There are only ten decimal digits (a digit is a single number, not to be confused with an integer of which there are an infinite number). So it is only possible to have 21 consecutive digits if they are arranged in ascending or descending order: 123456789876543210123. e is a transcendental number which has an infinite number of decimal places so every possible combination of digits will occur somewhere. Therefore if 123456789876543210123 is prime, then it may be the smallest 21-digit prime number with consecutive decimal digits. But this number is divisible by 3, so is not prime.
Note that if the number starts with an odd digit it also ends with one. So we can move 2 digits at a time. Proceeding like this we discover continued divisibility by 3 or 5, until we get to 101234567898765432101, which is not divisible by 3, and may be a prime number (assume it is, because it’s the only possible candidate). (Test for divisibility by 3 by summing the digits. If the sum is divisible by 3, then so is the number. Note that the sum from 1 to 8 is divisible by 3, therefore so is the sum from 1 to 9. This means we only have to sum some of the digits to check for divisibility by 3. And we can eliminate all numbers ending in 5 or an even digit. Note also that groups of 3 digits (012, 123, 234, 345, 456, 567, 678, 789) are all divisible by 3; but a change of direction is not (101, 898). A 21-digit number has 7 groups of 3 digits.)
[Sum of 3 consecutive digits can be represented by x-1+x+x+1=3x, that is, 3 times the central digit.]
