Harte Posted March 26, 2019 #276 Share Posted March 26, 2019 You don't need to assume anything about the phrase "the first boy that can tell me." You have to assume three damn intelligent boys that all need a dollar bad enough to stop and think. You've got the first step Duck - without assuming plurality in "first boy." Harte 1 Link to comment Share on other sites More sharing options...
danydandan Posted March 26, 2019 Author #277 Share Posted March 26, 2019 (edited) 24 minutes ago, Harte said: You don't need to assume anything about the phrase "the first boy that can tell me." You have to assume three damn intelligent boys that all need a dollar bad enough to stop and think. You've got the first step Duck - without assuming plurality in "first boy." Harte Let me see if I can't get it. We need to put ourselves into one of the kids shoes. Step One: If only one of the kids had the smudge, then the one who had the smudge wouldn't notice any smudge and would not raise there hand. Since all of them didn't raise their hands at once and they can see each other, they all realise that there must be at least two smudges. Step Two: If only two of them had smudges, those with the smudges would notice no smudge on the third and realise that they themselves must have a smudge. They both would have raised their hand at the same time. This didn't occur because the third kid also has a smudge on their heads. Step Three: Since one and two didn't occur a smart kid realises all three have smudges and raises his hands. Hence why I said the others didn't raise their hands. Case solved, and I can't think of another way that any of the kids could have 'known', and not assumed, they had a smudge on their head. Edited: Probably should have elaborated on my initial post. Edited March 26, 2019 by danydandan 1 Link to comment Share on other sites More sharing options...
Harte Posted March 26, 2019 #278 Share Posted March 26, 2019 (edited) Essentially solved, with caveats. 1 hour ago, danydandan said: Let me see if I can't get it. We need to put ourselves into one of the kids shoes. Step One: If only one of the kids had the smudge, then the one who had the smudge wouldn't notice any smudge and would not raise there hand. Since all of them didn't raise their hands at once and they can see each other, they all realise that there must be at least two smudges. Each boy sees two smudges so no reason for this. Also, every boy knows that the case of all 3 NOT having smudges is false. Quote Step Two: If only two of them had smudges, those with the smudges would notice no smudge on the third and realise that they themselves must have a smudge. They both would have raised their hand at the same time. This didn't occur because the third kid also has a smudge on their heads. This requires elaboration. Boy A assumes he has no smudge in the interest of arriving at a proof by contradiction. Boy A then knows that Boy B must then see that Boy A has no smudge and Boy C has a smudge. Boy A allows that Boy B considers whether he has a smudge, and assumes not (like Boy A did.) If Boy A's assumption of no self-smudge is correct, then Boy A knows that Boy B would know that Boy C sees no other smudges (if Boy B's assumption is correct) and Boy C would therefore know that he himself was smudged. Boy B would then know that his assumption was incorrect, since Boy C has not answered the challenge, and Boy B would know he was smudged. Since Boy B hasn't answered the man, Boy A knows his assumption of no self-smudge has been contradicted and that he therefore has a smudge on his forehead and is a dollar richer. Harte Edited March 26, 2019 by Harte or by farte 1 Link to comment Share on other sites More sharing options...
Harte Posted March 26, 2019 #279 Share Posted March 26, 2019 Now that that's over with, here's a little light reading. Sum-of-Three-Cubes Problem Solved for ‘Stubborn’ Number 33 Harte 2 Link to comment Share on other sites More sharing options...
danydandan Posted March 27, 2019 Author #280 Share Posted March 27, 2019 7 hours ago, Harte said: Essentially solved, with caveats. Each boy sees two smudges so no reason for this. Also, every boy knows that the case of all 3 NOT having smudges is false. This requires elaboration. Boy A assumes he has no smudge in the interest of arriving at a proof by contradiction. Boy A then knows that Boy B must then see that Boy A has no smudge and Boy C has a smudge. Boy A allows that Boy B considers whether he has a smudge, and assumes not (like Boy A did.) If Boy A's assumption of no self-smudge is correct, then Boy A knows that Boy B would know that Boy C sees no other smudges (if Boy B's assumption is correct) and Boy C would therefore know that he himself was smudged. Boy B would then know that his assumption was incorrect, since Boy C has not answered the challenge, and Boy B would know he was smudged. Since Boy B hasn't answered the man, Boy A knows his assumption of no self-smudge has been contradicted and that he therefore has a smudge on his forehead and is a dollar richer. Harte That's basically what I said.......... Since nobody got it Harte, you can set a new one or pass the buck if you want too. 1 Link to comment Share on other sites More sharing options...
