First observation is that the number must be between 100^2 and 999^2 or in the range of [10000, 998001].
As the majority of numbers has 6 digits and we're looking for the largest, we ignore 5 digits numbers.
Based on this, we can construct a palindromic number as:
'abccba' =100000a+10000b+1000c+100c+10b+a
=100001a+10010b+1100c
=11(9091a+910b+100c) = p.q
This equation shows us, that either p or q, but not both must have a factor of 11!!!
def palindrome_number(x,y):
my_list =[]
for i in range(x,y):
for j in range(x,y):
num = i * j
if num % 11 ==0 and str(num) == str(num)[::-1]:
my_list.append(num)
my_list.sort()
print(my_list[-1])
First observation is that the number must be between 100^2 and 999^2 or in the range of [10000, 998001].
As the majority of numbers has 6 digits and we're looking for the largest, we ignore 5 digits numbers.
Based on this, we can construct a palindromic number as:
'abccba' =100000a+10000b+1000c+100c+10b+a
=100001a+10010b+1100c
=11(9091a+910b+100c) = p.q
This equation shows us, that either p or q, but not both must have a factor of 11!!!
def palindrome_number(x,y):
void solve()
{
fast;
ll n,largepalin=0;
cin>>n;
for(ll i=999;i>=100;i--)
{
for(ll j=999;j>=100;j--)
{
if((i*j)%11==0 && palin(i*j) && (i*j)<n) largepalin=max(i*j,largepalin);
}
}
cout<<largepalin<<endl;
return;
}