This one was my favorite, probably because it’s been the least mathy so far. I was also tickled pink by the idea of numbers being palindromes. I saw some other solutions on-line but I was really happy with how mine turned out.
Problem #4
A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
Find the largest palindrome made from the product of two 3-digit numbers.
My Solution:
def get_highest_palindrome high, low
highest = 0
high.downto low do |i|
high.downto low do |j|
sum = i * j
if sum <= highest
break
end
if is_palindrome(sum.to_s)
highest = [highest, sum].max
end
end
end
highest
end
def is_palindrome str
return str == str.reverse
end
puts get_highest_palindrome 999, 100
In this particular (and probably only) case I liked my solution better than Olathe's. I found that by counting down from 999 and breaking after any number less than my current max, I was able to save a lot of iterations.
max = 0
100.upto(999) { |a|
a.upto(999) { |b|
prod = a * b
max = [max, prod].max if prod.to_s == prod.to_s.reverse
}
}
puts "Maximum palindrome is #{ max }."
However, just before I smugged off to bed I saw a post by Begoner in Python. His explanation:
The palindrome can be written as:
abccba
Which then simpifies to:
100000a + 10000b + 1000c + 100c + 10b + a
And then:
100001a + 10010b + 1100c
Factoring out 11, you get:
11(9091a + 910b + 100c)
So the palindrome must be divisible by 11. Seeing as 11 is prime, at least one of the numbers must be divisible by 11. So brute force in Python, only with less numbers to be checked:
def c():
max = maxI = maxJ = 0
i = 999
j = 990
while (i > 100):
j = 990
while (j > 100):
product = i * j
if (product > max):
productString = str(product)
if (productString == productString[::-1]):
max = product
maxI = i
maxJ = j
j -= 11
i -= 1
return max, maxI, maxJ
I got PWNED!
