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An infinitely stretchable rubber-band has one end attached to the wall and the other end it attached to a scooter being driven Prof. Templeton. The good Prof. drives away from the wall at a constant speed of 1 meter per second. Starting from the wall and walking out onto the rubber-band is a bug. The bug travels at a constant speed of 1 cm per second. Both start moving at the same time in the same direction. The Prof. moves away from the wall perpendicularly and the bug moves along the stretching rubber-band away from the wall toward the Prof. Let’s assume the rubber-band is perpendicular to the wall with no angle of incidence (all along one plane) and the bug is a point on a line initially at distance 0 from the wall.

Will the bug ever reach Prof. Templeton, and if so can you tell how long it will take?

Edited by Prof. Templeton
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Is that a typo, or is it really 1 km per second? If I'm understanding the wording right, it all looks something like this...

WALL|

| |

| |

| |

| ^ |BUG---------------------------------------SCOOTER

| V |1 cm per second -> 1 km per second ->

| |

| |

WALL|

Is that the proper layout? If so, and the speeds are correct, the bug can reach the good Professor when he runs out of gas or circles the world and laps the bug

Edited by Mumbles140
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Is that a typo, or is it really 1 km per second? If I'm understanding the wording right, it all looks something like this...

|

|

|

|BUG---------------------------------------SCOOTER

|

|

|

Is that the proper layout? If so, and the speeds are correct, the bug can reach the good Professor when he runs out of gas or circles the world and laps the bug

I edited it to 1 meter. Sorry for the confusion, but that would be one heck of a scooter.

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Yes, the bug will reach the professor. Since the bug is on the same rubber band, travelling at the same speed (think airplane passenger), he has a relative speed 1 cm per second greater than the Professor (no relative velocity)...the time it takes will be the distance from the rubber band attachment to the Professor at a speed of 1 cm per second

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Yes, the bug will reach the professor. Since the bug is on the same rubber band, travelling at the same speed (think airplane passenger), he has a relative speed 1 cm per second greater than the Professor (no relative velocity)...the time it takes will be the distance from the rubber band attachment to the Professor at a speed of 1 cm per second

After 1 second the Prof. is 1 meter from the wall after 2 seconds he's 2 meters, and so on...while the bug has traveled 1 cm after 1 second and 2 cm after 2 seconds, and so on... So, initially they were at the same spot, but when the timer starts they travels in the same direction. Will they be at the same spot again, and how long from now? Does this clarify at all?

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Yes--it confirms my drawing was correct. But what if the stretching of the rubber band comes from the wall side, whereas the bug is riding on the Scooter side. Hold your two thumbnails together and call them Distance-Zero (the location of the Scooter and Bug at 0 Time). Then pull one of them away in a straight line. The bug could still be attached to the moving thumbnail (Scooter) as opposed to the stationary one (wall), thus he is always 'with' the Professor

At the very least, assuming equal stretching came from both ends of the rubber band, the bug would wind up in the very middle of the rubber band plus the distance traveled in that amount of seconds. (i.e. at 10 seconds, the bug would be at 5 meters and 10 centimeters). Ultimately, he will always be at 51% of the distance between the wall and the Professor.

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Yes--it confirms my drawing was correct. But what if the stretching of the rubber band comes from the wall side, whereas the bug is riding on the Scooter side. Hold your two thumbnails together and call them Distance-Zero (the location of the Scooter and Bug at 0 Time). Then pull one of them away in a straight line. The bug could still be attached to the moving thumbnail (Scooter) as opposed to the stationary one (wall), thus he is always 'with' the Professor

At the very least, assuming equal stretching came from both ends of the rubber band, the bug would wind up in the very middle of the rubber band plus the distance traveled in that amount of seconds. (i.e. at 10 seconds, the bug would be at 5 meters and 10 centimeters). Ultimately, he will always be at 51% of the distance between the wall and the Professor.