Harte Posted March 27, 2019 #281 Share Posted March 27, 2019 Actually, You solved it. Just wanted to be clear why Boys B and C would realize they had a smudge if Boy A didn't. Harte Link to comment Share on other sites More sharing options...
danydandan Posted March 27, 2019 Author #282 Share Posted March 27, 2019 21 minutes ago, Harte said: Actually, You solved it. Just wanted to be clear why Boys B and C would realize they had a smudge if Boy A didn't. Harte Not sure I explained it correctly or clearly enough. But if you think that my simple elaboration was sufficient I'll try think of a new question. Link to comment Share on other sites More sharing options...
danydandan Posted March 27, 2019 Author #283 Share Posted March 27, 2019 If 1/2 of 5 equals 3. Using the same proportions, what does 1/3 of 10 equal? Link to comment Share on other sites More sharing options...
Ozymandias Posted March 27, 2019 #284 Share Posted March 27, 2019 (edited) 1 hour ago, danydandan said: If 1/2 of 5 equals 3. Using the same proportions, what does 1/3 of 10 equal? These recent problems (with answers) are readily found on the internet. It kinda defeats the purpose of the thread - in my view. Anyway, to answer the current teaser: 1/2 of 5 = 3 > 5 = 2(3) = 6 > 10 = 2(6) = 12 > 1/3 of 10 = 1/3 of 12 = 4 Edited March 27, 2019 by Ozymandias Link to comment Share on other sites More sharing options...
danydandan Posted March 27, 2019 Author #285 Share Posted March 27, 2019 3 minutes ago, Ozymandias said: These recent problems (with answers) are readily found on the internet. It kinda defeats the purpose of the thread - in my view. Anyway, to answer the current teaser: 1/2 of 5 = 3 > 5 = 2(3) = 6 > 10 = 2(6) = 12 > 1/3 of 10 = 1/3 of 12 = 4 That sucks, I got it from one of my uncle's old engineering books. Applied Advanced Mathematics. Link to comment Share on other sites More sharing options...
Ozymandias Posted March 27, 2019 #286 Share Posted March 27, 2019 2 minutes ago, danydandan said: That sucks, I got it from one of my uncle's old engineering books. Applied Advanced Mathematics. I was just sayin', Dan, - not trying to be negative or undermining. Anyway, here's my new problem. There is nothing advanced about it. As agreed under the rules of the thread, Second-level (High School) maths is enough to solve it. A walker must make a journey of 238km (miles, if you prefer). The first day he walks 30km and every day thereafter he walks 2km less than the previous day. How long will it take him to complete his journey? Link to comment Share on other sites More sharing options...
