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0 3 = 0; 1 3 = 1 up 1; 2 3 = 8 down 3; 3 3 = 27 down 1; 4 3 = 64 down 3; 5 3 = 125 up 1; 6 3 = 216 up 1; 7 3 = 343 down 3; 8 3 = 512 down 1; 9 3 = 729 down 3; 10 3 = 1000 up 1; There are two steps to extracting the cube root from the cube of a two-digit number. For example, extracting the cube root of 29791. Determine the one's place (units) of ...
Figure 2 is used for the multiples of 2, 4, 6, and 8. These patterns can be used to memorize the multiples of any number from 0 to 10, except 5. As you would start on the number you are multiplying, when you multiply by 0, you stay on 0 (0 is external and so the arrows have no effect on 0, otherwise 0 is used as a link to create a perpetual cycle).
One half is the rational number that lies midway between 0 and 1 on the number line. Multiplication by one half is equivalent to division by two, or "halving"; conversely, division by one half is equivalent to multiplication by two, or "doubling".
Squaring is the same as raising to the power 2, and is denoted by a superscript 2; for instance, the square of 3 may be written as 3 2, which is the number 9. In some cases when superscripts are not available, as for instance in programming languages or plain text files, the notations x ^2 ( caret ) or x **2 may be used in place of x 2 .
The digital root pattern for triangular numbers, repeating every nine terms, as shown above, is "1, 3, 6, 1, 6, 3, 1, 9, 9". The converse of the statement above is, however, not always true. For example, the digital root of 12, which is not a triangular number, is 3 and divisible by three.
where f (2k−1) is the (2k − 1)th derivative of f and B 2k is the (2k)th Bernoulli number: B 2 = 1 / 6 , B 4 = − + 1 / 30 , and so on. Setting f ( x ) = x , the first derivative of f is 1, and every other term vanishes, so [ 15 ]
Take each digit of the number (371) in reverse order (173), multiplying them successively by the digits 1, 3, 2, 6, 4, 5, repeating with this sequence of multipliers as long as necessary (1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5, ...), and adding the products (1×1 + 7×3 + 3×2 = 1 + 21 + 6 = 28). The original number is divisible by 7 if and only if ...
if a number is divisible neither by 2 nor by 3, its square ends in 1, and its preceding digit must be even; if a number is divisible by 2, but not by 3, its square ends in 4, and its preceding digit must be 0, 1, 4, 5, 8, or 9; and; if a number is not divisible by 2, but by 3, its square ends in 9, and its preceding digit must be 0 or 6.