Golden Duck Posted March 14, 2019 #201 Share Posted March 14, 2019 10 hours ago, Ozymandias said: Numbers in a sequence are generated by multiplying successive terms by the same factor. The first four numbers are 24, X, Y, 375 What is X and Y? It's a Geometric Progression. So, we need to find the the factor or the Geometric Mean (GM). GM (in Pythonic form) >>> (375/24)**(1/3) 2.5 X is the result of the first multiplication; ie, raise the factor (GM) to the power of 1 >>> 24*2.5**1 60.0 Y is the result of the second multiplication; ie raise the factor (GM) to the power of 2 >>> 24*2.5**2 150.0 X = 60, Y = 150 1 1 Top Link to comment Share on other sites More sharing options...

Ozymandias Posted March 15, 2019 #202 Share Posted March 15, 2019 1 hour ago, Golden Duck said: It's a Geometric Progression. So, we need to find the the factor or the Geometric Mean (GM). GM (in Pythonic form) >>> (375/24)**(1/3) 2.5 X is the result of the first multiplication; ie, raise the factor (GM) to the power of 1 >>> 24*2.5**1 60.0 Y is the result of the second multiplication; ie raise the factor (GM) to the power of 2 >>> 24*2.5**2 150.0 X = 60, Y = 150 Lovely hurling, Golden Duck! It was a geometric progression with a common ratio r = 2.5 and you are correct, X = 60 and Y =150. The four numbers were: 24, 60, 150, 375, ..... Your turn to set a question. 1 Top Link to comment Share on other sites More sharing options...

Golden Duck Posted March 15, 2019 #203 Share Posted March 15, 2019 (edited) 20 minutes ago, Ozymandias said: Lovely hurling, Golden Duck! It was a geometric progression with a common ratio r = 2.5 and you are correct, X = 60 and Y =150. The four numbers were: 24, 60, 150, 375, ..... Your turn to set a question. OK, I was thinking about a thread in the Alternative History Forum and the 23 Enigma. It might be easy to get to 23 because many numbers have a factor of 2 or 3. What fraction of counting numbers DO NOT have a factor of 2 or 3? Edited March 15, 2019 by Golden Duck Link to comment Share on other sites More sharing options...

danydandan Posted March 15, 2019 Author #204 Share Posted March 15, 2019 7 hours ago, Ozymandias said: Lovely hurling, Golden Duck! It was a geometric progression with a common ratio r = 2.5 and you are correct, X = 60 and Y =150. The four numbers were: 24, 60, 150, 375, ..... Your turn to set a question. I'm glad we have another participant. @Golden Duck I went to bed not expecting anyone else to answer Ozymandis's question. Good to have you on board. 1 Top Link to comment Share on other sites More sharing options...

Golden Duck Posted March 15, 2019 #205 Share Posted March 15, 2019 1 minute ago, danydandan said: I'm glad we have another participant. @Golden Duck I went to bed not expecting anyone else to answer Ozymandis's question. Good to have you on board. I'll try to keep up. I'm not really good with setting questions. I did encounter a question while looking at some metrics for Cricket over the Australian summer. Is finding limits beyond the scope of this thread? PS. Has anyone dabbled in Project Euler? Link to comment Share on other sites More sharing options...

danydandan Posted March 15, 2019 Author #206 Share Posted March 15, 2019 (edited) 7 hours ago, Golden Duck said: OK, I was thinking about a thread in the Alternative History Forum and the 23 Enigma. It might be easy to get to 23 because many numbers have a factor of 2 or 3. What fraction of counting numbers DO NOT have a factor of 2 or 3? Can we include negative factors? 13 minutes ago, Golden Duck said: I'll try to keep up. I'm not really good with setting questions. I did encounter a question while looking at some metrics for Cricket over the Australian summer. Is finding limits beyond the scope of this thread? PS. Has anyone dabbled in Project Euler? Yeah I used to contribute to it when it started. Haven't been able to contribute lately with life getting in the way. And yeah the limits is not beyond the scope. As long as it mathematical it's fine. Edited March 15, 2019 by danydandan Link to comment Share on other sites More sharing options...

Golden Duck Posted March 15, 2019 #207 Share Posted March 15, 2019 3 minutes ago, danydandan said: Can we include negative factors? I was only thinking of integers greater than zero. More or less in the spirit of the Fundamental Theorem Of Arithmetic. But, an expansion to negative numbers would be interesting to see. 3 minutes ago, danydandan said: Yeah I used to contribute to it when it started. Haven't been able to with life getting in the way. And yeah the limits is not beyond the scope. As long as it mathematical it's fine. IIRC I got to about 97 problems completed and likewise, time was spent elsewhere. Link to comment Share on other sites More sharing options...

