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The sum of the ones digit, double the tens digit, and four times the hundreds digit is divisible by 8. 34,152: 4 × 1 + 5 × 2 + 2 = 16. 9: The sum of the digits must be divisible by 9. [2] [4] [5] 2,880: 2 + 8 + 8 + 0 = 18: 1 + 8 = 9. Subtracting 8 times the last digit from the rest gives a multiple of 9. (Works because 81 is divisible by 9)
Next one repeats step 2, using the small digit concatenated with the next digit of the dividend to form a new partial dividend (15). Dividing the new partial dividend by the divisor (4), one writes the result as before — the quotient above the next digit of the dividend, and the remainder as a small digit to the upper right. (Here 15 divided ...
This allows one to predict that certain forms of numbers will always lead to a smaller number after a certain number of iterations: for example, 4a + 1 becomes 3a + 1 after two applications of f and 16a + 3 becomes 9a + 2 after four applications of f.
This is denoted as 20 / 5 = 4, or 20 / 5 = 4. [2] In the example, 20 is the dividend, 5 is the divisor, and 4 is the quotient. Unlike the other basic operations, when dividing natural numbers there is sometimes a remainder that will not go evenly into the dividend; for example, 10 / 3 leaves a remainder of 1, as 10 is not a multiple of 3.
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
The harmonic mean of a set of positive integers is the number of numbers times the reciprocal of the sum of their reciprocals. The optic equation requires the sum of the reciprocals of two positive integers a and b to equal the reciprocal of a third positive integer c. All solutions are given by a = mn + m 2, b = mn + n 2, c = mn.
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Next, the 1 is multiplied by the divisor 4, to obtain the largest whole number that is a multiple of the divisor 4 without exceeding the 5 (4 in this case). This 4 is then placed under and subtracted from the 5 to get the remainder, 1, which is placed under the 4 under the 5.