<p>in a basket, there are three electronic toys. One toy beeps every 8 minutes, another toy beeps every 10minutes,and the third toy beeps every 12 minutes. All the toys beeped together at 1:00 am. When is the next time the three toys will beep together?</p>
<p>Please explain your answer. Thanks guys.</p>
<p>120 mins.. so 3 am? i’ll explain if its correct</p>
<p>You have to find the least common multiple between 8, 10, and 12. That happens to be 120. So every 120 minutes the toys will beep simultaneously. Since there are 60 minutes in an hour, the toys will beep together every 2 hours. So the next time they will beep at the same time will be 3:00 AM</p>
<p>Prime factoring: 2x2x2x3x5=120. In other words, at 3 o’clock. </p>
<p>Explanation:
To find when they will all beep together you have to find the least common multiple, or LCM. To do this, find the prime factorization of each number. For 8, it’s 2x2x2, for 10, it’s 5x2, for 12 it’s 3x2x2. Once you find the prime factors multiply them together. There are 3 2s in 8, one 5 in 10, and one 3 in 12. Disregard the 2s in 10 and 12 because there are more of them in the prime factorization of 8. Sorry for the convoluted explanation; I am not a good teacher :(</p>
<p>I also got 120 but i can’t quite grasp the concept for some reason. I understand we have to find the LCM for these types of problems but i don’t see how starting 3 beens, each with different rates, can beep all at once eventually. I know it works but my head isnt taking in the concept. How exactly are these 3 going to beep at once when they are all going at different rates? Can someone give me another example? Every time i see these problems i think about me racing a car. When are we going to eventually meet up? The answers never, the cars just going to get further and further. This is killing me.</p>
<p>hmm good point. racing car.. u almost lost me there. but if u think of the race track as a circle…the one goes faster will evetually come back to to its original starting pt, then bypass the 2 slower ones slowly… soon enough, few hrs later, they meet at the same spot, but the fastest car has traveld a bjiaosdllion circles already, while the slowest one only a few circles~~</p>
<p>So in a hypothetical situation, if the track length is infinite- ill never catch up. So this only works for things that moves in circles? What about the LCM’s then, are there only a set number of LCMS for 2 numbers of different rates before one gets too far?</p>
<p>Bump-help me out.</p>
<p>NVM. I got it now. When we speak in terms of me vs a racing car, its a different story because i cant just appear next to the car whenever it hits its multiples. But when we talk about alarms, it works because the alarms just beep at those multiples-its not trying to catch up with anything. Right?</p>
<p>Yes (10char)</p>
<p>We aren’t talking physics here..this isn’t speed or velocity. If a clock beeps every 8 mins, every 8 minutes it will beep (obviously). Same thing with the other two clocks. Now, isn’t that the same as adding 8 a bunch of times, adding 10 a bunch of times, and adding 12 a bunch of times? Won’t there be a time when all of them will equal each other? So isn’t that the same as just finding a common multiple since in essence you are multiplying each of those 3 numbers by a number? It will take 8 more “tries” to “catch up” to 12, but it will happen. For example, 8x3 = 24 whereas 12x2=24. I’m not sure if it helped.. that last bit might have made it more complex but eh…I tried</p>
<p>I’m not sure if it “catches up”</p>
<p>Since A beeps every 8 minutes, it’ll beep on 0 mins, 8 mins, 16 mins, 24 minutes.
Since B keeps every 12 minutes, it’ll beep on 0 minutes, 12 minutes, 24 minutes.</p>
<p>Whenever the numbers line up, thats when both will beep, so at 24 minutes A and B beeps together.
LCM allows you to find out the last amount of time it takes for all three to beep together.</p>
<p>Wantivy your overcomplicating it.. I got it bro, thanks.</p>