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The first Project Euler problem is Multiples of 3 and 5. If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000. It is a 5% rated problem, indicating it is one of the easiest on the site.
For example, when d=4, the hash table for two occurrences of d would contain the key-value pair 8 and 4+4, and the one for three occurrences, the key-value pair 2 and (4+4)/4 (strings shown in bold). The task is then reduced to recursively computing these hash tables for increasing n , starting from n=1 and continuing up to e.g. n=4.
For instance, the first counterexample must be odd because f(2n) = n, smaller than 2n; and it must be 3 mod 4 because f 2 (4n + 1) = 3n + 1, smaller than 4n + 1. For each starting value a which is not a counterexample to the Collatz conjecture, there is a k for which such an inequality holds, so checking the Collatz conjecture for one starting ...
Prizes are often awarded for the solution to a long-standing problem, and some lists of unsolved problems, such as the Millennium Prize Problems, receive considerable attention. This list is a composite of notable unsolved problems mentioned in previously published lists, including but not limited to lists considered authoritative, and the ...
A solution to Kirkman's schoolgirl problem with vertices denoting girls and colours denoting days of the week [1] Kirkman's schoolgirl problem is a problem in combinatorics proposed by Thomas Penyngton Kirkman in 1850 as Query VI in The Lady's and Gentleman's Diary (pg.48). The problem states:
By restricting ourselves to reversible actions only, we can construct the solution to the problem from the desired result. From the point [4,4,0], there are only two reversible actions: transferring 3 liters from the 8 liter jug to the empty 3 liter jug [1,4,3], and transferring 3 liters from the 5 liter jug to the empty 3 liter jug [4,1,3].
Although the partition problem is NP-complete, there is a pseudo-polynomial time dynamic programming solution, and there are heuristics that solve the problem in many instances, either optimally or approximately. For this reason, it has been called "the easiest hard problem". [2] [3]
Langley's Adventitious Angles Solution to Langley's 80-80-20 triangle problem. Langley's Adventitious Angles is a puzzle in which one must infer an angle in a geometric diagram from other given angles. It was posed by Edward Mann Langley in The Mathematical Gazette in 1922. [1] [2]