Smallest Integer #223
What is the smallest integer greater than 0 that can be written entirely with 0s and 1s and is exactly divisible by 223?
I don't know how difficult this will be for others, but it took me 4 seconds.
No binary numbers
I don't know how difficult this will be for others, but it took me 4 seconds.
No binary numbers
Labels: mathschallenge
posted by Ragknot at
6:06 PM
16 Comments:
Chris,
I hope you like this one, I don't know any tricks to help.
Is octal allowed :)
Ok, Octal is allowed, but still have to use only 0's and 1's.
"tricky". "it might take me some time". (Not sure if I got the quotes right).
Nah, I'm not going to do it in octal.
got it :) simple
no, not 87 and not 42.
You can cross two numbers off the list.
The answer is 1001111001, but I can't think of a nice way to prove it.
Was proof required?
How 'bout
4,489,287 * 223 = 1,001,111,001?
9/21/2009 8:05:39 PM
4489287 1001111001
9/21/2009 8:05:43 PM
How did you do it? Brute Force?
Hi Ragknot. I'd nodded off.
For me the "mathematical" proof is far more interesting. Why is it 6 1s for instance?
I do like the pretty pattern though.
I've got to get some sleep. It's nearly 4:30 am here. Thank you for the problem.
Just saw your post in time. Lazy brute force, I nicked your code from the 225 problem, and changed 1 whole digit. I wanted to see if the pattern would give me a clue. I really just wanted to know how many 1s were in it. 'nite
Chris,
I had the thought that there was no trick to this, but I thought you
might have some trick. I thought someone might try my brute force code. I suspected no one would try but you.
Thank you for the 225. And I am glad you found the solution for 223.
I thought you might figure that 225 took 57 seconds, and 223 took 4 seconds so maybe the 223 solution might be about 1/11 of the 225. Actually it is about 1/14. But that's not close enough to help much. Somewhere between 4 million
and 5 million is not a narrow window.
Hi Ragknot. I had noticed that the 223 code ran a lot faster, etc.
"Technique" is the word I should have used. Unfortunately (or though my change into fortunately) 223 is a prime.
The only things I've worked out are that the last digit must be 1, because 0 implies all that would have been done, is to have pushed the original problem one place to the left. In turn that means the other factor (4489287) must be end with 7 (as xx3*yyyyN=zzzz1 is only satisfied by N=7). Some progress, but not a lot.
Gotta go. Busy day, but I'll probably sneak back afew times. Greetz.
Of course, in binary it's 11011111 :)
I've given up. I had got some nice residue table of 10^n mod 223. But can see no way to select the optimum residues to add up to 0.
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