Would it matter if the scooter started 1 meter from the wall at time=0. Than the bug would be at the wall and the good Prof. would be 1 meter away.

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Well that would make an obvious difference, so acting under the assumption that they both moved at the rate of the rubber band being pulled, plus the rate of the bug moving relative to the scooter, it would take 100 seconds.

If you are looking for some other solution, then it all depends on how the rubber band moves the bug (not just how the bug moves on the rubber band)

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Well that would make an obvious difference, so acting under the assumption that they both moved at the rate of the rubber band being pulled, plus the rate of the bug moving relative to the scooter, it would take 100 seconds.

If you are looking for some other solution, then it all depends on how the rubber band moves the bug (not just how the bug moves on the rubber band)

I think the OP would be better if I had worded it with a headstart as you pointed out, but I also think the answer may be a lot larger than what you said above.

it would take the bug 100 seconds to travel a 1 meter band, but the band is now 101 meters long after 100 seconds (and headstart).

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An infinitely stretchable rubber-band has one end attached to the wall and the other end it attached to a scooter being driven Prof. Templeton. The good Prof. drives away from the wall at a constant speed of 1 meter per second. Starting from the wall and walking out onto the rubber-band is a bug. The bug travels at a constant speed of 1 cm per second. Both start moving at the same time in the same direction. The Prof. moves away from the wall perpendicularly and the bug moves along the stretching rubber-band away from the wall toward the Prof. Let’s assume the rubber-band is perpendicular to the wall with no angle of incidence (all along one plane) and the bug is a point on a line initially at distance 0 from the wall.

Will the bug ever reach Prof. Templeton, and if so can you tell how long it will take?

1. I can walk faster, than Prof. T. rides his scooer.

2. The initial length of the string is not mentioned, nor the length of the bug. If the bug can reach the rear fender from start, it can simply hang on to it.

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Imagine you are on an escalator and a person is 100 steps above you (very long escalator). The escalator begins moving under you, but since you are on it, you move at the same speed it does. In addition to this, you walk at a rate of 1 step per second. You will reach the other person in 100 seconds.

Translating this to the problem at hand...

Scooter: With the 1m headstart, the scooter will be 101 meters from the wall after 100 seconds (100 X 1 mps)

Bug: The bug will be 101 meters from the wall after 100 seconds. The rubber band will carry it the 100 meters in 100 seconds, while simultaneously, he will move 100 cms in 100 seconds.

This can all be seen by the formula below, where the bug is the left side and the scooter is the right:

1m(100s) + 1cm (100s) = 1m (100s) + 1m (0s)

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1. I can walk faster, than Prof. T. rides his scooer.

2. The initial length of the string is not mentioned, nor the length of the bug. If the bug can reach the rear fender from start, it can simply hang on to it.

Yeah, the scooters a little beat up, but it gets great gas milage. I told Mumbles140 I should have put a headstart in the OP. I don't want people to assume the bug can hang on (especially at the break-neck speed of 1 meter / second).

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But if the bug doesn't hang on, that means the rubber band would be a frictionless surface, and the bug would be unable to walk on it in the first place ;)

Hypothetically, if no rubber band exists and one object moves along a straight line at 1m per second, and another object moves along that very same line at 1/100 of that speed, then the time it would take for them to meet would be 101% of the time it takes for the first object to completely orbit the sphere in which they are moving

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Are we allowed the use of double integral and natural logarithm for solving this puzzle?

Do we allow puzzles that require the knowledge of advanced calculus on this forum?

Without use of advanced calculus, I can give you only half solution for now. I'll try to find simple non-calculus way to solve the puzzle. But if you are not aware of such way yourself, please let us know. <_<

Strange as it sounds, Bug will catch up to the scooter, if it doesn't die from exhaustion and exhaust fumes first.

