999 and 326998 and 327
997 and 328
and so on to
663 and 662.
11 and 18
21
89 x 90 = 8010 which are two consecutive two digit numbers that equal 8010.
One of them is twelve and the other is a dozen. They both equal 12.
A 3 or 4 digit number.
There are 28706 such combinations. 5456 of these comprise three 2-digit numbers, 19008 comprise two 2-digit numbers and two 1-digit numbers, 4158 comprise one 2-digit number and four 1-digit numbers and 84 comprise six 1-digit numbers.
I'm sure there are more than 2 prime numbers that are 400 digits long.
There are 9 1-digit numbers and 16-2 digit numbers. So a 5 digit combination is obtained as:Five 1-digit numbers and no 2-digit numbers: 126 combinationsThree 1-digit numbers and one 2-digit number: 1344 combinationsOne 1-digit numbers and two 2-digit numbers: 1080 combinationsThat makes a total of 2550 combinations. This scheme does not differentiate between {13, 24, 5} and {1, 2, 3, 4, 5}. Adjusting for that would complicate the calculation considerably and reduce the number of combinations.
There is no such ratio that applies for all single-digit and double-digit integers.
No digit that is 5 places long is equal to 6 times 2 other than 12.000
-2
72. (with the range of two digit numbers being from 10 to 99).