The following iterative sequence is defined for the set of positive integers: n n/2 (n is even) n 3n + 1 (n is odd) Using the rule above and starting with 13, we generate the following sequence: 13 40 20 10 5 16 8 4 2 1 It can be seen that this seque…

(CodeForces - 5C)Longest Regular Bracket Sequence time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output This is yet another problem dealing with regular bracket sequences. We should remind you t…

The nth term of the sequence of triangle numbers is given by, tn = ½n(n+1); so the first ten triangle numbers are: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ... By converting each letter in a word to a number corresponding to its alphabetical position and…

The number 145 is well known for the property that the sum of the factorial of its digits is equal to 145: 1! + 4! + 5! = 1 + 24 + 120 = 145 Perhaps less well known is 169, in that it produces the longest chain of numbers that link back to 169; it tu…

Consider the fraction, n/d, where n and d are positive integers. If nd and HCF(n,d)=1, it is called a reduced proper fraction. If we list the set of reduced proper fractions for d  8 in ascending order of size, we get: 1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1…

There are exactly ten ways of selecting three from five, 12345: 123, 124, 125, 134, 135, 145, 234, 235, 245, and 345 In combinatorics, we use the notation, 5C3 = 10. In general, It is not until n = 23, that a value exceeds one-million: 23C10 = 114406…

The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the 4-digit numbers are permutations of one another. There are no arithmetic seq…

The first two consecutive numbers to have two distinct prime factors are: 14 = 2  7 15 = 3  5 The first three consecutive numbers to have three distinct prime factors are: 644 = 2²  7  23 645 = 3  5  43 646 = 2  17  19. Find the first four consecutiv…

Consider all integer combinations ofabfor 2a5 and 2b5: 22=4, 23=8, 24=16, 25=32 32=9, 33=27, 34=81, 35=243 42=16, 43=64, 44=256, 45=1024 52=25, 53=125, 54=625, 55=3125 If they are then placed in numerical order, with any repeats removed, we get the f…

Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows: 21 22 23 24 2520  7   8   9  1019  6   1   2  1118  5   4   3  1217 16 15 14 13 It can be verified that the sum of the numbers on the di…