Ozymandias Posted March 15, 2019 #208 Share Posted March 15, 2019 9 hours ago, Golden Duck said: What fraction of counting numbers DO NOT have a factor of 2 or 3? I'm approaching this using native wit and blunt reasoning. The percentage is certainly less than 50% as all even natural numbers (N) are divisible by 2 and odd numbers are not. This leaves all of the additional odd numbers, elements of N, that are divisible by 3. There is no discernible pattern here. For 1<N<100 these are 3, 9,15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99 Clearly, prime factors and their density are an increasingly important factor here but what percentage of the odd numbers are divisible by 3? However, since 1/3 of all N are divisible by 3 and half of them are even (divisible by 2) then I will argue that 1/6 of all odd numbers have 3 as a factor. This is supported by my list above - 17 out of the first 100 natural numbers (approx 1/6) were divisible by 3. So, what fraction of the counting numbers DO NOT have a factor of 2 or 3? - the answer is 1 - 1/2 - 1/6 = 1/3 or 33.3%. 2 Top Link to comment Share on other sites More sharing options...

Golden Duck Posted March 15, 2019 #209 Share Posted March 15, 2019 1 hour ago, Ozymandias said: I'm approaching this using native wit and blunt reasoning. The percentage is certainly less than 50% as all even natural numbers (N) are divisible by 2 and odd numbers are not. This leaves all of the additional odd numbers, elements of N, that are divisible by 3. There is no discernible pattern here. For 1<N<100 these are 3, 9,15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99 Clearly, prime factors and their density are an increasingly important factor here but what percentage of the odd numbers are divisible by 3? However, since 1/3 of all N are divisible by 3 and half of them are even (divisible by 2) then I will argue that 1/6 of all odd numbers have 3 as a factor. This is supported by my list above - 17 out of the first 100 natural numbers (approx 1/6) were divisible by 3. So, what fraction of the counting numbers DO NOT have a factor of 2 or 3? - the answer is 1 - 1/2 - 1/6 = 1/3 or 33.3%. Correct! One third (1/3) is the answer I came up with. I will make a comment on the bolded part of your reply. I first figured that it was something like: 1 - (1/2 + 1/3 - 1/[2*3]) == 1 - (1/2 + 1/3 - 1/6) Your subset represents an arithmetic progression with a common difference of 6. Take the ordered subset of all counting numbers divisible by 3 (or it's complement) and you should observe that every second number is divisible by 2. One sixth (1/6) of all numbers are divisible by 2 and 3. They are in both subsets but we don't want to count them twice. I still feel like I struggle to prove what seems to be a simple problem. Please set your problem. Link to comment Share on other sites More sharing options...

Ozymandias Posted March 15, 2019 #210 Share Posted March 15, 2019 8 hours ago, Golden Duck said: Correct! One third (1/3) is the answer I came up with. I will make a comment on the bolded part of your reply. I first figured that it was something like: 1 - (1/2 + 1/3 - 1/[2*3]) == 1 - (1/2 + 1/3 - 1/6) Your subset represents an arithmetic progression with a common difference of 6. Take the ordered subset of all counting numbers divisible by 3 (or it's complement) and you should observe that every second number is divisible by 2. One sixth (1/6) of all numbers are divisible by 2 and 3. They are in both subsets but we don't want to count them twice. I still feel like I struggle to prove what seems to be a simple problem. Please set your problem. I will. I've been extremely busy today and will not get time until tomorrow. 1 Top Link to comment Share on other sites More sharing options...

Ozymandias Posted March 16, 2019 #211 Share Posted March 16, 2019 Sorry about the delay in setting a new problem. So, staying with ratio but choosing a different context, here it is:- A right-circular cone, a hemisphere and a cylinder all have the same circular base of radius r and all have the same height above their base. What is the ratio of their volumes to each other? Link to comment Share on other sites More sharing options...

danydandan Posted March 16, 2019 Author #212 Share Posted March 16, 2019 (edited) 14 minutes ago, Ozymandias said: Sorry about the delay in setting a new problem. So, staying with ratio but choosing a different context, here it is:- A right-circular cone, a hemisphere and a cylinder all have the same circular base of radius r and all have the same height above their base. What is the ratio of their volumes to each other? 1:1:1. Volume of a cone = (1/3) πr^22r = (2/3) πr^3 Volumeof cylinder = 2πr^3 Volume of hemisphere = (2/3) πr^3. Edit: No wait 1:3:1. Not 1:1:1. I read that as the ratio of the radius to each other. Edited March 16, 2019 by danydandan 1 Top Link to comment Share on other sites More sharing options...

