”. And so a week of shameless ethical hedonism ensued.I wouldn’t call myself social media-addicted, but over the past few months in quarantine my phone screen time report tells a different story: One of late night Snapchat binges, multi-hour FaceTime marathons, and Instagram stalking sessions. The funny thing is that all ended when I went through a brief depressive episode over the past two weeks. I stopped posting daily to my Snap, didn’t reply to texts or FaceTime, and even gave up my beloved hobby of Instagram stalking. Well perhaps that last one was a good thing.(Exhibit A: I got up at 5:30am to take this photo while the sun was rising 😳)Eventually I started getting texts and Snaps from friends pouring in asking “Are you dead?”, “Stop turtling”, “For [expletive]’s sake Ari, reply to us!”. My friends are quite gifted in the art of subtlety, no? Eventually three of them decided the solution was to drive down to my house and haul my sad butt out of bed at 8am for a day of hiking. At 8am on Monday, my friends (Dev, Annie, and Aarón) rolled up to my house and said what amounted to, “Good, you’re not dead. Now get in loser, we’re going hiking”. And so a week of shameless hedonism ensued.(The road above Silicon Valley our to the boonies)Ordinarily I hate driving and being inside at all, but as Dev wove in and out of the 101, with me tucked comfortably in the passenger seat, for the first time in a week I realized how tired I was and began to weave in and out of consciousness. Between the golden California hills I rambled incoherently about the taste and color of music while my friends tried not to giggle and tried to urge me to get some sleep. At some point I clocked out, finally at peace among the three of them, only to be awakened by a gentle nudge on my shoulder. Dev was pointing at our surprise destination- Redwood National Park. For the first time in two weeks, I didn’t have a panic attack right as I woke up and instead placidly gazed out at the giants before us.(Annie at the wheel after Dev and I got popped into the backseat due to excessive rambling)If my friends know anything about me, it’s that I’m like a dog- I need to be walked/run twice daily and I can get wildly excited about almost anything. The second we rolled into a vacant lot, I jumped out of the car (which was still rolling), and eagerly ran in circles around a sun faded park sign. “Isn’t it amazing?” I shouted, oblivious to the fact that the trio were on the other side of the parking lot, and then promptly chased after a squirrel yelling “Let me pet you!”. I was just a tad sleep deprived. A hike in the middle of giant redwoods was exactly what I needed to get out of my head.(Us in the middle of the woods)The night before Aarón had texted me saying, “Could you make your ganache brownies for tomorrow- the ones you made for Annie’s 21st?”. Then he dropped a little hint about the possibility of a long hike. So that’s how we wound of with gooey softball sized brownies, wrapped in tinfoil, on a log in the middle of the Redwoods at 11am. It had been days since I had the drive to eat, and sitting there on the log, it finally hit me how hungry I was. I didn’t realize that I had been starving for meaningful interaction and connection for so long. Needless to say I pounded the entire brownie for lunch with absolutely zero regrets.Then we laid back on the log and watched the Forrest go by upside down. While the blood rushes to our heads, words flowed freely from our lips- about the things we feared, and what we found beautiful in this brave new world. How we longed for the comfort of human touch and how there was comfort in hiding from the world now.The next day, and the day after, and the day after was saturated with companionship and the wanton revelry in youth and semi-intact knees. We rode horses in mountains, played board games in meadows, had dumb balancing competitions in decaying buildings (ex: who can run along a beam backwards with their eyes closed the fastest), wandered through the cattails in the lakes, clapped for fishermen when they caught fish, and had picnics in parking lots.So my friends, from 6 feet apart, are the sun that keeps me rising everyday in quarantine. And, as I’ve learned, it’s impossible to feel sad in the California sun among people you love.

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Assume [math]\sqrt{n}[/math] is irrational, and take [math]r - \sqrt{n} = \sqrt{n + 2013}[/math]. But then [math]r^2 + n - 2r\sqrt{n} = n + 2013[/math], a contradiction (the left side is irrational; the right side is not). So both [math]\sqrt{n}, \sqrt{n + 2013}[/math] are rational.The short answer is there are infinitely many such [math]n[/math]: if you let [math]n = \frac{1}{q^2}[/math], then you're looking for solutions to [math]2013q^2 + 1 = s^2[/math], or the well-known Pell's equation [math]s^2 - 2013q^2 = 1[/math], which has a family of infinitely many solutions (just check the link).If you want to know how many families of solutions there are, you want a solution to the more general equation [math]s^2 - 2013q^2 = p^2[/math], or, I suspect more helpfully, [math]s^2 - p^2 = ([/math][math]s[/math][math] + p)(s - p) = 2013q^2[/math]. I don't immediately see any great way to attack this...taking it mod 3 quickly tells you that [math]s,p[/math] aren't divisible by 3, though either their sum or difference is. (The right side is divisible by 3, so the left must be, too, so [math]s^2[/math] and [math]p^2[/math] are both 0 or 1 mod 3. If they both were 0, then the left side would be divisibly by 9, but the right side only has one explicit factor of 3, so [math]q[/math] must also be divisibly by 3, but then our system wasn't reduced to start with so we can throw this case out. This also works for 11, 61 since these are the other prime factors of 2013.) Mod 4 gives you that exactly one of [math]p,q,s[/math] is even (so it's not much help). I don't see why there wouldn't be infinitely many families, though.Edit: it is in fact not difficult to construct new solutions to this equation, which we can show just by considering [math]q[/math] odd. Let [math]d[/math] be any prime factor of the RHS (I've used up all the letters I like to use for primes :( ) other than 3, 11, 61. If [math]d[/math] divides both terms on the left (in the factored version), then it divides their difference, [math]2p[/math]; but then [math]d[/math] divides [math]p[/math] (recall that [math]d[/math] is odd) and so [math]\dfrac{p}{q}[/math] was not reduced; therefore each odd prime factor (other than our three exceptions) of the right side is found in exactly one of the terms on the left. For each [math]d[/math] of 3, 11, 61: if it doesn't divide [math]q[/math], we can just throw the one factor of [math]d[/math] onto either term on the left. If it does, the same argument as before tells us, again, that all the factors of [math]d[/math] are in exactly one of the terms on the left. Thus we can rewrite the left side as [math]3*11*61*a^2 b^2 = 2013 a^2 b^2[/math], so [math]ab = q[/math]. (The case for [math]q[/math] even is very much the same, except you need to take some time to deal with the factors of 2.)For example, let [math]q = 35 = 5 * 7[/math]. Since 3, 11, and 61 don't divide [math]q[/math], we know [math](s+p)(s-p)[/math] is really just [math](6+1)^2 (6-1)^2 = (49)(25)[/math] with the factor of 2013 thrown in somewhere; for fun let's put the 3 and 11 in the first factor and 61 in the second. This gives us [math](1617)(1525) = (1571 + 46)(1571 - 46)[/math], so [math]1571^2 - 46^2 = 2013(35^2)[/math], and [math]n = \dfrac{46^2}{35^2}[/math]. You can construct as many solution you want along these lines just by picking an odd integer for [math]q[/math] and splitting it into two factors (including 1 and itself).