Mathematics - Printable Version +- World Beyblade Organization by Fighting Spirits Inc. (https://worldbeyblade.org) +-- Forum: Off-Topic Forums (https://worldbeyblade.org/Forum-Off-Topic-Forums) +--- Forum: General Discussion (https://worldbeyblade.org/Forum-General-Discussion) +--- Thread: Mathematics (/Thread-Mathematics) |
RE: Mathematics - Ga' - Mar. 31, 2011 HAH! This is right, I'm positive: 9867312, and is the largest I can get. RE: Mathematics - Dan - Mar. 31, 2011 (Mar. 31, 2011 10:21 PM)GaHooleone Wrote: Hah, didn't even notice there wasn't a topic on this. I'm actually a 'prodigy' in math, so to speak. I'm in 9th grade and taking 11Honors math (Pre-Calc). Now, most people would think, "That's not big of an achievement. I've seen better." So have I. But what makes me different is that while others went to CTY (Center for Talented Youth) and took math courses there, therefore giving them the credit they need to pass a course, or LISG (Long Island School for the Gifted), where they were accelerated multiple times with heavy-duty education and thrown back into the curriculum with accelerated in almost all of their courses, or some other program. I didn't. I did so well in the public school system where I go that I was recommended to go to a higher level, which is why I'm known as a prodigy. Plus, I'm a grade ahead in English. But let's see about that problem... (will be edited with answer that I get eventually) I had a big essay ready for you, but I'll just say this: if people just put effort into something they like, it shouldn't be too hard to excel past high school mathematics; The motivation/drive/interest has to be there though, glad you did something with it. 9867312? :V Yeah math can be pretty cool. RE: Mathematics - Rustled Jimmies - Mar. 31, 2011 (Mar. 31, 2011 10:21 PM)GaHooleone Wrote: GOT IT! Here: 9867312, this is the largest I can get. It IS indeed divisible by 7, 6, AND 42. I don't know how I skipped this one, but I was pretty close last night. I should've thought about switching the 6 and 7 than switching the 8 and 7 after 7 failed to divide... But can you explain how you came across this? RE: Mathematics - SaMaster14 - Mar. 31, 2011 (Mar. 31, 2011 10:30 PM)Kai-V Wrote:(Mar. 31, 2011 8:38 PM)SaMaster14 Wrote: Actually, my dad works as a business manager, and grosses about $1million a year, personal salary ... sure he has to do basic arithmetic, but in no way was he ever good at math, he never even took calculus... My godfather, a strong businessman and entrepreneur owns 11 businesses, and one of his best friends and partners is worth $3 billion ... never even took any form of higher math. Same goes for some other family members I have... Doing well in business, being a CEO, business management, etc etc... if you have the right calculator or computer, or if your brain is just programmed to run a business, you really don't need math to succeed. Sure, as I said, its a great thing to have, but the most successful people didn't get to where they are today solely through math... and yes, some did, but its not the only way. I agree that math is great and is amazing for intelligence... thats why I enjoy it haha, but the sad thing is, unless you are going into research and development and the such, most of the highest paying jobs don't really revolve around math... especially with the technology that allows computers to do the math for us. At least where I live, my family is the norm... As I've said, I like math, and do believe that if you can master all subjects you will go farther with yourself, but lawyers and businessmen and, of course actors, singers, and entertainers, are 'successful' in a money sense and probably never even took advanced math. RE: Mathematics - Serotonin - Mar. 31, 2011 you're still missing her point completely RE: Mathematics - PlayMadeSimple - Apr. 01, 2011 Excellent Ga'Hooleone, you are correct in believing that 9867312 is the largest possible number given what I said. Since I planned to have a new problem every Monday, a tournament in July and to have explained how to achieve the conclusion to the problem on weekends, your timing is perfect. I will have a new problem up by Monday, or if preferred, I'll put it up now as an exception to the roster? Ah, I can't help waiting, I'll post it now. On some other