Search results
Results From The WOW.Com Content Network
For six-digit numbers, there are two solutions that satisfy equations (1) and (2). [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.
The computer may also offer facilities for splitting a product into a digit and carry without requiring the two operations of mod and div as in the example, and nearly all arithmetic units provide a carry flag which can be exploited in multiple-precision addition and subtraction. This sort of detail is the grist of machine-code programmers, and ...
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.
5 is halved (2.5) and 6 is doubled (12). The fractional portion is discarded (2.5 becomes 2). The figure in the left column (2) is even, so the figure in the right column (12) is discarded. 2 is halved (1) and 12 is doubled (24). All not-scratched-out values are summed: 3 + 6 + 24 = 33. The method works because multiplication is distributive, so:
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.
If it is divisible by 2 continue by adding the digits of the original number and checking if that sum is a multiple of 3. Any number which is both a multiple of 2 and of 3 is a multiple of 6. Example. 324 (The original number) Final digit 4 is even, so 324 is divisible by 2, and may be divisible by 6. 3 + 2 + 4 = 9 which is a multiple of 3.
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 ...
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.