This is my implementation, but it not efficient when given 6 digit number.
Input : n = 2
Output : 9009
9009 is the largest number which is product of two
2-digit numbers. 9009 = 91*99.
def isPali(x):
n = str(x)
for i in range(len(n)):
if not n[i] == n[-i-1]:
return False
return True
def isProduct(x,A):
counter = A
while counter > 1:
if x // counter <= A and x % counter == 0:
return True
else:
counter-=1
return False
def largestProduct(A):
for i in range(A*A,1,-1):
if isPali(i) and isProduct(i,A):
return i
return False
largestProduct(999999)
Let x and y be the two n-digit factors of the palindrome number.
You can iterate over them in a descending number.
Key is to stop as soon as possible, which mean, once a first solution has been found, you don't check any product below that solution.
def get_max_palindrome(n):
res = 0
for x in range(10 ** n - 1, 1, -1):
for y in range(10 ** n - 1, 1, -1):
p = x * y
if res > p:
break
if str(p) == str(p)[::-1]:
res = p
break
if (x - 1) ** 2 < res:
break
return res
print(get_max_palindrome(6))
Exec in 0.378s on my laptop.
Codewise, this is not too difficult:
n = 999999
max_pali =0
t = ()
for i in range(1,n+1):
for j in range(i,n+1):
m = i*j
s = str(m)
if s == s[::-1] and m > max_pali:
max_pali = m
t = (i,j)
print(max_pali,t)
However, this is a brute force approach. For numbers with 6 digits, this will not terminate in a reasonable amount of time. Even if it will, I could ask you the same question for 7 or 42 digits. I suggest you look for some structure, or property, of those numbers whose multiple is a palindrome. Could such a pair be any pair of numbers? Is the case 91*99 = 9009 a mere coincidence, or is there a pattern?
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