 Imagine you have nine balls, I wouldn't be able to walk, and one is a little heavier. Oh man, I'm in trouble, I gotta go to the doctor. Sorry, you have to say it. You have a beam scale and are allowed to wait three times to find the heavier ball. Let's check this out. Let's read the other one too. The two digit AB, we've got a couple of riddles going on. Oh my god, I love Spider-Man. Math is fun, math is fun. Hey, quick question. Serious medical aid right there. Imagine you have nine balls and one is a little heavier. So you've got nine balls, let's do it. One, two, three, four, five, six, seven, eight, nine. I love Spider-Man, we'll read your thing after this one. Can you explain all the calculus to me? Sure, calculus is an introduction of time into mathematics, so it's the rate of change. So going into calculus, if you're studying calculus, just think of it as time, rate of change. And the only absolute in life is change, so calculus is pretty damn important, right? A little heavier. You have a beam scale and are allowed to wait three times to find the heavier ball. What would you do? So we have a scale, we've got three times we could weight things. What would you do? And we've got nine balls total. I mean the ideal thing would be you would break it in. What would I do? I would do this. Let's see if this is going to work. I would weigh five balls in one, four balls in another, right? And you would calculate per unit ratio. So you would weigh this. You would put five balls here without work and then divide whatever you get by five. Put four balls here. But three times doesn't work. And you get this number, right? If this is bigger than that, you know your big balls in here. And then what do you got to do? If you could do it four times, you could do this easy. Break this in half again. Wham. And then you know you can do the next one and you're done, right? Like a seesaw. Moving around. Oh, like a seesaw. So it's like this. Oh my bad. So that one we could do with four weigh-ins. Pretty sure. When does calculus turn to physics? Mathematics is just language, base language of everything. It's a beam scale. That's not what you call it. So we got this guy. So we got nine balls. So what are we going to do? Nine balls. So what would we do? Do we have to have all nine balls on the beam scale at the same time? We have a left and a right side of that scale. How is that scale? How is that? I don't know what it's called. I don't know what it's called. Just pit it. So you can't, you have to weigh all the balls at the same time. So you can't take eight balls and put them. Always count two of them. Pit it up and weigh the heavier half each time. You're free to do what you want. You're free to do what you want. So I would put four balls here. Four balls here, right? If this thing is level, whatever ball that was left out is your big ball. I'm just cracking up saying that, right? Twice you get the answer until you have the last. So you weigh it once. Four here, four here. I'm just exercising. I don't know if this is the answer. So if these are totally level, then that one is the ball you're looking for. If this tilts, then obviously the heavier one is going to be on the side where it's tilting down, right? So if this thing goes down this way, you eliminate these guys. Then you put two here, oops, two here and two here. If this one tilts, then you put one here, one here, whichever one tilts is the big ball, right? Twice you got the answer until you have the last two. Same solution, best case check. Cool, cool. You can just add two balls evenly. And when it tilts, you know that seems like cheating. That's true, Dante, you could do that too. Two balls here, two balls here. If it tilts, then one of them is to have your contains the heavier. So you could actually theoretically do it in two measurements. This one, if you're lucky, you could do it in one on the first go. You just put one ball on either side. If they weigh the same, you discard them. Repeat three. The kicker is. So one ball, once, twice, you could only do it three times. You could only weigh three times, right? Because it would be multiple measurements, so to speak. Okay, I got it wrong. You're allowed to weigh two times only now. Hey, you're changing the rules of the game. No, I mean you could start with one each and keep adding. But that should count as multiple measurements. Yeah, that would count as multiple measurements. That's what I would think too, right? So if you added those guys, if it stays level, then you gotta add more. You add those ones, keep track of what you're adding. If it stays level, you gotta add more. But that's already passed the two measurements, right? I swear, there's a similar problem about weighing 13 people with identifying one person being heavier. Yeah, I mean to use two on one side and don't need four at one start. So if you use two, if one is heavier, you got on the second one. If you use two, one is not heavier. How would you do it with two weights? I don't know how you would do it with two weights. Split the ball into three groups, Yucho. Three groups. Oh, split the ball into three groups. That's not a bad idea. Let's check it out. But then how do you isolate the one that's the heaviest one? Oh, because it's three. So split them into three, right? One, two, three, one, two, three, one, two, three. Take this one and that one, put them here. Right? If it stays level, then you know the heavy one is in this one. Right? If it tilts, discard this one. And then take two of these guys, put them here. The one that tilts is the heavy ball. If it stays level is this guy. Great. Break them into threes. Nice. Good. Awesome. Right? Oh, logic takes place. Yeah, break them into threes. Awesome. That's better. With the other way, if you're lucky, you get it in one way. Victory. Victory. Awesome. That was great. What was Spider-Man's little riddle? Spider-Man's real or Spider-Man's question. Let's check it out. I love Spider-Man. I love Spider-Man. Let's check it out. The two digit number AB stands for 10A plus B. Since the first digit represents tens and the second represents units. I'm already confused. If 10A plus B equals 7A plus B, then 10A plus B equals 7A, 7B. And if 3A equals 6B or more simply A, that is, the second digit must be twice the first. The smallest such number is 21. I'm just going to write down the equations. That's what makes sense to me. 10A plus B. 10A plus B is equal to 7A plus B.