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An odd number does not have the prime factor 2. The first: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23 (sequence A005408 in the OEIS). All integers are either even or odd. A square has even multiplicity for all prime factors (it is of the form a 2 for some a). The first: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144 (sequence A000290 in the OEIS).
An example where it does not is given by the isolated singularity of x 2 + y 3 z + z 3 = 0 at the origin. Blowing it up gives the singularity x 2 + y 2 z + yz 3 = 0. It is not immediately obvious that this new singularity is better, as both singularities have multiplicity 2 and are given by the sum of monomials of degrees 2, 3, and 4.
The additive persistence of 2718 is 2: first we find that 2 + 7 + 1 + 8 = 18, and then that 1 + 8 = 9. The multiplicative persistence of 39 is 3, because it takes three steps to reduce 39 to a single digit: 39 → 27 → 14 → 4. Also, 39 is the smallest number of multiplicative persistence 3.
Also, a monomial is a multiset of indeterminates; for example, the monomial x 3 y 2 corresponds to the multiset {x, x, x, y, y}. A multiset corresponds to an ordinary set if the multiplicity of every element is 1. An indexed family (a i) i∈I, where i varies over some index set I, may define a multiset, sometimes written {a i}.
A number is called "even" if it is an integer multiple of 2. As an example, the reason that 10 is even is that it equals 5 × 2. In the same way, zero is an integer multiple of 2, namely 0 × 2, so zero is even. [2] It is also possible to explain why zero is even without referring to formal definitions. [3]
Specifically, the 2-order of a nonzero integer n is the maximum integer value k such that n/2 k is an integer. This is equivalent to the multiplicity of 2 in the prime factorization. A singly even number can be divided by 2 only once; it is even but its quotient by 2 is odd.
Meanwhile, every number larger than 1 will be larger than any decimal of the form 0.999...9 for any finite number of nines. Therefore, 0.999... cannot be identified with any number larger than 1, either. Because 0.999... cannot be bigger than 1 or smaller than 1, it must equal 1 if it is to be any real number at all. [1] [2]
231 is also a multiple of 3: one has 231 = 3 · 77, and thus n = 2 · 3 2 · 77. Continue with 77, and 3 as a first divisor candidate. 77 is not a multiple of 3, since the sum of its digits is 14, not a multiple of 3. It is also not a multiple of 5 because its last digit is 7. The next odd divisor to be tested is 7. One has 77 = 7 · 11, and ...