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disco wrote: ↑Thu Jan 13, 2022 12:18 am I found something more boring than actual math problems. Men arguing over how to do math problems.

What are the odds?

stonedmegman wrote: ↑Thu Jan 13, 2022 12:37 am

disco wrote: ↑Thu Jan 13, 2022 12:18 am I found something more boring than actual math problems. Men arguing over how to do math problems.

What are the odds?

37.46/1

Hyperv12 wrote: ↑Thu Jan 13, 2022 12:30 pm

stonedmegman wrote: ↑Thu Jan 13, 2022 12:37 am

disco wrote: ↑Thu Jan 13, 2022 12:18 am I found something more boring than actual math problems. Men arguing over how to do math problems.

What are the odds?

37.46/1

That's probably right.

Said before I get shafted

Hyperv12 wrote: ↑Thu Jan 13, 2022 12:30 pm

stonedmegman wrote: ↑Thu Jan 13, 2022 12:37 am

disco wrote: ↑Thu Jan 13, 2022 12:18 am I found something more boring than actual math problems. Men arguing over how to do math problems.

What are the odds?

37.46/1

I'll take the over.

necronomous wrote: ↑Thu Jan 13, 2022 2:03 pm

Hyperv12 wrote: ↑Thu Jan 13, 2022 12:30 pm

stonedmegman wrote: ↑Thu Jan 13, 2022 12:37 am

disco wrote: ↑Thu Jan 13, 2022 12:18 am I found something more boring than actual math problems. Men arguing over how to do math problems.

What are the odds?

37.46/1

That's probably right.

Said before I get shafted

Animal wrote: ↑Thu Jan 13, 2022 3:00 pm

necronomous wrote: ↑Thu Jan 13, 2022 2:03 pm

Hyperv12 wrote: ↑Thu Jan 13, 2022 12:30 pm

stonedmegman wrote: ↑Thu Jan 13, 2022 12:37 am

disco wrote: ↑Thu Jan 13, 2022 12:18 am I found something more boring than actual math problems. Men arguing over how to do math problems.

What are the odds?

37.46/1

That's probably right.

Said before I get shafted

Animal wrote: ↑Thu Jan 13, 2022 3:00 pm

necronomous wrote: ↑Thu Jan 13, 2022 2:03 pm

Hyperv12 wrote: ↑Thu Jan 13, 2022 12:30 pm

stonedmegman wrote: ↑Thu Jan 13, 2022 12:37 am

disco wrote: ↑Thu Jan 13, 2022 12:18 am I found something more boring than actual math problems. Men arguing over how to do math problems.

What are the odds?

37.46/1

That's probably right.

Said before I get shafted

necronomous wrote: ↑Thu Jan 13, 2022 2:03 pm

Hyperv12 wrote: ↑Thu Jan 13, 2022 12:30 pm

stonedmegman wrote: ↑Thu Jan 13, 2022 12:37 am

disco wrote: ↑Thu Jan 13, 2022 12:18 am I found something more boring than actual math problems. Men arguing over how to do math problems.

What are the odds?

37.46/1

That's probably right.

Said before I get shafted

stonedmegman wrote: ↑Thu Jan 13, 2022 2:08 pm

Hyperv12 wrote: ↑Thu Jan 13, 2022 12:30 pm

stonedmegman wrote: ↑Thu Jan 13, 2022 12:37 am

disco wrote: ↑Thu Jan 13, 2022 12:18 am I found something more boring than actual math problems. Men arguing over how to do math problems.

What are the odds?

37.46/1

I'll take the over.

With those two I’d take the spread.

A string is wrapped around a rod exactly four times end to end. The rod is 12 units long and has a circumference of 4 units. What is the length of the string?

Edit: the string wraps around the rod exactly like the red strip around the barber's pole, screw thread like. Just so we're clear, since I haven't included a picture.

By the way, wrong answers so far. Hint: split the rod and thread length-wise and unfold.

5.09

QillerDaemon wrote: ↑Fri Mar 11, 2022 5:04 am A string is wrapped around a rod exactly four times. The rod is 12 units long and has a circumference of 4 units. What is the length of the string?

