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First, take any number (for this example it will be 492) and add together each digit in the number (4 + 9 + 2 = 15). Then take that sum (15) and determine if it is divisible by 3. The original number is divisible by 3 (or 9) if and only if the sum of its digits is divisible by 3 (or 9).
If the digit 9 is ignored when summing the digits, the effect is to "cast out" one more 9 to give the result 12. More generally, when casting out nines by summing digits, any set of digits which add up to 9, or a multiple of 9, can be ignored. In the number 3264, for example, the digits 3 and 6 sum to 9.
The number 18 is a harshad number in base 10, because the sum of the digits 1 and 8 is 9, and 18 is divisible by 9. The Hardy–Ramanujan number (1729) is a harshad number in base 10, since it is divisible by 19, the sum of its digits (1729 = 19 × 91). The number 19 is not a harshad number in base 10, because the sum of the digits 1 and 9 is ...
Digit sums and digital roots can be used for quick divisibility tests: a natural number is divisible by 3 or 9 if and only if its digit sum (or digital root) is divisible by 3 or 9, respectively. For divisibility by 9, this test is called the rule of nines and is the basis of the casting out nines technique for checking calculations.
In mathematics, a square-free integer (or squarefree integer) is an integer which is divisible by no square number other than 1. That is, its prime factorization has exactly one factor for each prime that appears in it. For example, 10 = 2 ⋅ 5 is square-free, but 18 = 2 ⋅ 3 ⋅ 3 is not, because 18 is divisible by 9 = 3 2. The smallest ...
The numbers 8 and 9 are coprime, despite the fact that neither—considered individually—is a prime number, since 1 is their only common divisor. On the other hand, 6 and 9 are not coprime, because they are both divisible by 3. The numerator and denominator of a reduced fraction are coprime, by definition.
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d() is the number of positive divisors of n, including 1 and n itself; σ() is the sum of the positive divisors of n, including 1 and n itselfs() is the sum of the proper divisors of n, including 1 but not n itself; that is, s(n) = σ(n) − n
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