Double the Last Number

[BitBuster]

268435456

[IceyLava108]

536,870,912

[micronexer97]

1061753824

Hmm... I can see 7-5-3 which is a nice string, and the appearance of each even digit once, pretty cool number.

[BitBuster]

2123507648



I like how the digits are sort of clustered with low ones at the beginning and higher ones at the end (save for that deviant 5). I think of the 0 as acting as the break between the two classes.

[IceyLava108]

4247015296

Umm.. no pattern.

[BitBuster]

8494030592



Say someone messed up with this whole doubling thing...at this point, would anyone notice?

[ajmiam]

8494030592



Say someone messed up with this whole doubling thing...at this point, would anyone notice?

Well, as a CS geek, I know that 2^31-1=2147483647 is the largest number that an int can hold in Java, so I know 2147483648=2^31 is a power of 2.

The number 2123507648, is not a power of 2, so someone must have messed up along the way...

Perhaps I can restore the correct sequence...

2147483648*2=4294967296.

A 10-digit number with only 5 distinct digits (2, 4, 6, 7, 9)! Also, each block of 2 digits in the first number (21, 47, 48, 36, and 48) got perfectly doubled (to 42, 94, 96, 72, and 96) without any carry from the previous block messing them up! What are the odds of that?

[mathgrant]

8589934592

Did I do it right?

[BitBuster]

17179869184



There's always this cheat sheet, which should keep us safe for at least a few more pages.

http://www.tsm-resources.com/alists/pow2.html

[mathgrant]

536,870,912

1061753824

Aha, here's the problem!

P.S. 34359738368