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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.
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]
0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24, 8, 25, 43, 62, ... "subtract if possible, otherwise add" : a (0) = 0; for n > 0, a ( n ) = a ( n − 1) − n if that number is positive and not already in the sequence, otherwise a ( n ) = a ( n − 1) + n , whether or not that number is already in the sequence.
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.
The digital root (also repeated digital sum) of a natural number in a given radix is the (single digit) value obtained by an iterative process of summing digits, on each iteration using the result from the previous iteration to compute a digit sum. The process continues until a single-digit number is reached.
A number which is a harshad number in every number base is called an all-harshad number, or an all-Niven number. There are only four all-harshad numbers: 1 , 2 , 4 , and 6 . The number 12 is a harshad number in all bases except octal .
I felt like I was 5 and learning how to subtract triple digit numbers all over again. #98 "Fifthteen" ended up on a worksheet I made for Korean school children while I was teaching public school ...
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 ...