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Math game...

We've done logic puzzles and a mystery, but the title of the thread is "Math game".

:topic

There are two train stations, which we shall call "East Station" and "West Station", with a perfectly straight track running between them. The distance between the stations in exactly 78 miles. A train leaves East Station going at 60 mph. At the exact same moment, a bird leaves West Station going at 70 mph. When it meets the train, it instantly turns around and goes back to the station, and then turns around and goes back to the train. It keeps going back and forth until both bird and train are at West Station. How many miles does the bird travel?

For extra bonus points, how many miles does the bird travel if it is going at 60 mph and the train is going at 70 mph?

The TOG​
 
Because trains can not instantly achieve speeds of 60 miles per hour, and instead they gradually build up to that speed from zero miles per hour, we simply can not resolve the problem. Even if we were to assume that the train traveled through the East Station while going exactly 60 miles per hour and maintained constant speed thereafter (so that the problem didn't involve acceleration rates) we'd still have the problem of the bird. For the bird, it's speed is stated as 70 mph but there is no information given about the head or tail winds. Air speed is not the same as land speed.

Please don't assume that just because I'm trying to point out flaws (tongue in cheek) that it means that I will try to solve the math problem. For me? The train stops here.
 
Okay... To make it simpler, we'll assume that the train goes through the starting station at 60 mph and continues at a constant speed, and stops at it's destination when it runs into a brick wall and stops instantly, instead of decelerating. We will also assume that there is no wind along the entire track, and that turbulance from the train doesn't affect the flight of the bird. The bird flies in a perfectly straight line, without any fluctuations virtically or horizontally.

The TOG​
 
TOG said:
Okay... To make it simpler, we'll assume that the train goes through the starting station at 60 mph and continues at a constant speed, and stops at it's destination when it runs into a brick wall and stops instantly, instead of decelerating. We will also assume that there is no wind along the entire track, and that turbulance from the train doesn't affect the flight of the bird. The bird flies in a perfectly straight line, without any fluctuations virtically or horizontally.

The TOG​
Click to expand...

Does the bird also have instantaneous max speed :biggrin

I'll show you my working until I realized the easy way ( You must be laughing )

East - West = 78m , Let Train = t , Bird = b, Time = T, distance traveled = D

T ( 60 +70 ) = 78
T=0.6

Dt = 60*0.6 = 36m Db = 70*0.6 = 42m

Bird returns to W in 0.6hr @ 42 m while train travels 36m leaving 8 m to go. Db = 84m

T(130) = 8
T=0.06154 ( now starting to think this is getting messy )

Dt = 60*0.06154 = 3.7m Db= 70*0.06154 = 4.3m ( now starting to think angon how do I know when to stop the train can only travel 78 m which will take 1.3 hrs. Scratch head there must be a trick . Light bulb comes on !)

If the train travels for a total time of 1.3 hrs to get to West the Bird can only travel for 1.3 hrs @ 70mph = 91 miles. :biggrin

If the Bird is traveling at 60 and the train at 70 the Bird will travel 36 m * 2 ( he's now sitting on the train ) = 72 m

Ops i just looked up and saw a silly math mistake it should have bee 6 mile left after the Bird returned first time. Glad I didn't keep going with that :D
 
Correct! :clap

I checked the site where I found this, to confirm your answer (mine was a little different, but not much), and I realized that I had nearly every number wrong. I have no idea how I did that. I could swear I had the right numbers earlier. They must change them randomly each time a person tries the puzzle. That has to be it... Let me check...

Yep. It was different the next time I did it. That's not fair. You can't learn the answers when they do that. :nonono

Here's another one...

You have $84 in one dollar bills. Place the bills in a number of bags in such a way that I can ask you for any amount from $1 to $84 and you will be able to give it to me by handing me the appropriate bags and without having to modify the amount in any bag. You may put the same amount in more than one bag. What is the minimum number of bags you would have to use, and what amounts do they contain.

The TOG​
 
TOG said:
Correct! :clap

I checked the site where I found this, to confirm your answer (mine was a little different, but not much), and I realized that I had nearly every number wrong. I have no idea how I did that. I could swear I had the right numbers earlier. They must change them randomly each time a person tries the puzzle. That has to be it... Let me check...

Yep. It was different the next time I did it. That's not fair. You can't learn the answers when they do that. :nonono

Here's another one...

You have $84 in one dollar bills. Place the bills in a number of bags in such a way that I can ask you for any amount from $1 to $84 and you will be able to give it to me by handing me the appropriate bags and without having to modify the amount in any bag. You may put the same amount in more than one bag. What is the minimum number of bags you would have to use, and what amounts do they contain.

The TOG​
Click to expand...
7 bags - 1, 2, 4, 8, 16, 32, 64?
 
Barbarian said:
So, 13 is next?
Click to expand...
$13? As in:

You have $13 in one dollar bills. Place the bills in a number of bags in such a way that I can ask you for any amount from $1 to $13 and you will be able to give it to me by handing me the appropriate bags and without having to modify the amount in any bag. You may put the same amount in more than one bag. What is the minimum number of bags you would have to use, and what amounts do they contain.
 
It would take 4 bags, containing 1, 2, 4 and 6 dollars. The amounts would be as follows...

Dollars - Bags
1 - 1
2 - 2
3 - 2+1
4 - 4
5 - 4+1
6 - 6 or 4+2
7 - 6+1 or 4+2+1
8 - 6+2
9 - 6+2+1
10 - 6+4
11 - 6+4+1
12 - 6+4+2
13 - 6+4+2+1

The TOG​
 
I think using the same sequence eng35 used in the last one will work too.

1,2,4,8
 
I have one that isn't about the answer, but how you get it. (Remember "new math"?)

Anyone can add a few numbers together. For example...

1+2=3
1+2+3=6
1+2+3+4=10

... and so on. But when you get up to larger numbers, adding them all can get to be tiresome. Is there a simpler way to find the sum of all the numbers between 1 and any given positive integer, without having to add them all together? (Note: There may be different answers depending on whether the highest number in the series is odd or even.)

The TOG​
 
Oh man now this is gonna get messy. I remember doing this before but I don't remember the formula or the proof.

sum 1 to n

1 + ( 1+ 1 ) + ( 1+1+1) + (1+1+1+1) + ( 1+1+1+1+ ... n-1 ) etc.
 
Ok using logic and a little brute force I can see that if we take the mean of the numbers. ( 1 + n) / 2 and multiply that by the number of numbers n we get.

(1+n) / 2 * n

This seems to work. Why does it matter if the largest number is even or odd ?
 
agua. said:
Ok using logic and a little brute force I can see that if we take the mean of the numbers. ( 1 + n) / 2 and multiply that by the number of numbers n we get.

(1+n) / 2 * n

This seems to work.
Click to expand...

Let's test your theory, first for the numbers 1-4.

1+2+3+4=10

(1+4)/2*4 = 5/2*4 = 2.5*4 = 10

It works :)

Now for the numbers 1-5.

1+2+3+4+5=15

(1+5)/2*5 = 6/2*5 = 3*5 =15

That also works :)

Why does it matter if the largest number is even or odd ?
Click to expand...

I didn't say you did need a different function for odd numbers, I said it might be different. That was just to get you thinking.

Next puzzle... Can you modify the formula so that it applies to series that start with something other than 1, for example the sum of all numbers from 3 to 24 or from 16 to 87? I didn't see this one on a website (thought of it just now) and I haven't figured out the answer myself.

The TOG​
 
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