note, in regard to opening doors,bit of a cliche overuse, eh? Bewildered, Forgetful and not so fail, Bob. Bob has an unreliable memory. When asked for his phone number on six different occasions, it wasn't surprising that he failed to give out his real number on each of his different attempts. #1: 913916 #2: 917931 #3: 917601 #4: 914544 #5: 919772 #6: 911805 Although, his memory is not a complete failure since: a) his telephone number does in fact consist of 6 digits. b) in every attempt, the first two digits are correct. (91) c) in each of his attempts, he also had exactly one of the last four digits in the correct place. Question: What is his telephone number provided that the last four digits are all different? What I have seen is that people consider the guide to be for losers, when in reality, it isnt. I actually encourage you to make use of the guide anyway. If yu can solve it on your own then thats great, but it is better to show working and use the guide rather to prove just the answer, or complete it yourself then answer the questions? [The guide shall be posted soon enough.] RE: Mathematics - venkai - Apr. 01, 2011 in an equilateral triangle each side is 12 cm.if circle is incribed touching the sides. root 3 value=1.73,pi=3.14 then raduis of circle? area of incircle? the area of portion not included in the circle? draw triangle and circle in which circle is incribed in triangle touching three sides. RE: Mathematics - Rustled Jimmies - Apr. 01, 2011 (Apr. 01, 2011 8:01 AM)Mushy Wrote: Excellent Ga'Hooleone, you are correct in believing that 9867312 is the largest possible number given what I said. Since I planned to have a new problem every Monday, a tournament in July and to have explained how to achieve the conclusion to the problem on weekends, your timing is perfect. I will have a new problem up by Monday, or if preferred, I'll put it up now as an exception to the roster? Ah, I can't help waiting, I'll post it now. On some other note, in regard to opening doors,bit of a cliche overuse, eh? You say that the last four digits are ALL different. Meaning that within that last four, there shouldn't be any repeating numbers. Therefore, your choices are then limited to... #1: 913916 #2: 917931 #3: 917601 #6: 911805 After that, ONE of the last four digits are in the correct place. Even though that digit is unknown, you can rule actually rule out a set based on the number of times he repeated a number in a row. #1: 913916 #2: 917931 #3: 917601 Out of the three, #2 is either followed or preceded by a repeating number. HOWEVER. While I want to stand by my reasoning for 2, the 3rd hint, "in each of his attempts, he also had exactly one of the last four digits in the correct place," throws me off. Set 3 has 2 numbers "in the correct place" (in my opinion). SO.......I'm kind lost. RE: Mathematics - Ga' - Apr. 01, 2011 (Mar. 31, 2011 10:58 PM)Kaji Motomiya Wrote:(Mar. 31, 2011 10:21 PM)GaHooleone Wrote: GOT IT! Here: 9867312, this is the largest I can get. This required more common sense then anything. If you had a set of numbers, how do you make the largest one? say, 1 4 2 5 3 6? You put them in descending order: 654321. Since you had to account for 7, 2 had to be the last one. So I switched the 2 numbers that would affect the size the least: 1 and 2. So now we have 9876312. However, by putting 8 and 7 next to each other, you get some divisibility problems, which I'm sure most of you noticed. So we have to switch one of the lower numbers (preferably 6 or 3 to maintain size) with 7 (again to maintain size). Now, when you try to get two fractions with different denominators add together, you would take the LCM of the 2 numbers and add. Since the LCM of 8 and 7 is 56, and really the only place value that matters is the ones value (especially since 5 isn't in here), the 6 should be between the 8 and 7 to make it divisible by both, no? Sorry if it's hard to understand, that's how my mind works when it comes to math, and how I reasoned it out. XD But you come up with 9867312, with only a slight variation from the largest number possible with 1, 2, 3, 6, 7, 8, and 9: 9876321. That's how I did it. EDIT: I just realized that Mushy explained the 8 and 7 rule better upon looking through past posts. Just look at that one for a better explanation if you have trouble