Assume the string stretches enenly. The end point attached to the scooter is moving at 1m/sec (100cm/sec); the mid-point of the string is moving at 50cm/sec away from the wall. (Each second, the half of the string from the scooter to the mid-point stretches 50cm and the other half from mid point to the wall stretches another 50cm. Thus the mid point moves away from the wall at 50cm/sec).

Similarly, the point at 1/4 current length of the string from the wall moves at 25cm/sec; the point at 1/100 string length from the wall -- at 1cm/sec; 2/100 -- 2cm/sec and so on.

The instanteneous speed of Bug with respect to the wall is the speed of the string point which it is on, plus Bug's own speed (1cm/sec). For example, if Bug made it to the mid-point somehow, it would be moving at 50 + 1 = 51 (cm/sec).

To even start catching up to some point on the string ahead, Bug must travel faster than that point. Thus to start catching up to the scooter, Bug must travel faster than 100cm/sec. Therefore, Bug must be past 99/100 of the length of the string from the wall where its combined speed with the point of the string it is on will exceed 100 cm/sec.

Let's examine now how Bug can catch up with different points on the string:

Initially, Bug is moving at 1cm/sec plus whatever speed of the point it is on. That is faster than than the point at 1/100 length of the string, which moves at constant speed of 1cm/sec. Thus Bug will catch up to that point. But once it passes that point, it'll moves faster than 2cm/sec, which is faster than the 2/100 point. And so on.

THUS BUG CAN CATCH UP TO ANY POINT ON THE STRING.

How long -- is another question. I can tell you off the back of my head, it would be more than 200 seconds. (Bug's acceleration is less than 1cm/sec2.)

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I thought Mumbles might have gotten it with the "Prof. Templeton would have to catch up to the bug by traveling around the world" part, cause if he's moving faster than the bug, and the rubber band doesn't move the bug while it stretches, then there's really no way the bug could catch up...

BTW that would be like the coolest rubber band ever... I'll bet it would make an awesome rubber band ball xD

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I thought Mumbles might have gotten it with the "Prof. Templeton would have to catch up to the bug by traveling around the world" part, cause if he's moving faster than the bug, and the rubber band doesn't move the bug while it stretches, then there's really no way the bug could catch up...

BTW that would be like the coolest rubber band ever... I'll bet it would make an awesome rubber band ball xD

I think the trick is that the rubber band does move the bug as it stretches. If the bug is standing still at the halfway point of the rubber band, it will stay at the halfway point as the rubber band stretches. Basically, the distance in front of the bug and the distance behind it increase at the same rate (if it were standing still).

If this weren't true, then the bug could obviously never catch up.

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Yes, I'm quoting myself...

At the very least, assuming equal stretching came from both ends of the rubber band, the bug would wind up in the very middle of the rubber band plus the distance traveled in that amount of seconds. (i.e. at 10 seconds, the bug would be at 5 meters and 10 centimeters). Ultimately, he will always be at 51% of the distance between the wall and the Professor.

Is my logic wrong in the above thoughts? If we are saying the bug is always halfway between the scooter and the wall at all times, then that still means ever second that passes equates to the bug travelling 1 cm towards the scooter, but the scooter moving 50 cm away. The bug is actually losing ground at the rate of 49 cms per second, and if he's losing ground, he can't "catch up". If after 1 second he is 49 cms behind the scooter (which is at 100 cms), that puts him at 51%

Of course, this all assumes that there would be no 1m 'headstart'

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I think the trick is that the rubber band does move the bug as it stretches. If the bug is standing still at the halfway point of the rubber band, it will stay at the halfway point as the rubber band stretches. Basically, the distance in front of the bug and the distance behind it increase at the same rate (if it were standing still).

If this weren't true, then the bug could obviously never catch up.

Well then that poses the question 'What did Prof. Templeton mean when he said the bug moves at 1 cm per second?'

Did he mean that the bug walked 1 cm/s or that the rubber band moved him at 1 cm/s?

It's impossible if the rubber band moved him 1 cm/s because the end of a rubber band doesn't move if you stretch it.

