I've discussed this with my mathematics teacher, and she's told me that this is completely legitimate.
3 is an easy number to manipulate, so that is why I've chosen it.
Take all the numbers 1-10. There is one 3 between that limit. The proportion is 1/10.
Take all the numbers 1-100. There is now 19 between that limit. The proportion is 19/100. The percentage has increased.
Take all the numbers 1-1000. There is now 271 between that limit. The proportion is 271/1000. The percentage is still increasing.
If we were to continue this hundreds of thousands of times (millions), we could write a proof that said almost all numbers contain the digit 3.
The equation for this is... 1-(9/10)^n. n is whatever number you chose to find how many of whatever number is in the number you chose (so, if you use 10,000,000, you would place it in the n place).
The 9/10 part is because if you see the digit 0, you can see that the 0 can be any digit from 0,1,2,3,4,5,6,7,8,9. Or ten numbers. So if you take a 3 out of that list, there is 9 numbers remaining. So you find whatever is left.
1-(9/10)^10,000,000 = 1. 1=100%, 100% of all numbers contain a digit 3. Of course we know this is "nonsense" because I can just as easily say that the digit 2 is in every number. That is why it is said "almost all" digits contain a number 3.
I owe credit to Brady Haran and his YouTube channel Numberphile http://www.youtub...umberphile for helping with the equation bits.
Can you explain a bit more carefully, please? Sentences like "n is whatever number you chose to find how many of whatever number is in the number you chose" are really hard to follow.
Revenge wrote:
Take all the numbers 1-10. There is one 3 between that limit. The proportion is 1/10.
Take all the numbers 1-100. There is now 19 between that limit. The proportion is 19/100. The percentage has increased.
Take all the numbers 1-1000. There is now 271 between that limit. The proportion is 271/1000. The percentage is still increasing.
I agree with these figures. To be clear, you're finding how many of the numbers from 1 to n contain at least one 3.
The equation for this is... 1-(9/10)^n. n is whatever number you chose to find how many of whatever number is in the number you chose (so, if you use 10,000,000, you would place it in the n place).
The 9/10 part is because if you see the digit 0, you can see that the 0 can be any digit from 0,1,2,3,4,5,6,7,8,9. Or ten numbers. So if you take a 3 out of that list, there is 9 numbers remaining. So you find whatever is left.
The equation for what, the proportion? Can you explain the derivation?
1-(9/10)^10,000,000 = 1. 1=100%, 100% of all numbers contain a digit 3. Of course we know this is "nonsense" because I can just as easily say that the digit 2 is in every number. That is why it is said "almost all" digits contain a number 3.
1-(9/10)^10,000,000 is close to 1, but not exactly 1.
With n = 10,000,000 I calculate there to be 5,217,031 numbers containing at least one 3. This is not much more than half.
I am in a bit on a rush this morning, you should watch the video! You can look up "3 is everywhere" and should find it. A bit of the calculating is lost to me, so they'll explain it better.
Revenge wrote:
I am in a bit on a rush this morning, you should watch the video! You can look up "3 is everywhere" and should find it. A bit of the calculating is lost to me, so they'll explain it better.
It sounds really weird, but I guess it sort of makes sense. When you have very large numbers (with lots of digits), at least one of those digits is likely to be a 3.