Ozymandias Posted March 16, 2019 #213 Share Posted March 16, 2019 For all three shapes their height equals their base radius; i.e. h = r. Volume of Cone = 1/3(pi)(r^2)h = 1/3(pi)r^3 Volume of Hemisphere = 1/2(4/3)(pi)r^3 = 2/3(pi)r^3 Volume of Cylinder = (pi)(r^2)h = (pi)r^3 Therefore, Cone : Hemisphere : Cylinder = [1/3(pi)r^3] : [2/3(pi)r^3] : [(pi)r^3] = 1 : 2 : 3 1 1 Top Link to comment Share on other sites More sharing options...

Ozymandias Posted March 17, 2019 #214 Share Posted March 17, 2019 Danydandan gave a correct solution although he made a small computational error, so I'll let him set ghe next problem. Hope that's OK with you, Golden Duck. 2 Top Link to comment Share on other sites More sharing options...

Golden Duck Posted March 17, 2019 #215 Share Posted March 17, 2019 13 hours ago, Ozymandias said: Danydandan gave a correct solution although he made a small computational error, so I'll let him set ghe next problem. Hope that's OK with you, Golden Duck. Carry on @danydandan Link to comment Share on other sites More sharing options...

danydandan Posted March 18, 2019 Author #216 Share Posted March 18, 2019 Coolio @Golden Duck and @Ozymandias. This might be a bit simple, but it might get a few others involved What must be added to each term of the ratio 2:3, so that it may become equal to 4:5? Link to comment Share on other sites More sharing options...

Golden Duck Posted March 19, 2019 #217 Share Posted March 19, 2019 16 hours ago, danydandan said: Coolio @Golden Duck and @Ozymandias. This might be a bit simple, but it might get a few others involved What must be added to each term of the ratio 2:3, so that it may become equal to 4:5? OK, when you ask what must be added to each term of the ratio, I'm assuming the ratio is, in fact, a vector. For example, you may have conducted an experiment and recorded two (2) success from three (3) trials. After a subsequent running of the experiment, you find you have recorded a total of four (4) successes from five (5) trials. What were the results of the second iteration of the experiment? If this is the case we will use what is occasionally called the freshman sum, the mediant, or the vector sum. U + X == V X == V - U A pythonic calculation: >>> U = (2, 3) >>> V = (4, 5) >>> X = (V[0] - U[0], V[1] - U[1]) >>> X (2, 2) The second running of the experiment resulted in two (2) successes from two (2) trials. To answer the original question - 2 must be added to each term to get 4:5. 1 Top Link to comment Share on other sites More sharing options...

danydandan Posted March 19, 2019 Author #218 Share Posted March 19, 2019 (edited) 3 hours ago, Golden Duck said: OK, when you ask what must be added to each term of the ratio, I'm assuming the ratio is, in fact, a vector. For example, you may have conducted an experiment and recorded two (2) success from three (3) trials. After a subsequent running of the experiment, you find you have recorded a total of four (4) successes from five (5) trials. What were the results of the second iteration of the experiment? If this is the case we will use what is occasionally called the freshman sum, the mediant, or the vector sum. U + X == V X == V - U A pythonic calculation: >>> U = (2, 3) >>> V = (4, 5) >>> X = (V[0] - U[0], V[1] - U[1]) >>> X (2, 2) The second running of the experiment resulted in two (2) successes from two (2) trials. To answer the original question - 2 must be added to each term to get 4:5. Yeap. Interesting way to achieve the correct result. I was thinking something a bit less technical. (2 + x) : (3 + x) = 4 : 5 (2 + x)/(5 + x) = 4/5 5(2 + x) = 4(3 + x) 10 + 5x = 12 + 4x 5x - 4x = 12 - 10 x = 2 Your turn to set a question. Edited March 19, 2019 by danydandan 1 Top Link to comment Share on other sites More sharing options...

