Since P is

This is problem 3 from the Project Euler. Compute and display the average of the two highest numbers. while(y=0 && y-x 999), means that j would be a four digit or above, and thus this would be true for any i smaller than the current one I am testing and then we can break. Initialise variables and common functions: Since all even numbers can be divided by 2, we shall return 2 for every even input number. It differs a bit in the nature of the problem from the first 3 we have looked at so far. I havent found a better solution to this problem than the link below: http://spectre92z.wordpress.com/2014/08/25/eulers-problem-4-the-largest-palindrome-made-from-the-product-of-two-3-digit-numbers/, Implemented a brute force solution in Java and optimized it down to 5ms.

while b >= a Today it is time to look at the solution to Problem 4 of Project Euler. break; Project Euler – Problem 4. Please feel free to […].

cout<0 a = a-1

P=100000x+10000y+1000z+100z+10y+x. Palindromes in Table 1 are divisible by 11 and, therefore, contain an even number of digits. you probably meant 998001 instead of 9998001. maior=i; } Since 11 is prime, at least one of the integers a or b must have a factor of 11. if a mod 11 = 0 return 0; It will be fun and we can learn a thing or two by solving this problem in different ways. if a*b <= largestPalindrome Thanks for sharing. f=(((i%10000)%1000)%100)/10; Brute-force is really disappointing from a mathematical standpoint, isn’t it? break; The number of divisors of a natural number \(n\) is given by tau(n) or \(\tau(n)\) or sometimes \(\delta(n)\) as mentioned here already.

The problem.



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