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Any even natural number is of the form 2n, where n is a natural number. Therefore, let us assume that we have two even numbers which we will denote by 2k and 2l. Their product is (2k)(2l) = 4(kl) = 2(2kl). Since 2kl is also a natural number, the product is even. Note: In this case, both exhaustivity and exclusivity were needed.
Landau's fourth problem asked whether there are infinitely many primes which are of the form = + for integer n. (The list of known primes of this form is A002496 .) The existence of infinitely many such primes would follow as a consequence of other number-theoretic conjectures such as the Bunyakovsky conjecture and Bateman–Horn conjecture .
The Ages of Three Children puzzle (sometimes referred to as the Census-Taker Problem [1]) is a logical puzzle in number theory which on first inspection seems to have insufficient information to solve. However, with closer examination and persistence by the solver, the question reveals its hidden mathematical clues, especially when the solver ...
For example, if s=2, then ๐(s) is the well-known series 1 + 1/4 + 1/9 + 1/16 + …, which strangely adds up to exactly ๐²/6. When s is a complex number—one that looks like a+b๐, using ...
The misdirection in this riddle is in the second half of the description, where unrelated amounts are added together and the person to whom the riddle is posed assumes those amounts should add up to 30, and is then surprised when they do not โ — โ there is, in fact, no reason why the (10 โ − โ 1) โ × โ 3 โ + โ 2 โ = โ 29 sum should add up to 30.
The case = of this problem was used by Bjorn Poonen as the opening example in a survey on undecidable problems in number theory, of which Hilbert's tenth problem is the most famous example. [57] Although this particular case has since been resolved, it is unknown whether representing numbers as sums of cubes is decidable.
In addition to S(2,3,9), Kramer and Mesner examined other systems that could be derived from S(5,6,12) and found that there could be up to 2 disjoint S(5,6,12) systems, up to 2 disjoint S(4,5,11) systems, and up to 5 disjoint S(3,4,10) systems. All such sets of 2 or 5 are respectively isomorphic to each other.
G(3) is at least 4 (since cubes are congruent to 0, 1 or −1 mod 9); for numbers less than 1.3 × 10 9, 1 290 740 is the last to require 6 cubes, and the number of numbers between N and 2N requiring 5 cubes drops off with increasing N at sufficient speed to have people believe that G(3) = 4; [17] the largest number now known not to be a sum of ...