Golden Duck Posted March 19, 2019 #219 Share Posted March 19, 2019 (edited) 3 hours ago, danydandan said: Yeap. Interesting way to achieve the correct result. I was thinking something a bit less technical. (2 + x) : (3 + x) = 4 : 5 (2 + x)/(5 + x) = 4/5 5(2 + x) = 4(3 + x) 10 + 5x = 12 + 4x 5x - 4x = 12 - 10 x = 2 Your turn to set a question. A further post mortem on the previous problem: Spoiler Damn! Your method brings back memories of call centre reporting. (s + c)/(t + c) = 80%. I found trouble with this formula when s/t was greater than 80%. I never really found a use for it in reports. I did try representing periods as vectors to try to preserve the magnitude and show the effects of busy periods. I'd have to adjust my vector method if s/t was of a larger magnitude; eg, 20/30. The Problem I want to build a quadrangular chicken yard against a 5 metre wall of my house. So, I purchase 5 metres of fencing to make the remaining 3 sides of the yard. What dimensions will maximise the area of the yard? Edited March 19, 2019 by Golden Duck Link to comment Share on other sites More sharing options...

Ozymandias Posted March 19, 2019 #220 Share Posted March 19, 2019 5 hours ago, Golden Duck said: The Problem I want to build a quadrangular chicken yard against a 5 metre wall of my house. So, I purchase 5 metres of fencing to make the remaining 3 sides of the yard. What dimensions will maximise the area of the yard? Fundamental calculus is all that is required here:- Let X be the length of each of the two sides coming out from the existing wall of your house. Let Y be the length of the other side opposite the wall of your house. Area (A) of Yard = XY = A 3-sided Length of Fence => 5 = 2X + Y => Y = 5 - 2X A = XY = X(5 - 2X) = 5X - 2X^2 = A A(max) occurs at dy/dx = 0 dy/dx = 5 - 4X => 5 - 4X = 0 => X = 5/4 = 1.24 Perimeter Length 5 = 2X + Y 5 = 2(1.25) + Y => Y = 5 - 2(1.25) = 2.5 Your yard should be 1.5m × 2.5m giving a max area of 3.75m^2. 2 Top Link to comment Share on other sites More sharing options...

Ozymandias Posted March 19, 2019 #221 Share Posted March 19, 2019 (edited) [A few typos in my last solution: X = 1.25m and Y = 2.5m] Problem What is the percentage change in the volume of a cylinder whose radius is reduced by 10% and whose height is increased by 15%? Edited March 19, 2019 by Ozymandias 1 Top Link to comment Share on other sites More sharing options...

danydandan Posted March 19, 2019 Author #222 Share Posted March 19, 2019 (edited) 37 minutes ago, Ozymandias said: Problem What is the percentage change in the volume of a cylinder whose radius is reduced by 10% and whose height is increased by 15%? V=πr^2h. V=π.(1^2).1 V=3 Take away the amounts above. V=2 Percentage change is around 33.33333333333333333% Give or take. Edit: Bloody increase in height. Let me start again. Percentage change, needs to be more decimal points now, (you snake Ozy) It's around 6-7%. I actually edited my original response which said 7, but I was taking the p***. Maybe my subconscious read the problem correctly? Edited March 19, 2019 by danydandan 1 Top Link to comment Share on other sites More sharing options...

Ozymandias Posted March 19, 2019 #223 Share Posted March 19, 2019 3 hours ago, danydandan said: V=πr^2h. V=π.(1^2).1 V=3 Take away the amounts above. V=2 Percentage change is around 33.33333333333333333% Give or take. Edit: Bloody increase in height. Let me start again. Percentage change, needs to be more decimal points now, (you snake Ozy) It's around 6-7%. I actually edited my original response which said 7, but I was taking the p***. Maybe my subconscious read the problem correctly? So, Dan. Do you want to give the solution with the exact answer? Link to comment Share on other sites More sharing options...

danydandan Posted March 19, 2019 Author #224 Share Posted March 19, 2019 18 minutes ago, Ozymandias said: So, Dan. Do you want to give the solution with the exact answer? I'll have to wait till tomorrow, too late in the evening now. Link to comment Share on other sites More sharing options...

Golden Duck Posted March 19, 2019 #225 Share Posted March 19, 2019 (edited) V_0 (r = 1, h = 1) == π(r^2)(h) == π(1)(1) == π V_1 (r = 0.9, h = 1.15) == π(r^2)(h) == π(0.9^2)(1.15) == π(0.81)(1.15) = π(0.9315) V_1 is 93.15 per cent of V_0. A difference of 6.85. Pretty good estimation by @danydandan. I would've just averaged 0.88 and 0.96 to get 0.92. Edited March 19, 2019 by Golden Duck 1 Top Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now