For example, take a rubber band and cut it so its not in a circle, then draw a line at the end with a pen or something and stretch the rubber band out. You'll see that the black line doesn't move...

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Alright, I went through the differential equation, but I think I might have messed up in one or more places. Anyone with a better memory of this stuff want to try it and see what they get?

OK, let t be the time in seconds, s(t) be the distance of the scooter from the wall at time t, and b(t) be the distance of the bug from the wall at time t.

We know:

s(t)=1+t because it starts at 1m from the wall and has a speed of 1 m/s going away

ds/dt=1, because that is the speed of the scooter

db/dt=.01+ds/dt*b(t)/s(t), because its speed will be .01 + the speed it gets from the band stretching. I think the speed it gets from the band stretching is the speed the band is stretchign (ds/dt) times the portion of the band the bug has crossed (b/s). In other words, when it is a quarter of the way down the band, its speed (relative to the wall) will be .01+ .25=.26

b(0)=0

Substituting in, we can eventually get to db/dt=.01+b(t)/(1+t), which we rewrite:

db/dt-1/(1+t)*b(t)=.01

Here's where my memory gets fuzzy. If I did this right, b(t) comes out to .005t. I'm pretty sure I didn't do that right, though, because it doesn't make any sense.

Anyways, if I did it right, you would then set b(t)=s(t) to find when they meet, and you get about -1 seconds, meaning that the bug would never catch the scooter.

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If, rather than being at x=0, the bug was in fact half a centimeter away from the wall, it makes things slightly easier to visualise. Break the rubber band up into 100 sections in your mind. At t=0, those sections are 1cm long, and the bug is standing at the center. After 1 second, that section is now 2cm long (1/100 of the band's length), but the bug's displacement from the wall has increased as well: even if stationary the bug will have advanced 1cm from the wall; adding on the 1cm speed of the bug, Florence is now 2cm away from the wall.

From that perspective, things are looking up. Unfortunately, When Florence began, he was 99.5cm from the scooter. Though he has advanced 2cm, the distance he needs to conquer has also increased: it is now 198cm.

In the next second, the stretching of the band increases the length of the sections to 3cm in length. Adding on his 1cm of movement, Florence is now 4cm from the wall. However, the band is now 300cm long; thus, he is 296cm from the scooter.

Since the band will never reach maximum tension, the distance between Florence and the scooter will continue to increase, and he'll never be able to catch up.

That's a simplified visualisation - the stretching is a curve, rather than a set of linear hops - but I'm pretty sure the curve would form an asymptote.

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This isn't going to be pretty, but I think I have a solution with one small modification to the problem:

The Prof needs to start a distance d from the wall initially, ie. the prof needs a head start. Otherwise, the bug has already caught up and my equations won't work as they are undefined for d=0 at t=0.

The speed of the bug is his starting speed, plus a proportion of the Prof's speed:

(I define s as the distance from the wall, and s' as the speed)

s'_bug = 0.01 + s_bug/(t+d)

so, the acceleration (s''_bug) is

s''_bug = d/dt [ s_bug/(t+d) ]

s''_bug = s'_bug/(t+d) - s_bug/(t+d)²

s''_bug = 0.01/(t+d), substituing the equation for s'_bug above

This is positive for all t>0, as long as d>0. Therefore, the bug is always accelerating (though at a decreasing rate!). As the Prof has a fixed speed, the bug must catch up the Prof at some t.

But the question says 'how long'....

For the general solution, we define the problem as a function of the distance from the wall, s, wrt time, t:

s_prof = At+d, where A is the speed of the scooter.

s_bug = ?

s'_prof = A (' denote derivates, ie speed)

s'_bug = B + s'_prof x (s_bug / s_prof), where B is the speed of the bug in his frame of reference.

s''_prof = 0, cleary!

s''_bug = B/t

the s'_bug is simply his personal speed, B, plus a proportion of the prof's speed as a faction of his distance along the band.

the s'_bug is our differential equation which needs solving, so I have dropped the 'bug' bit for clarity (ie s = s_bug)

s' = B + s / (t + d/A)

s' = B + s / u, where u = t +d/A

using u will simply make everything else easier and has no effect since u' = t'.

