This is the fourth post in my Project Euler series. The third problem is here.
A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is
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Find the largest palindrome made from the product of two 3-digit numbers.
.I was a little disappointed at the simplicity of this problem. There’s just not really much to it. First off, we need to know how to determine if a number is a palindrome. To do this, we need to be able to calculate a number’s reverse. We can obtain the digits of a number 1 at a time by using the modulo operator. Example: 
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long reverseLong(long x) { long y = 0; int digit; while (x > 0) { digit = x % 10; y = (y * 10) + digit; x /=10; } return y; } |
This is simple enough. As we know from the description of the problem, a palindrome is something that reads the same forward and backward. So from here it’s a pretty obvious step to know if a number is a palindrome.
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BOOL isPalindrome(long n) { return n == reverseLong(n); } |
From here, there’s nothing left but looping through to find the largest palindromic number. There’s not really much room for optimization, however I did find some. Assuming that you always calculate 
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long foo() { long max = 0; for(int x = 100; x <= 999; x++) { for(int y = x; y <= 999; y++) { long product = x * y; if(product > max && isPalindrome(product)) { max = product; } } } return max; } |