Problem 235. Project Euler: Problem 4, Palindromic numbers

Sorry for cheating but I couldn't find a solution that doesn't crash the servers

A double for loop that performs a brute force scan from x down to 0.9*x in each loop - doesn't crash the servers :)

hi, my cod gives me true ans but its worked too slow because of for loop but could not figure it out without for loop. could u send me our code or check mines ? function u = euler004(n)
function L = ispalindrome(n)
s=num2str(n);
l=length(s);
for i=1:ceil(l/2)
if str2double(s(i)) == str2double(s(l-i+1))
continue
else
L=false;
return
end
end
L=true;
end

c=numel(num2str(n));
switch c
case 3
l=9;
case 4
l=99;
end

g=l:n;
m=l:n;
z=g'*m;
b=z(:)';
T=[];

for vz=1:numel(b)
if ispalindrome(b(vz))>0
T=b(vz);
end
end

u=max(T);

end

can anyone explain what is meaning of products of two inputs ,here input is only one number ,does it mean the square of input

? cheat

More like exploiting the limited test suite by hard coding. The test suite has been updated now.

Thanks for the note Aurelien. I added to the test suite.

the largest palindrome made from the product of 9999 is 99977999， not 99000099！
and x=999，y=997799, not 906609.
x=2500, y=6249246, not 6167616
x=100, y=999, not 9009

zd c, 99977999 cannot be written as the product of two numbers less than or equal to 9999.

Size 42 ! wow should be the best solution