Euler problem 61

Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygonal) numbers and are generated by the following formulae:

Triangle		P3,n=n(n+1)/2		1, 3, 6, 10, 15, ...
Square		P4,n=n2		1, 4, 9, 16, 25, ...
Pentagonal		P5,n=n(3n1)/2		1, 5, 12, 22, 35, ...
Hexagonal		P6,n=n(2n1)		1, 6, 15, 28, 45, ...
Heptagonal		P7,n=n(5n3)/2		1, 7, 18, 34, 55, ...
Octagonal		P8,n=n(3n2)		1, 8, 21, 40, 65, ...

The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three interesting properties.

1. The set is cyclic, in that the last two digits of each number is the first two digits of the next number (including the last number with the first).
2. Each polygonal type: triangle (P_3,127=8128), square (P_4,91=8281), and pentagonal (P_5,44=2882), is represented by a different number in the set.
3. This is the only set of 4-digit numbers with this property.

Find the sum of the only ordered set of six cyclic 4-digit numbers for which each polygonal type: triangle, square, pentagonal, hexagonal, heptagonal, and octagonal, is represented by a different number in the set.

analyse:

Start with n = 19. This is the minimal number to generate a four digit Octagonal number.

Problem 61 =       28684 elapsed time:    1 ms. Test Passed.

Euler problem 60

The primes 3, 7, 109, and 673, are quite remarkable. By taking any two primes and concatenating them in any order the result will always be prime. For example, taking 7 and 109, both 7109 and 1097 are prime. The sum of these four primes, 792, represents the lowest sum for a set of four primes with this property.

Find the lowest sum for a set of five primes for which any two primes concatenate to produce another prime.

Problem 60 =       26033 elapsed time:  236 ms. Test Passed.