So where is the mistake?

As usual, watch the video for a solution.

Proof that 2 does not exist. Where is the mistake?

Or keep reading.

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Answer To Proof that 2 does not exist. Where is the mistake?

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

The mistake in reasoning happens when saying the probability of a given prime is even is zero.

It is not possible to assign a uniform distribution to a countably infinite set. The reason is if you set the probability to be 0 for each event, then the total probability is 0 (less than 1). If you set the probability to be greater than 0, then summing over the infinite set will become unbounded (greater than 1). So it’s impossible to set a uniform distribution to a countably infinite set.

So yes, the number 2 does exist after all.

Reference

X
https://x.com/Math_files/status/2039328376247484763

What is the value of x?

As usual, watch the video for a solution.

What Is The Width? Difficulty 97.5 Percent

Or keep reading.

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Answer To What Is The Width? Difficulty 97.5 Percent

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

I first want to go over a common mistake. It is correct to connect the centers of the circles for a length of 8 + 5 = 13. Then consider the vertical and horizontal distances between the centers. It appears the center of the 8m circle is aligned with the bottom of the 5m circle, so many people incorrectly give a vertical distance of 5m. This will make for a 5-12-13 right triangle. Thus the rectangle base is equal to 8 + 12 + 5 = 25m.

The problem is the vertical length is not 5, but rather equal to 20 – 5 – 8 = 7. Using the same construction we get a right triangle with a hypotenuse of 13 and one leg of 7, so the last leg is √(13² – 7²) = 2√30.

Thus the rectangle base is 8 + 2√30 + 5 = 13 + 2√30.

Reference

Original puzzle
https://x.com/codek_tv/status/2036304500135252072
My modified puzzle
https://x.com/preshtalwalkar/status/2036555257069896073

x⁵ + x⁴ + x³ + x² + x + 1 = 0

I share 2 different methods to solve this problem in a new video.

As usual, watch the video for a solution.

The Math Problem Young Sherlock Holmes Couldn’t Solve, But You Can

Or keep reading.

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Answer To The Math Problem Sherlock Holmes Couldn’t Solve, But You Can

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Thanks Don for alerting me of typos!

In the episode Sherlock is confused by the problem. However, the same blackboard contains a derivation of the solution.

Method 1: polynomial division

Let f(x) = x⁵ + x⁴ + x³ + x² + x + 1

By the Rational Root Theorem, the only candidates for rational roots are ±1/1 = ±1. Testing values we find f(1) = 6 and f(-1) = 0. Thus x = -1 is a root, and x + 1 is a factor of f.

Let’s divide f by x + 1.

Thus we have:

f(x) = (x + 1)(x⁴ + x² + 1)

We can further factor the quartic by adding and subtracting x².

x⁴ + x² + 1
= x⁴ + x² + x² – x² + 1
= x⁴ + 2x² + 1 – x²
= (x² + 1)² – x²
= (x² + 1 + x²)(x² + 1 – x²)

Thus we have:

f(x) = (x + 1)(x² + 1 + x)(x² + 1 – x)

The roots of this equation correspond to the zeroes of the factored equations. The linear factor gives the root of x = -1, and the quadratic equations are easily solvable by the quadratic equation. Thus we get the 5 roots:

x = 1
x = -1/2 ± i√3/2
x = 1/2 ± i√3/2

Method 2: roots of unity

There is a well-known formula for n an integer greater than 2:

xⁿ – 1 = (x – 1)(x^n-1 + x^n-2 + … + 1)

This is easily verifiable by expanding and comparing coefficients. In our problem we have:

x⁶ – 1 = (x – 1)(x⁵ + x⁴ + … + 1)

The roots of the equation correspond to the roots of the factors. The linear factor has a root of x = 1, which we already verified is not a factor of the 5th degree equation. So the 5 roots of this equation are the 6 roots of the equation x⁶ – 1 excluding x = 1.

x⁶ – 1 = 0
x⁶ = 1

We want to find the 6 roots of 1 (also called unity). This is easily done by putting 1 in the complex plane in polar form as e^2πk for integers k.

1 = e^2πk
1^1/6 = [e^2πk]^1/6
1^1/6 = e^2πk/6

We will get 6 distinct values from k = 0, 1, 2, …, 5, giving:

e⁰ = 1
e^2π/6 = 1/2 + i√3/2
e^4π/6 = -1/2 + i√3/2
e^6π/6 = -1
e^8π/6 = -1/2 – i√3/2
e^10π/6 = 1/2 – i√3/2

We know x = 1 is not a root of x⁵ + x⁴ + x³ + x² + x + 1, so we have its 5 roots are:

e^2π/6 = 1/2 + i√3/2
e^4π/6 = -1/2 + i√3/2
e^6π/6 = -1
e^8π/6 = -1/2 – i√3/2
e^10π/6 = 1/2 – i√3/2

What a fun problem! Credit to the math consultants on the show Young Sherlock.

References

The Mathematical Crimes of Young Sherlock
https://theconversation.com/the-mathematical-crimes-of-the-young-sherlock-holmes-series-278812

Christopher Scott Vaughen
https://www.youtube.com/watch?v=tRCkPqQTg8k

IGN clip
https://www.youtube.com/watch?v=WPC8mAcl0iw

Prime Video clip
https://www.youtube.com/watch?v=7DOsYvn3S7c

Young Sherlock Prime Video
https://www.amazon.com/gp/video/detail/B0G2BTDG5G/

WolframAlpha
https://www.wolframalpha.com/input?i=1%5E%281%2F6%29