Question

From Project Euler, problem 45:

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

Triangle T_(n)=n(n+1)/2 1, 3, 6, 10, 15, ...

Pentagonal P_(n)=n(3n−1)/2 1, 5, 12, 22, 35, ...

Hexagonal H_(n)=n(2n−1) 1, 6, 15, 28, 45, ...

It can be verified that T_(285) = P_(165) = H_(143) = 40755.

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

[ http://projecteuler.net/problem=45 ]

Now to solve them I took three variables and equated the equations to A.

n(n + 1)/2 = a(3a - 1)/2 = b(2b - 1) = A

A = number at which the threee function coincide for values of n, a, b

Resultant we get 3 equations with n and A. Solving with quarditic formula, we get 3 equations.

 (-1 + sqrt(1 + 8*A ) )/2
 ( 1 + sqrt(1 + 24*A) )/6
 ( 1 + sqrt(1 + 8*A ) )/4

So my logic is to test for values of A at which the three equation give a natural +ve value. So far it works correct for number 40755 but fails to find the next one upto 10 million.

(Edit): Here is my code in python

from math import *

i=10000000
while(1):
    i = i + 1
    if(((-1+sqrt(1+8*i))/2).is_integer()):
        if(((1+sqrt(1+24*i))/6).is_integer()):
            if(((1+sqrt(1+8*i))/4).is_integer()):
                print i
                break

How is my logic wrong? (Apologies for a bit of maths involved. 🙂 )

Answer 1

Your logic is not wrong, your program just takes a long time to run (by my estimate it should provide an answer in about an hour). I know the answer and tested your program by setting i to a value just below it. Your program then popped out the right answer at once.

Heed the advice of ypercube.