Golden Duck Posted March 28, 2019 #287 Share Posted March 28, 2019 20 hours ago, Ozymandias said: I was just sayin', Dan, - not trying to be negative or undermining. Anyway, here's my new problem. There is nothing advanced about it. As agreed under the rules of the thread, Second-level (High School) maths is enough to solve it. A walker must make a journey of 238km (miles, if you prefer). The first day he walks 30km and every day thereafter he walks 2km less than the previous day. How long will it take him to complete his journey? The distance per day (speed), for the walker is decreasing at a constant rate; ie, -2km per day. A graph representing the walker's speed can be represented by a straight line graph with a gradient of m = -2. f(x) == c - 2x The area under the line equals the total distance travelled by the walker; and, will either be a triangle or a trapezoid. The area under the graph for the first day is 30. A == 30 == int[0>1] (b - 2x)dx This represents a trapezoid with the base at x=0 and top at x=1 and the area A=30. If we let b=f(0), a=f(1) and h=x-1 we can calculate the y-intercept using the formula for the area of a trapezoid. A == h(b + a)/2 30 == 1(b + a)/2 60 = b + a We also know that because the gradient m=-2 that a-b=-2. We can use a system of equations to find b (the y-intercept). b + a == 60 b - a == 2 b + a + b - a == 60 + 2 2b == 62 b == 31 Our equation to model the speed of the walker is: f(x) == y == 31 - 2x The time taken to walk 238 will be the value of x for a trapezoid bound by the x-axis the y-axis and the function f(x) - 31 - 2x. A == 238 == int[0>x] (31 - 2x)dx == [31x - x^2 + k][0:x] == 31x - x^2 + k - k x^2 - 31x == -238 x^2 - 31x + (31/2)^2 == (15.5)^2 - 238 (x - 15.5)^2 == 240.25 - 238 x - 15.5 == (2.25)^(1/2) x == 15.5 +/- 1.5 x == 14; or, x == 17 Since f(17) is negative x = 14. Therefore, it took the walker 14 days to complete the journey of 238km. 1 Link to comment Share on other sites More sharing options...
Ozymandias Posted March 28, 2019 #288 Share Posted March 28, 2019 1 hour ago, Golden Duck said: The distance per day (speed), for the walker is decreasing at a constant rate; ie, -2km per day. A graph representing the walker's speed can be represented by a straight line graph with a gradient of m = -2. f(x) == c - 2x The area under the line equals the total distance travelled by the walker; and, will either be a triangle or a trapezoid. The area under the graph for the first day is 30. A == 30 == int[0>1] (b - 2x)dx This represents a trapezoid with the base at x=0 and top at x=1 and the area A=30. If we let b=f(0), a=f(1) and h=x-1 we can calculate the y-intercept using the formula for the area of a trapezoid. A == h(b + a)/2 30 == 1(b + a)/2 60 = b + a We also know that because the gradient m=-2 that a-b=-2. We can use a system of equations to find b (the y-intercept). b + a == 60 b - a == 2 b + a + b - a == 60 + 2 2b == 62 b == 31 Our equation to model the speed of the walker is: f(x) == y == 31 - 2x The time taken to walk 238 will be the value of x for a trapezoid bound by the x-axis the y-axis and the function f(x) - 31 - 2x. A == 238 == int[0>x] (31 - 2x)dx == [31x - x^2 + k][0:x] == 31x - x^2 + k - k x^2 - 31x == -238 x^2 - 31x + (31/2)^2 == (15.5)^2 - 238 (x - 15.5)^2 == 240.25 - 238 x - 15.5 == (2.25)^(1/2) x == 15.5 +/- 1.5 x == 14; or, x == 17 Since f(17) is negative x = 14. Therefore, it took the walker 14 days to complete the journey of 238km. Correct, Golden Duck. But there is a simpler solution: The distance walked each day is 30, 28, 26, 24, ...... This is an arithmetic progression with a common difference d = -2 and a = 30. Using the sum of n terms of an AP formula Sn = (n/2)[2a + (n-1)d] > 238 = (n/2)[2(30) + (n-1)(-2)] > 238 = (n-2)[60 - 2n + 2] > 238 = (n/2)[62 - 2n] > 238 = n[31 - n] > 238 = 31n - n^2 > n^2 - 31n + 238 = 0 (a quadratic equation: a = 1, b= -31 and c = 238) n = {-(-31) +/- rt[(-31)^2 - 4(1)(238)]}/2(1) n = [31 +/- 3]÷2 = 17 or 14. 14 days walking comes sooner than 17 so 14 gays is the answer. Your turn to set a problem. 1 Link to comment Share on other sites More sharing options...
