(then 11122113121113)
For universe.zip’s puzzle
method spoiler here
[spoiler] You have to call out the number of instances of each number.
First line is 12. So there’s 1 One and 1 Two. So that’s 1112. In this second line, there’s 3 Ones and 1 Two. So that’s 3112. In the final line given, it’s 212223, so that’s 1 One, 4 Twos and 1 Three. Making it voila → 114213 [/spoiler]
Ah, I was almost right.
As usual with these puzzles I spent too little time examining it; I saw the first two lines and (stupidly) assumed that the next line described the sequence of the previous one; if this were the case, the next line would have been 12113213 One two, one one, three twos, one three, but it isn’t the case.
Damn, that’s the second time today that perfect logic and lazy reading has made me give the wrong answer to a puzzle
Without lookingWell, for a5 to still be under 6 digits, a must be 1.
Then for 3f to end in 1, then f must be 7.
Then 27 ends in 4, giving us b=4.
57 ends in 5, giving e=5,
and 47 ends in 8, which gives d=8.
So, we know that the number is 14c857.
since 214c857 begins with 2, c must be 2.
So, 142857.
Which are the repeating 6 digits that you get whenever you divide a number by 7 so, if you realized that in the first place, you wouldn’t have had to do the rest of the math. Should have seen it, didn’t. Cute puzzle
Great explanation, btw, Gyan9, concise and logical.
Alescus, how did you do it?
amarinth - really? I didn´t know that…and I was wondering how much time whoever made the puzzle had spent looking for a number with which it works…
anyone got any more?
From fabcde = 5 * abcdef, a must be 0 or 1, and f is not 0 (unless b is 0 or 1, and on closer inspection must be 0, a line of reasoning that eventually leads to the solution of 000000). From f>0 and bcdefa = 3 * abcdef, a is therefore 1. This equation also means b is either 3 or 4, and from cdefab = 2 * abcdef, b must be 4, and knowing this the same equation means c = 2. Knowing a, b, c and the fact that fabcde = 5 * abcdef, f must be 7. e must be 0 or 5, and using the f that we’ve just found, we can say e = 5. I actually found d = 8 by knowing that a+b+c+d+e+f = (a multiple of 3), and eliminating the possibilities of d = 2 and d = 5. Using efabcd = 4 * abcdef would have made more sense though.
As for other puzzles, it might be possible to make a similar one using the digits that repeat after dividing by 13.