In prime numbers, what is the most common individual digit?

Answer 1

Shame on everyone who says you can't say because it is infinite. Don't you know about limits?

I will make some guesses because this should be tough to prove, but they seem reasonable.

First, since all primes (except 2) end in an odd number, I would say the odd digits will tend to outweight the even digits., so the answer should be an odd digit.

Second, look at Benford's Law. It states that the initial digit of numbers are distributed logarithmically: as a consequence, 30% of all numbers begin with 1. (if you look at lists like height of mountains in inches, population statistics, or any other list of numbers you can see this.) See reference at http://www.rexswain.com/benford.html)

While this is most apparent with initial digits, it also holds for first 2 digits: 10 will be very slightly more common than 11, and so on.

As a result, I would claim that the most common digit in prime numbers is 1.

In fact, any random sequence of numbers should contain more 1s than any other digit (except possibly 0?)

Answer 2

It is true that just because there are an infinite number of numbers does not mean you can't calculate odds. For example, 10% of all numbers end in 7 but there still are an infinite numbers that end in 7. The comment that the first digit of a set of number follows a logrithmic distribution depends on how the numbers are distributed. In the case of the sequence of primes, they are equally distributed in a linear sense so all initial digits are equally probable. The only real skew in the distribution is that an even digit cannot end a number. So it would seem that the odd digits outnumber the even ones. However, the fraction by which the odd ones lead must be infintesimally small since prime numbers with an infinite number of digits do exist. In all, it seems that all digits are equally probable, but for any lower set of primes, the odds should outweigh the evens by a factor of 1/(2*n) where n is the limit on the number of digits. Its a pretty simple matter to write a program to see if this seems to be correct.

Answer 3

To answer your question, I will state a few facts about numbers.
Given: There are an infinite number of numbers on either side of the number line.
Prove (Find (w/e)): The most common individual digit in prime numbers.

1. There are an infinite number of numbers on either side of the number line. (Given)
2. Since there are an infinite number of numbers on either side of the number line, there are also an infinite number of prime numbers (and composite numbers. Prime and composite numbers make up the numbers on the number line).
3. Prime numbers will always end in either a 1, 3, 7, or 9.

Since there are an infinite number of prime numbers, there must be an infinite number of prime numbers that end in either 1, 3, 7, or 9.
Therefore, you cannot calculate the most common individual digit in prime numbers because there is an infinite number of numbers that end in 1, 3, 7, 9, because it's infinite.

If you mean the most common individual digit in each prime number, then you still cannot calculate it, because any some digit can be put into a number and make it prime. For example, you have some number that is composite, like 353 (it's not composite, but for an example), and put in a 5, to make it 3553 (I don't know if 3553 is prime or not, but again it's just an example) and all of a sudden your number is prime, and has no most common individual digit. Therefore, it is impossible to calculate this with sheer brain or calculator power known to man.

Answer 4

there is no "most common individual digit" for the sequence of prime numbers. the sequence of primes is transcendental (no pattern). the distribution of digits is flat (all occur with the same frequency).

Answer 5

There are infinite numbers of prime numbers.

Answer 6

I'm not sure that question is meaningful since theoretically the number of prime numbers is infinite. You can't calculate a percentage based on a set with an infinite number of elements.