Problem 004

[MaJJ]

Hi,
I am stuck again. !@#$%

I have done code that finds if number in a variable is palindromic, but there is probably an error with the code that tries all combinations of numbers.
I am doing it with WHILE cycle - it repeats the code until there is a palindromic number or one of numbers that I am multipling, is < 100.

When I tried code that lowers only one of the numbers, it found

Expand

698896

, but it's not enough.
When I tried code that lowest one of the numbers and if it's below 100, it returns it back to 999 and lowers second number. It found

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580085 (995*583)

, which is even lower. But this code (if done right) should try all combinations, so ... it must be wrongly written ...

I tried to edit that "palindromic requirement" for 4-digit palindroms and other code for 2-digit numbers (if below 10, return to 99), and it found that 9009 (result of example problem). So maybe this code is right, maybe not.

Any ideas? Should I try some completely different algorithm to multiply numbers?

[Georg]

"I am doing it with WHILE cycle - it repeats the code until there is a palindromic number "
"But this code (if done right) should try all combinations"

This is a contradiction.

[uws8505]

It would be better if you test (999,999) (999,998) (999,997) (998,998) (999,996) (998,997) and so on
rather than (999,999),...,(999,100),(998,999),...

[MaJJ]

2 georg - this code goes from higher numbers, so it should stop at the highest, or not? EDIT: It shouldn't, now I know that I will try to make it go to the end

Expand

(100,100)

[MaJJ]

Ha! Got it I just had to let the program show me all the palindroms.

Thanks!

[badperson]

Hi,

I'm just going thru the problems, just getting started out, and my answer to number 4 was rejected, but I'm sure it's right. I double checked it on the calculator, and the same algorithm works with the 2-digit example given in the problem description.

Is it some kind of trick answer I'm not getting?
bp

[quilan]

No trick on this one; it's fairly straight forward. If you want to PM your answer, I can let you know if there's anything incorrect about it.

[Georg]

The main problem is to identify the maximum correctly.

[jessicasco]

I am a new member, and I think I have a lot to learn.
Now I want to say that the overview of problem 004 has a mistake.
It's the function "reverse",
may be the point of the problem is not at this function, but I still have to say
this so it can be changed to a better version.

If n is 1000;
the return value of the reverse function is 1;
surely not the intended answer. the author made this little mistake.

[TripleM]

Why would the reverse of 1000 not be 1? When I reverse 1000, I get 0001, which I can't see to be any number other than 1.

[hk]

Just to make it clear:
Integers are written without leading zeros. This applies to all Project Euler problems unless explicitly stated otherwise.

[jessicasco]

got it.

now i realize i am wrong.

[MoHSalim]

This is the code I got:

(code snipped)

//The Exceptions I receive are below

----jGRASP exec: java ProblemFour

Exception in thread "main" java.lang.NumberFormatException: For input string: "-"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
at java.lang.Integer.parseInt(Integer.java:474)
at java.lang.Integer.parseInt(Integer.java:497)
at Palindromic.isPalindromic(Palindromic.java:26)
at ProblemFour.main(ProblemFour.java:16)

----jGRASP wedge2: exit code for process is 1.
----jGRASP: operation complete.


//What's wrong with my life/program?

[Georg]

This line

Code: Select all

for(int k = 999; 999>=100; i--)

and more.

[MoHSalim]

Sorry. I already fixed it a while ago and forgot to post here. Thanks anyways, you're awesome.

[Georg]

You should edit your post and delete the source code.

[pbear]

What's wrong with this code?

Code: Select all

39   for( x = 999; x >= 100; x-- ) {
40     for( y = 999; y >= x; y-- ) {
41       r = x * y;
42       printf("%u * %u = %u\n", x, y, r);
43
44       if( is_palindrome( r ) ){
45         printf("I has it: %d\n", r);
46         return 0;
47       }
48     }
49   }

It does yield a palindrome of six digits, and I can't see what is wrong here

[TripleM]

If it yields a palindrome of 6 digits, and it tells you you have the wrong answer, then logically it cannot be the largest.

[daniel.is.fischer]

Exactly, there are 80 6-digit palindromes larger than what that code returns which are the product of two 3-digit numbers.

[theslimzmassive]

Would i be wrong in saying the actual answer to problem 4 is 997799? The accepted answer is < snip >. The question is:

Find the largest palindrome made from the product of two 3-digit numbers.

999*999 = 998001 <- upper limit

so basically

< snip ><997799<(999*999)

ala < snip > is the wrong answer.