48 units if wrapped end to end 16 if wrapped around it. Since string diameter not given I did not allow to offset of the turns

16 if wrapped around the diameter
(12+12+4+4)*4 =32*4 = 128 units if wrapped end to end

The length of the string is 20.

Animal wrote: ↑Fri Mar 11, 2022 3:06 pm The length of the string is 20.

And how'd you figure that?

QillerDaemon wrote: ↑Fri Mar 11, 2022 3:25 pm

Animal wrote: ↑Fri Mar 11, 2022 3:06 pm The length of the string is 20.

And how'd you figure that?

we use spiral bars in bridge columns. L = n * SQRT( C^2 + P^2). In this problem, n = number of turns (4). C = circumference (4). P = pitch or the distance between each turn (12/4 = 3).

So L = 4 * Sqrt (4^2 + 3^2) = 4 * 5 = 20

53.09

the solution looks much more complicated than it is.

Basically, for each wrap the string has to make a hypotenuse of a right triangle formed by the Circumference and the Pitch (Length/# Turns).

In this case that is the hypotenuse of a right triangle with leg 4 and 3. That hypotenuse is 5. Now how many times does it do that? In this case it makes 4 turns. so that is 4 x 5 = 20.

Animal wrote: ↑Fri Mar 11, 2022 3:28 pm

QillerDaemon wrote: ↑Fri Mar 11, 2022 3:25 pm

Animal wrote: ↑Fri Mar 11, 2022 3:06 pm The length of the string is 20.

And how'd you figure that?

we use spiral bars in bridge columns. L = n * SQRT( C^2 + P^2). In this problem, n = number of turns (4). C = circumference (4). P = pitch or the distance between each turn (12/4 = 3).

So L = 4 * Sqrt (4^2 + 3^2) = 4 * 5 = 20

Ah, the "professional's" solution! I used calculus to solve it, and no I'm not about to try to post that solution. I don't think BBcode can handle it.

Turns out, if you pretend the rod is a tube (like for toilet paper), and cut it lengthwise including the string and lay it out flat, you come up with four triangles in a row, with the cut string segments representing the hypotenuses. Find the segment length, and since there's four of them, they add up to 20 units. That's likely where your professional's equation came from, even. That is the correct answer. (edit: leeroyed! )

There is a combination lock with a three number combination. You have to solve the combination with one try. 5 people have tried ahead of you and you got to watch their attempts and results. Now its your turn.

Here are what the first 5 people tried:

2 9 1 - 1 number is correct and it is in the right place.
2 4 5 - 1 number is correct, but it is in the wrong place.
4 6 3 - 2 numbers are correct, but both are in the wrong place.
5 7 8 - all numbers are wrong.
5 6 9 - one number is correct, but in the wrong place.

What is your guess.

Animal wrote: ↑Fri Mar 11, 2022 4:01 pm There is a combination lock with a three number combination. You have to solve the combination with one try. 5 people have tried ahead of you and you got to watch their attempts and results. Now its your turn.

Here are what the first 5 people tried:

2 9 1 - 1 number is correct and it is in the right place.
2 4 5 - 1 number is correct, but it is in the wrong place.
4 6 3 - 2 numbers are correct, but both are in the wrong place.
5 7 8 - all numbers are wrong.
5 6 9 - one number is correct, but in the wrong place.

What is your guess.

Can you have a double number?

Reservoir Dog wrote: ↑Fri Mar 11, 2022 4:13 pm

Animal wrote: ↑Fri Mar 11, 2022 4:01 pm There is a combination lock with a three number combination. You have to solve the combination with one try. 5 people have tried ahead of you and you got to watch their attempts and results. Now its your turn.

Here are what the first 5 people tried:

2 9 1 - 1 number is correct and it is in the right place.
2 4 5 - 1 number is correct, but it is in the wrong place.
4 6 3 - 2 numbers are correct, but both are in the wrong place.
5 7 8 - all numbers are wrong.
5 6 9 - one number is correct, but in the wrong place.

What is your guess.

Can you have a double number?

No.

this is like wordle but with numbers.

642

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