Golden Duck Posted March 30, 2019 #289 Share Posted March 30, 2019 On 27/03/2019 at 10:05 PM, Ozymandias said: I was just sayin', Dan, - not trying to be negative or undermining. Anyway, here's my new problem. There is nothing advanced about it. As agreed under the rules of the thread, Second-level (High School) maths is enough to solve it. A walker must make a journey of 238km (miles, if you prefer). The first day he walks 30km and every day thereafter he walks 2km less than the previous day. How long will it take him to complete his journey? I need some time to pose a new problem I'm happy to forfeit my turn Link to comment Share on other sites More sharing options...
danydandan Posted March 30, 2019 Author #290 Share Posted March 30, 2019 This might be too niche, but I was messing around with capacitor banks in work yesterday so if I can interject with this question as Golden Duck needs some time. What mathematical method should I employ to calculate a power factor or the efficiency of building or device? Once the power factor is calculated, how can I calculate the corrections needed to achieve an improved efficiency. If that's too niche of a question someone else can set a question. Link to comment Share on other sites More sharing options...
Ozymandias Posted March 30, 2019 #291 Share Posted March 30, 2019 1 hour ago, danydandan said: This might be too niche, but I was messing around with capacitor banks in work yesterday so if I can interject with this question as Golden Duck needs some time. What mathematical method should I employ to calculate a power factor or the efficiency of building or device? Once the power factor is calculated, how can I calculate the corrections needed to achieve an improved efficiency. If that's too niche of a question someone else can set a question. Not being an electrical (but a mechanical) engineer I doubt I will be of any assistance to you. However, as far as I know the determination of a power factor or its correction is a standard calculation. Its essentially just an efficiency ratio (factor, percentage); i.e. a measure of the energy (power) gainfully used by a device or system compared to the energy supplied to it. In other words, just compare the output to the imput. Link to comment Share on other sites More sharing options...
danydandan Posted March 30, 2019 Author #292 Share Posted March 30, 2019 (edited) 2 hours ago, Ozymandias said: Not being an electrical (but a mechanical) engineer I doubt I will be of any assistance to you. However, as far as I know the determination of a power factor or its correction is a standard calculation. Its essentially just an efficiency ratio (factor, percentage); i.e. a measure of the energy (power) gainfully used by a device or system compared to the energy supplied to it. In other words, just compare the output to the imput. Your basically correct but the method involves, Pythagorean theorem and applying Cos^-1 and Tan conversion's of said efficiencies. So you need to acquire the apparent power (kva), the power (kw) and the reactive power (kvar). When you do it with DC it's slightly different. You should set a particularly little known mechanical engineering question. Edited March 30, 2019 by danydandan Link to comment Share on other sites More sharing options...
Ozymandias Posted March 31, 2019 #293 Share Posted March 31, 2019 On 30/03/2019 at 11:48 AM, danydandan said: Your basically correct but the method involves, Pythagorean theorem and applying Cos^-1 and Tan conversion's of said efficiencies. So you need to acquire the apparent power (kva), the power (kw) and the reactive power (kvar). When you do it with DC it's slightly different. You should set a particularly little known mechanical engineering question. Sorry for the delay in responding to this. I was away on family business and was offline due to a faulty iphone. Came back home this evening to my laptop. I could set a mechanical engineering problem - that is my daily activity as a lecturer - but I doubt the wisdom of doing it here when more general problems involving second level maths are not garnering much interest. After all, besides the mathematical skills, engineering requires specialist knowledge of the context as well as an ability to handle units. 1 Link to comment Share on other sites More sharing options...
danydandan Posted March 31, 2019 Author #294 Share Posted March 31, 2019 13 minutes ago, Ozymandias said: Sorry for the delay in responding to this. I was away on family business and was offline due to a faulty iphone. Came back home this evening to my laptop. I could set a mechanical engineering problem - that is my daily activity as a lecturer - but I doubt the wisdom of doing it here when more general problems involving second level maths are not garnering much interest. After all, besides the mathematical skills, engineering requires specialist knowledge of the context as well as an ability to handle units. No worries, I'll set one tomorrow. Or if anyone else wants to pop off a question they are more than welcome. I'm finally fit enough to start back working more than two days a week, so tomorrow is my first Monday in work in two years. So I might have to leave off until after 5 or 6 pm. 1 Link to comment Share on other sites More sharing options...
