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For any integer n, n ≡ 1 (mod 2) if and only if 3n + 1 ≡ 4 (mod 6). Equivalently, n − 1 / 3 ≡ 1 (mod 2) if and only if n ≡ 4 (mod 6). Conjecturally, this inverse relation forms a tree except for the 1–2–4 loop (the inverse of the 4–2–1 loop of the unaltered function f defined in the Statement of the problem section of ...
Using the example above: 16,499,205,854,376 has four of the digits 1, 4 and 7 and four of the digits 2, 5 and 8; Since 4 − 4 = 0 is a multiple of 3, the number 16,499,205,854,376 is divisible by 3. Subtracting 2 times the last digit from the rest gives a multiple of 3.
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.
For example, there are six divisors of 4; they are 1, 2, 4, −1, −2, and −4, but only the positive ones (1, 2, and 4) would usually be mentioned. 1 and −1 divide (are divisors of) every integer. Every integer (and its negation) is a divisor of itself. Integers divisible by 2 are called even, and integers not divisible by 2 are called odd.
Thus the fraction 3 / 4 can be used to represent the ratio 3:4 (the ratio of the part to the whole), and the division 3 ÷ 4 (three divided by four). We can also write negative fractions, which represent the opposite of a positive fraction. For example, if 1 / 2 represents a half-dollar profit, then − 1 / 2 represents ...
For instance, the numeral for 10,405 uses one time the symbol for 10,000, four times the symbol for 100, and five times the symbol for 1. A similar well-known framework is the Roman numeral system . It has the symbols I, V, X, L, C, D, M as its basic numerals to represent the numbers 1, 5, 10, 50, 100, 500, and 1000.
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For example, using single-precision IEEE arithmetic, if x = −2 −149, then x/2 underflows to −0, and dividing 1 by this result produces 1/(x/2) = −∞. The exact result −2 150 is too large to represent as a single-precision number, so an infinity of the same sign is used instead to indicate overflow.