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Furthermore, it is clear that even-digits with greater than or equal to 8, [10] and with 9 digits, [11] or odd-digits with greater than or equal to 15 digits [12] have multiple solutions. Although 11-digit and 13-digit numbers have only one solution, it forms a loop of five numbers and a loop of two numbers, respectively. [13]
A number that has the same number of digits as the number of digits in its prime factorization, including exponents but excluding exponents equal to 1. A046758: Extravagant numbers: 4, 6, 8, 9, 12, 18, 20, 22, 24, 26, 28, 30, 33, 34, 36, 38, ... A number that has fewer digits than the number of digits in its prime factorization (including ...
6174 is a 7-smooth number, i.e. none of its prime factors are greater than 7. 6174 can be written as the sum of the first three powers of 18: 18 3 + 18 2 + 18 1 = 5832 + 324 + 18 = 6174, and coincidentally, 6 + 1 + 7 + 4 = 18. The sum of squares of the prime factors of 6174 is a square: 2 2 + 3 2 + 3 2 + 7 2 + 7 2 + 7 2 = 4 + 9 + 9 + 49 + 49 ...
There are no self-descriptive numbers in bases 2, 3 or 6. In bases 7 and greater, there is exactly one self-descriptive number: () + + +, which has b−4 instances of the digit 0, two instances of the digit 1, one instance of the digit 2, one instance of digit b – 4, and no instances of any other digits.
Since 7 October 2024, Python 3.13 is the latest stable release, and it and, for few more months, 3.12 are the only releases with active support including for bug fixes (as opposed to just for security) and Python 3.9, [55] is the oldest supported version of Python (albeit in the 'security support' phase), due to Python 3.8 reaching end-of-life.
But if exact values for large factorials are desired, then special software is required, as in the pseudocode that follows, which implements the classic algorithm to calculate 1, 1×2, 1×2×3, 1×2×3×4, etc. the successive factorial numbers. constants: Limit = 1000 % Sufficient digits.
If doubling a digit results in a value > 9, subtract 9 from it (or sum its digits). Sum all the resulting digits (including the ones that were not doubled). The check digit is calculated by (()), where s is the sum from step 3. This is the smallest number (possibly zero) that must be added to to make a multiple of 10.
Proof by exhaustion can be used to prove that if an integer is a perfect cube, then it must be either a multiple of 9, 1 more than a multiple of 9, or 1 less than a multiple of 9. [3] Proof: Each perfect cube is the cube of some integer n, where n is either a multiple of 3, 1 more than a multiple of 3, or 1 less than a multiple of 3. So these ...