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 45

Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:

Triangle		Tn=n(n+1)/2		1, 3, 6, 10, 15, ...
Pentagonal		Pn=n(3n1)/2		1, 5, 12, 22, 35, ...
Hexagonal		Hn=n(2n1)		1, 6, 15, 28, 45, ...

It can be verified that T₂₈₅ = P₁₆₅ = H₁₄₃ = 40755.

Find the next triangle number that is also pentagonal and hexagonal.

analyse:
1) Every Hexagonal is also a Triangle.
performance improvements: 
2) Start looping with Hexagonal numbers because they increase faster than other two number types.
3) H143 gives 40755 so start with n = 144.
4) use difference: H(n) - H(n-1) = n(2n-1) - ((n-1)(2(n-1)-1)) = 4 * n - 3

An alternate solution is to make use of the square root formula. x = (-b+- sqrt(b2-4ac))/2a

When we rewrite the pentagon formule P(n) = n(3n-1)/2 

we get 1.5n*n - 0.5n - P(n) = 0 where P(n) = H(n)

Problem45 =  1533776805 elapsed time:    1 ms. Test Passed.

Euler problem 44

Pentagonal numbers are generated by the formula, P_n=n(3n1)/2. The first ten pentagonal numbers are:

1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...

It can be seen that P₄ + P₇ = 22 + 70 = 92 = P₈. However, their difference, 70  22 = 48, is not pentagonal.

Find the pair of pentagonal numbers, P_j and P_k, for which their sum and difference is pentagonal and D = |P_k  P_j| is minimised; what is the value of D?


Not so proud on this one. Anybody a suggestion?

Problem44 =     5482660 elapsed time:    2 ms. Test Passed.