Puzzle 4

What's the largest palindrome number that is the product of two three digit numbers.

This time I have one of the better solutions:

def nextpal(low,high,reversed=False):
    assert low < high

    if reversed:
        candidate = high
        final = low
        increment = -1
    else:
        candidate = low
        final = high
        increment = 1
    while candidate != final:
        if str(candidate) == str(candidate)[-1::-1]:
            yield candidate
        candidate += increment

def getfactor(low,high,target):
    for test in range(low,high):
        if not target % test:
            residual = target/test
            if residual in range(low,high):
                return test, residual
    return None

if __name__ == "__main__":
    from time import time
    start = time()
    pal = nextpal(10000,1000000, reversed=True)
    while True:
        target = pal.next()
        result = getfactor(100,1000,target)
        if result:
            print target
            print result
            break
    end = time()
    print end - start

Most started from the factors to see if they built palindromes. Much better to go the other way; far fewer tests (the 94th palindrome succeeded, so less than 9400 calculations. The cast to string for the palindrome test is easy, but is it efficient?

The most interesting items on the discussion were the Pytho one-liner:

max(map(lambda s: bool(s) and max(s) or 0, [(filter(lambda n: n == int(str(n)[::-1]), map(lambda n: i*n, range(999, 99, -1)))) for i in range(999, 99, -1)]))

and the pen-and-paper solution

You can also do this with pen and paper. You have a number:

(100a + 10b + c)(100d + 10e + f)

Which is a palindrone. This factors out to:

100cd + 10ce + cf +
1000bd + 100be + 10bf +
10000ad + 1000ae + 100af

Assuming the first digit is 9, then cf must be equal to nine.
Also, both a and d must then be equal to nine. The only ways
to make the last digit of the product of two integers 9 would
be if those integers were either of:

1 and 9
3 and 3
7 and 7

So, both numbers must start with 9, end with either 1, 3, 7,
or 9, and one of them must be divisible by 11. The only
numbers divisible by 11 from 900 - 1000 are:

902
913
924
935
946
957
968
979
990

You can discard all of those that do not end in either 1, 3,
7, or 9, and you are left with:

913
957
979

So now the presumed answer is either:

(900 + 10 + 3)(900 + 10x + 3)
(900 + 50 + 7)(900 + 10x + 7)
(900 + 70 + 9)(900 + 10x + 1)

Factoring all those out, you get:

810000 + 9000x + 2700 + 9000 + 100x + 30 + 2700 + 30x + 9
824439 + 9130x

Now, for the first digit 824439 + 9130x to be 9, x must be 9
(if x were 8, then 824439 + 9130x = 897479, and the first
digit is 8). And so you have 913 * 993, which is the answer.
You can factor the others out to see if they produce a bigger
answer, which they don't.