With a bit of Maths I have largely forgotton (which is why this took me so long!)

Spoiler for complicated maths:

s' = B + s / u

s' + (-1/u)s = B [rearrange]

(1/u)s' + (-1/u²) s = (1/u)B [multiply through by an integrating factor of 1/u]

[(1/u)s]' = (1/u)B

s/u = B . ∫ 1/u .dt [integrating both sides]

s/u = B . (ln u + C)

s = Bu (ln u + C)

phew

we get

s = Bu(ln u + C), with C as a constant of integration.

Knowing that at u=d/A, s=0

s = Bk(ln d/A + C) = 0

C = -ln d/A

and leaving us with:

s = Bu (ln u - ln d/A)

s = Bu ln (Au/d)

to find the point where they meet, we solve

s_bug = s_prof

Bu ln (Au/d) = Au

ln (Au/d) = A/B

Au/d = e^(A/B)

u = d/A . e^(A/B)

so,

t = u + d/A

t = d/A [1 + e^(A/B)]

In the problem we are given that A = 1m and B = 0.01m and I will make an assumption that the prof is given a 1m headstart:

t = 1 + e^100

t = 2.7 x 10^43 seconds!

the Prof would have covered an equal distance in metres, which considering the Milky Way is about 10^21 m across (I think), is a little further than he is likely to have got before running out of petrol!

For interest, another couple of scenarios where the bug is going a little bit faster:

for B = 0.1m, 22027s (about 6 hours)

for B = 1m, 3.72s.

There is every chance that my maths is wrong here somewhere so please feel freee to correct, but it feels as though its about right....

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Are we allowed the use of double integral and natural logarithm for solving this puzzle?

Do we allow puzzles that require the knowledge of advanced calculus on this forum?

ooops. But I didn't use any double integrals!

Either way I agree with your argument and my answer is based on the same reasoning.

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This isn't going to be pretty, but I think I have a solution with one small modification to the problem:

The Prof needs to start a distance d from the wall initially, ie. the prof needs a head start. Otherwise, the bug has already caught up and my equations won't work as they are undefined for d=0 at t=0.

The speed of the bug is his starting speed, plus a proportion of the Prof's speed:

(I define s as the distance from the wall, and s' as the speed)

s'_bug = 0.01 + s_bug/(t+d)

so, the acceleration (s''_bug) is

s''_bug = d/dt [ s_bug/(t+d) ]

s''_bug = s'_bug/(t+d) - s_bug/(t+d)²

s''_bug = 0.01/(t+d), substituing the equation for s'_bug above

This is positive for all t>0, as long as d>0. Therefore, the bug is always accelerating (though at a decreasing rate!). As the Prof has a fixed speed, the bug must catch up the Prof at some t.

But the question says 'how long'....

For the general solution, we define the problem as a function of the distance from the wall, s, wrt time, t:

s_prof = At+d, where A is the speed of the scooter.

s_bug = ?

s'_prof = A (' denote derivates, ie speed)

s'_bug = B + s'_prof x (s_bug / s_prof), where B is the speed of the bug in his frame of reference.

s''_prof = 0, cleary!

s''_bug = B/t

the s'_bug is simply his personal speed, B, plus a proportion of the prof's speed as a faction of his distance along the band.

the s'_bug is our differential equation which needs solving, so I have dropped the 'bug' bit for clarity (ie s = s_bug)

s' = B + s / (t + d/A)

s' = B + s / u, where u = t +d/A

using u will simply make everything else easier and has no effect since u' = t'.

With a bit of Maths I have largely forgotton (which is why this took me so long!)

