Nooooooooooooo... a number cannot start with zero You are on the right track though But, it is not that simple Doh

Well if I would continue this sequence I ** will** find the number, or is my theory not correct.

Then it might get interesting after all !! haha

ok, continuing sequence:

00.000*7=0

0+40.579=40.579

400.000*7=2.800.000

2.800.000+40.579=2.840.579

8.000.000*7=56.000.000

56.000.000+2.840.579=58.840.579

80.000.000*7=560.000.000

560.000.000+58.840.579=618.840.579

100.000.000*7=700.000.000

700.000.000+618.840.579=1.318.840.579

3.000.000.000*7=21.000.000.000

21.000.000.000+1.318.840.579=22.318.840.579

20.000.000.000*7=140.000.000.000

140.000.000.000+22.318.840.579=162.318.840.579

600.000.000.000*7=4.200.000.000.000

4.200.000.000.000+162.318.840.579=4.362.318.840.579

pffff, now it gets boring.

I think i did something wrong

one hint please:

tell me,

does you number end with ......** 3.623.188.405.79**7 ???

(because multiply this with 7 and you get 25.** 362.318.840.579**)

and... your number starts with 11 , right ?

Now, about the sequence, that was really tough. I had to plot them on Matlab, and they looked like a quadratic function, and then out of nowhere, I saw it

0 = 0 * 1 * 2

6 = 1 * 2 * 3

24 = 2 * 3 * 4

60 = 3 * 4 * 5

120 = 4 * 5 * 6

210 = 5 * 6 * 7

which means the next number should be

6 * 7 * 8 = 336 YEAAAAAAAAAH !!! goooood job !!

one hint please:

tell me,

does you number end with ......** 3.623.188.405.79**7 ???

(because multiply this with 7 and you get 25.** 362.318.840.579**)

and... your number starts with 11 , right ?

Yes, you are absolutely right. These are the last 13 digits You are getting close too, it doesn't take much longer from now

But, no the number doesn't start with 11! Sorry

Tsk, as long as you can write a nice math equation for it, it can be solved with brute force:

Iterate over N: (N*10 + 7) * 7 == int(log10(N)+1)*7 + N

I'll leave the programming as an exercise for the reader

No. as you can see the number is larger than 13 digits. It can't even be represented as a long long! So, iterating is going to take billions of years

Tsk, as long as you can write a nice math equation for it, it can be solved with brute force:

Iterate over N: (N*10 + 7) * 7 == int(log10(N)+1)*7 + N

I'll leave the programming as an exercise for the reader

No. as you can see the number is larger than 13 digits. It can't even be represented as a long long! So, iterating is going to take billions of years

So what? You stated the problem cannot be solved with brute force, I just showed that it is possible -- you just need a bit of patience

(Or you could convert norakomi's pattern searching into an algorithm... I not so bored that I'm going to post *that*, but it is possible and will find the number faster than brute force)

Here's another one. How does this sequence continue?

0001001101011110...

Here's another one. How does this sequence continue?

0001001101011110...

Hmm...

0001001101011110

could be decomposed to

0001 0011 0101 1110

1 3 5 7

so, it should continue

0001001101011110 1001 1011...

And speaking of sequences, here is another one

4, 5, 9, 11, 12, 13, 14.....

How should it continue?

(Hint: This is called the "Even Steven" sequence!)

I not so bored that I'm going to post *that*, but it is possible and will find the number faster than brute force)

Ah well, what the heck. Here goes - in Java for a change

public class Norakomis7{ public static void main(String[] args) { int lastDigit = 7; int carry = 0; int value; boolean ready; Vector<Integer> digits = new Vector<Integer>(); digits.add(new Integer(lastDigit)); do { value = lastDigit * 7 + carry; lastDigit = value % 10; carry = value/10; ready = (lastDigit == 7 && carry == 0); if(!ready) digits.add(0, new Integer(lastDigit)); } while(!ready); for(Iterator<Integer> it = digits.iterator(); it.hasNext(); ) { System.out.print(it.next()); } System.out.println(); } }

The output is 1014492753623188405797

As for the sequence, your solution sounds valid, but is not what I anticipated:

0001 0011 0101 1110 0010 0110 1011 1100 ...

Now, how does the sequence continue?

Code cleanup

... value = lastDigit * 7 + carry; while (value != 7) { lastDigit = value % 10; carry = value/10; digits.add(0, new Integer(lastDigit)); value = lastDigit * 7 + carry; }; ...