understanding. RE: Mathematics - Rustled Jimmies - Apr. 01, 2011 I figured, but I think I forgot to switch the 6 and the 7 and went straight to switching the 8 and the 7. RE: Mathematics - Jynx - Apr. 01, 2011 (Mar. 31, 2011 10:21 PM)GaHooleone Wrote: Hah, didn't even notice there wasn't a topic on this. I'm actually a 'prodigy' in math, so to speak. I'm in 9th grade and taking 11Honors math (Pre-Calc). Now, most people would think, "That's not big of an achievement. I've seen better." So have I. But what makes me different is that while others went to CTY (Center for Talented Youth) and took math courses there, therefore giving them the credit they need to pass a course, or LISG (Long Island School for the Gifted), where they were accelerated multiple times with heavy-duty education and thrown back into the curriculum with accelerated in almost all of their courses, or some other program. I didn't. I did so well in the public school system where I go that I was recommended to go to a higher level, which is why I'm known as a prodigy. Plus, I'm a grade ahead in English. But let's see about that problem... (will be edited with answer that I get eventually) I'm doing the same my math skils scare my friends RE: Mathematics - Mr. N - Apr. 01, 2011 (Apr. 01, 2011 8:01 AM)Mushy Wrote: Excellent Ga'Hooleone, you are correct in believing that 9867312 is the largest possible number given what I said. Since I planned to have a new problem every Monday, a tournament in July and to have explained how to achieve the conclusion to the problem on weekends, your timing is perfect. I will have a new problem up by Monday, or if preferred, I'll put it up now as an exception to the roster? Ah, I can't help waiting, I'll post it now. On some other note, in regard to opening doors,bit of a cliche overuse, eh? I got Spoiler (Click to View) Spoiler (Click to View) RE: Mathematics - PlayMadeSimple - Apr. 01, 2011 Thats great that we have some early adopters, and those eager to solve the problems but I cannot reveal the answer until Friday. It's still just a concept at this point, but I'm sure it'll come along well. If you are the first to solve a problem correctly, then you are awarded a Math Credit. When you have accumulated enough of these credits, you get rewards, e.g if you obtain 20 credits you will get a free entry to my July tournaments. Depending on how difficult the problem was, you get a certain value. Easy: 1 Credit Medium: 2 Credits Hard: 4 Credits Pro: 8 Credits Hopefully, this idea can be implemented well. Ga, you are awarded one math credit. If the WBO staff are willing to co-operate, perhaps we can also have WBO faces for acheiving a certain amount of credits, and winning the July tournaments. Thats alright Kai, I see your reasoning.I guess we'll stick with Math Credits... RE: Mathematics - Kai-V - Apr. 02, 2011 (Apr. 01, 2011 11:42 PM)Mushy Wrote: Hopefully, this idea can be implemented well. Ga, you are awarded one math credit. If the WBO staff are willing to co-operate, perhaps we can also have WBO faces for acheiving a certain amount of credits, and winning the July tournaments. Unfortunately, that only revolves around one topic, has nothing to do with Beyblade, and does not particularly mean that the person contributed to the community in any way, so no. RE: Mathematics - PlayMadeSimple - Apr. 02, 2011 Thats fine Kai, I understand your reasoning. I was kind of expecting it, but I'll just satisfy myself with math credits. Those rules are quite interesting, I never thought the WBO only gave faces for contributing to the community. I guess if you look at the faces page you can make out that assumption but I usually don't think that far ahead, as to go to that page without any given hint. RE: Mathematics - NoodooSoup - Apr. 04, 2011 Speaking of Mathematics, has anyone here ever played Game 24? Really fun game. RE: Mathematics - PlayMadeSimple - Apr. 06, 2011 It took a while, but here's a few questions to answer, that will help you reach a conclusion. If you choose to use this method, post your answers to them. I encourage you to use this over simply working it out. This guide is for use with Bewildered Bob, the problem regarding the phone numbers... Why is 913642 NOT a possibility. Which condition, a, b, c or d is not satisfied? There are six options