Spoiler for complicated maths:

s' = B + s / u

s' + (-1/u)s = B [rearrange]

(1/u)s' + (-1/u²) s = (1/u)B [multiply through by an integrating factor of 1/u]

[(1/u)s]' = (1/u)B

s/u = B . ∫ 1/u .dt [integrating both sides]

s/u = B . (ln u + C)

s = Bu (ln u + C)

phew

we get

s = Bu(ln u + C), with C as a constant of integration.

Knowing that at u=d/A, s=0

s = Bk(ln d/A + C) = 0

C = -ln d/A

and leaving us with:

s = Bu (ln u - ln d/A)

s = Bu ln (Au/d)

to find the point where they meet, we solve

s_bug = s_prof

Bu ln (Au/d) = Au

ln (Au/d) = A/B

Au/d = e^(A/B)

u = d/A . e^(A/B)

so,

t = u + d/A

t = d/A [1 + e^(A/B)]

In the problem we are given that A = 1m and B = 0.01m and I will make an assumption that the prof is given a 1m headstart:

t = 1 + e^100

t = 2.7 x 10^43 seconds!

the Prof would have covered an equal distance in metres, which considering the Milky Way is about 10^21 m across (I think), is a little further than he is likely to have got before running out of petrol!

For interest, another couple of scenarios where the bug is going a little bit faster:

for B = 0.1m, 22027s (about 6 hours)

for B = 1m, 3.72s.

There is every chance that my maths is wrong here somewhere so please feel freee to correct, but it feels as though its about right....

Wow, diffeq was a long time ago for me. I'm glad you figured out how to rearrange that and do the integrating factor bit, because it was killing me that I couldn't remember this.

Cheers.

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I appreciate all of your guys' vast knowledge of calculus and differential equations, but it just seems like you are using these computer programs and advanced mathetmatical processes to prove solutions that can be explained using common sense and basic arithmetic...and until Prof. Templeton gives us an exact situation to work from, all of our proposed 'modifications' or 'adjustments' to make these things work are all for naught

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I appreciate all of your guys' vast knowledge of calculus and differential equations, but it just seems like you are using these computer programs and advanced mathetmatical processes to prove solutions that can be explained using common sense and basic arithmetic...and until Prof. Templeton gives us an exact situation to work from, all of our proposed 'modifications' or 'adjustments' to make these things work are all for naught

OK. My common sense answer:

As it is originally stated, the problem is trivial as they start on the wall together: When the Prof moves away from the wall, the bug goes with him since he is infinitely close to him already and so gets pulled along with the end of the band. The answer is 0 seconds. (In fact there is a singluarity here and you could argue that he never leaves the wall, since the band is of zero length)

The more interesting problem is where the Prof starts 'some' distance from the wall, because then the bug has some catching up to do.

In each second, the bug travels a fixed distance along the band. His difficulty is just that, every second, that distance becomes a smaller proportion of the distance he has to travel. Unfortunately, I can't prove without maths that the band doesn't stretch 'fast' enough. The simplest maths I can use is to imagine it like the series (1 + 1/2 + 1/3 + 1/4 .....). If you keep adding up the terms the number will get infiitely big (ie given any number, this series gets at least as big). However, the series (1 + 1/4 + 1/9 + 1/16 - the sum of reciprocal squares) is convergent and never gets larger than about 1.6.

This problem is like the first series in that the bug will always eventually catch up even though, at the end, he is only just overtaking the Prof.

Another way of thinking about it I have just thought of - he only has to get 99% of the way along the band to be going faster than the bike (because then his speed is 99% of 1 m/s + 0.01m/s). But to catch up with the '99%' bit of band, he only has to get 99% of the way there before he is also catching that up. We can repeat that all the way back to a point that is within the first 1cm that he reaches in the first second. I am not sure this quite works, but might make some sense to someone else....

Sorry Mumbles, I think this may still be a little Maths-heavy!

Edited by foolonthehill
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