Ozymandias Posted March 31, 2019 #295 Share Posted March 31, 2019 Just now, danydandan said: No worries, I'll set one tomorrow. Or if anyone else wants to pop off a question they are more than welcome. I'm finally fit enough to start back working more than two days a week, so tomorrow is my first Monday in work in two years. So I might have to leave off until after 5 or 6 pm. Best of luck with that, Dan. Hope your health returns to full strength soon. I might put up another maths question too if I can think of one. 1 Link to comment Share on other sites More sharing options...
danydandan Posted April 1, 2019 Author #296 Share Posted April 1, 2019 Ok Dude/ Dudettes. I've had this for a while I didn't want to use it for this thread as it's more formal logic than mathematical. But anyways.... You go for an interview at a major computer science manufacturer or company. During the interview you are given this problem to solve (I was actually asked this. I can't recall the exact wording but I remember the gist, I was also given three hours to solve it.) In the room in front of you is three AI's, on three desks. One AI always tells the truth, one always lies and the other is purely random with it's responses.( Truth, Lies and Random.) The AI's understand all languages but will only respond in 0 or 1. The issue is you don't know which AI is which, you don't know if 0 means yes or no, nor do you know if 1 means yes or no. You only have three closed ended questions to ask (yes or no questions), each question can only be directed at on AI. With your three question limit, how would you figure out which AI is which? Link to comment Share on other sites More sharing options...
danydandan Posted April 2, 2019 Author #297 Share Posted April 2, 2019 By the way, I'm not sure if I got it correct at that interview because....well I didn't get that job. 1 Link to comment Share on other sites More sharing options...
Ozymandias Posted April 10, 2019 #298 Share Posted April 10, 2019 On 01/04/2019 at 8:41 PM, danydandan said: Ok Dude/ Dudettes. I've had this for a while I didn't want to use it for this thread as it's more formal logic than mathematical. But anyways.... You go for an interview at a major computer science manufacturer or company. During the interview you are given this problem to solve (I was actually asked this. I can't recall the exact wording but I remember the gist, I was also given three hours to solve it.) In the room in front of you is three AI's, on three desks. One AI always tells the truth, one always lies and the other is purely random with it's responses.( Truth, Lies and Random.) The AI's understand all languages but will only respond in 0 or 1. The issue is you don't know which AI is which, you don't know if 0 means yes or no, nor do you know if 1 means yes or no. You only have three closed ended questions to ask (yes or no questions), each question can only be directed at on AI. With your three question limit, how would you figure out which AI is which? There are no takers on your last teaser, Dan, and it is not really a maths problem - more to do with logic, an field into which I rarely venture and certainly not on this occasion. If you don't mind I will pose another maths brain teaser: Three metal discs of radius 1metre, 2m and 3m respectively are laid flat in such a way that each disc touches the circumference of other two. What is the area trapped between the circumferences of the discs? 1 Link to comment Share on other sites More sharing options...
Harte Posted April 10, 2019 #299 Share Posted April 10, 2019 The disc centers form a 3-4-5 triangle. The right angle in that triangle sweeps out a fourth of the 1 meter circle area. The other two angles are not as clean (apprx. 53.13 degrees and it's complement.) The area of the triangle is 6. The sum of the areas areas of the sectors of the discs is approximately 5.59. The difference is 0.41 m^{2} . That's with rounding. I didn't feel like putting in the time to express it in terms of pi. Left my good calculator in my classroom. That's a nice problem and I'm certain there are several other ways to solve it. Harte Link to comment Share on other sites More sharing options...
purrrpetrator Posted April 10, 2019 #300 Share Posted April 10, 2019 (edited) accidental submission!!! Edited April 10, 2019 by purrrpetrator Link to comment Share on other sites More sharing options...
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