3 ms·
"A group of scientists will be sitting around the cafeteria, and one will idly wonder if there is an integer where, if you take its last digit and move it to th
by hhm 18y ago
"A group of scientists will be sitting around the cafeteria, and one will idly wonder if there is an integer where, if you take its last digit and move it to the front, turning, say, 112 to 211, it’s possible to exactly double the value. Dyson will immediately say, “Oh, that’s not difficult,” allow two short beats to pass and then add, “but of course the smallest such number is 18 digits long.”"
So how did he know that?
- byrneseyeview 18y agoSome people investigate stuff like that, and hold on to the answer for a long time. For this one, I guess you'd start by narrowing it down. The number must have the start-end pair of 1,2; 2,4; 3,6; or 4,8. That narrows it down to 4E16. Then you can narrow it down further by looking at what other digit-pairs are possible. You might assume it has a palindrome in the middle, but that digits near the end are not palindroms (e.g. it could have ...1001..., but couldn't be 101....102). At some point, you might have enough simple rules like that to just use a computer. Edit: I'm also interested in exactly how he figured it out, though.
- byrneseyeview 18y agoMaybe you'd lop off the last few digits to make it easier: Let A be the highest digit, B be the lowest, and X be the ones in the middle. Find X, A, and B, where 2(X-A*E16+A+BE16-B)=X That way, you can at least get each digit on one side, which might make it a more tractable problem. This is very frustrating! It's easy to narrow it down a whole lot, but it's still going to be intractable.
- shrughes 18y agoSpoiler Alert! The number in general (I said, Spoiler Alert!) can be written as _y_ _x_ where 0 <= x <= 9 and the value of the number is 10y + x. Then it must satisfy for some k 2(10y+x) = (10^k)x + y Simplify that and you can show that 10^k - 2 must be divisible by 19, which means k=17 (or 35, or 53, ...). Once you have that, pick x = 2, 3, or 4 and solve for y, and you have your 18-digit number. (Letting x = 1 gives you an 18-digit numbers whose first digit is 0, which doesn't count.)
- shrughes 18y agoBecause the decimal expansion of 19 repeats after 18 digits. That means all the fractions 1/19 .. 18/19 are just rotations of each other. And 20/19 = 1 + 1/19, which means that 2/19 is a single rotation away. If you've ever noticed that 142857 * [1..6] produces different rotations of the same number and wondered why, you might have thought about this. Dyson had probably already done all the hard work on the problem and knew immediately that the answers were rotations of some repeating fraction, and one where multiplying by 2 gives rotations by 1 digit.
- hhm 18y agoWow thanks!
- nkurz 18y agoThanks for the insight. I wouldn't have noticed any similarity to the multiples of 1/19. Accepting this, though, is there a reason it would have been obvious to him that there was no smaller number? The first such number, by the way, does indeed have 18 digits. I got there by considering every final digit and figuring out how long the string needed to be before the final digit reoccurred with no carry. (edit: oops, just read your explanation below which pretty much answers this)
- shrughes 18y agoWell, since you asked for an obvious reason, and not a completely separate derivation, here it is. The process of rotating a repeating decimal expansion x by 1 digit means adding some integer n and dividing by 10. (n + x) / 10 = 2x Simplifying, n = 19x, i.e. x = n/19. So you know the number has to correspond to a decimal expansion of n/19, which is 18 digits in length.