from Bobs list, but these must cover only four digits. WHat does this imply? Explain what happens when seven is placed as the third digit. hHat digit must then be placed as the last, and what options are left for the fourth and fifth digits. Using your answers above, determine Fred's phone number. Explain which digit you determine first, the implications of this, which digit you determine to be the second and so on. RE: Mathematics - NoodooSoup - May. 20, 2011 Necroing about a month, but this is important. OK, so I took yet another practice Algebra Regents (a standardized test) and I came across a question that I could just not solve. So here's what it says (or something like it): There are three hats. Hat A has 5 green marbles and 4 red marbles. Hat B has 6 red marbles and 5 blue marbles. Hat C has 5 green marbles and 5 blue marbles. Then it says for the second part: How many marbles can you add to each hat so that there is a 1/2 probability of picking green in every hat. Ok, so here's some info. You have to add the same amount to every hat, not different amounts. And, considering it's an Algebra test, you probably must solve it algebraically. Also, since this is an 8th/9th grade test, anything learned in higher grade levels probably can't be used. Also, the numbers might not be exactly correct, but I just need to know how to solve this type of question. So here's what I did: Let x = the amount of green marbles added Let y = the total amount of red and blue marbles added Then, I tried to solve it using an algebraic system, using: Hat A: 5 + x = 4 + y Hat B: x = 11 + y Hat C: 5 + x = 5 + y But, in the end, I ended up with y - 1 = y + 11, which makes no sense! RE: Mathematics - Kai-V - May. 20, 2011 Must the added marbles come from the other hats, or do you have an infinite amount elsewhere ? Because otherwise, you cannot always really "add" an amount to each hat. Furthermore, does it need to be the same amount of [colour] marble for each hat ? For instance, two more green marbles in each one ? That is what your system seems to suggest. RE: Mathematics - Ga' - May. 20, 2011 It should be 3 variables, one for red, one for green and one for blue. That could be why. RE: Mathematics - NoodooSoup - May. 20, 2011 (May. 20, 2011 2:50 AM)Kai-V Wrote: Must the added marbles come from the other hats, or do you have an infinite amount elsewhere ? Because otherwise, you cannot always really "add" an amount to each hat. Infinite amounts of marbles. Also, no, it doesn't need to be the same amount. But, half the hat has to be green. The amounts of red and blue don't matter, as long as half is green. And in order for it to be half, it would be 'green = red + blue' so that's what I made my equations on. RE: Mathematics - Ga' - May. 20, 2011 EDIT: Ack, NVM. Noodoo typed the same thing while I typed this. RE: Mathematics - NoodooSoup - May. 20, 2011 (May. 20, 2011 2:51 AM)GaHooleone Wrote: It should be 3 variables, one for red, one for green and one for blue. That could be why. Yeah, but it doesn't really matter how much red and blue you have as long as half of the bag is green. So I figured, I could solve my system and then proceed with finding the amount of blue and red marbles (separately) after I did that part. RE: Mathematics - Kai-V - May. 20, 2011 (May. 20, 2011 2:53 AM)NoodooSoup Wrote: Infinite amounts of marbles. OK, but right now you have : Hat 1 : green + x = red + blue + y Hat 2 : green2 + x = red2 + blue2 + y Etc. It will not be the same x in each case, so unless I am really missing way too much from my good, high school years of mathematics, you cannot blend them and use "x" as if it were always the same. The problem is that x1, x2 and x3 would not really be linked, nor would three y. It seems easy to solve without variables, so perhaps you could work backwards, hah. RE: Mathematics - NoodooSoup - May. 20, 2011 (May. 20, 2011 3:04 AM)Kai-V Wrote:(May. 20, 2011 2:53 AM)NoodooSoup Wrote: Infinite amounts of marbles. Well, x would have to be the same number because you have to add the same exact number of each marble into each bag, so if you add 1 green to Hat A, you would have to add it to all the hats. The same with the y. I tried working backwards, but gave up because it was taking waaay too long and then I tried this plan... which ultimately failed. |