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There is a highway that starts in the city of Savage. You must must place distance marker signs on this highway for the outgoing traffic. According to highway code, there must be a distance marker sign at least every 20 km, and every distance marker sign must be labelled with its distance from the start (the city of Savage).

Normally this isn't a problem for you but there is a snag. Your sign printing machine is broken and your only back-up plan is to steal a pack of stickers from your daughter. This pack of stickers contains ten of each digit, 0 through 9 (that's 100 total stickers). As luck would have it, using these stickers isn't against code and you have plenty of blank signs to put them on.

What is the furthest distance marker sign you can place without breaking highway code?

Note: This isn't meant to be a lateral-thinking question. Use no more than 10 of each digit across all signs, no more than a gap of 20 between signs, the answer is the last sign you place. You do not need leading zeros, so "004" can just be "4".

I do not claim to have the optimal answer (but it's probably pretty good). I did not use a computer program, but they are allowed. I guess if you want to answer you should also list all of your signs? Assuming somebody beats me I'll give out the checkmark after a couple of days.

Edit: my improved answer is

688 km


Stepping by 19 or 20 km gives


20 40 60 80 99 118 137 157 177 197 217 237 256 276 295 314 334 353 372 392 411 430 450 470 490 509 529 549 569 588 608 628 648 668 688

My (previous) answer is

488 km.


20 40 60 80 100 120 140 160 180 199 219 239 259 279 299 319 338 358 378 398 418 438 457 477 488


The signs go every 20km until I run out of 0s.

The next is after 19km, and again every 20 km until I run out of 9s.

The next is after 19km, and again every 20 km until I run out of 8s.

The next is after 19km, and again every 20 km until the furthest sign I can make within 20 km

— there are no 8s (48x) or 9s (49x) or 0s (50x) left.

646 km

I'm not sure this is the best answer, but I think it's close, and at the very least some decent headway for someone who didn't feel like writing a script to solve it.

To start:

I came up with a hard limit: what if you could just put signs every 20 km without worrying which digits were repeated? Then you'd spend 10 digits getting up to 100 km, and another 15 for each 100 km past that. That gets us to 700 km at the very most. We know the answer's not getting past that. (Actually you could probably get to 710 or 720 with some shenanigans involving high numbers with 1 fewer digit, like 9 and 99, but I digress. My answer doesn't really care about off-by-one errors like that).

So now that we know that

we have no chance of using lots of 7's, 8's, 9's, and 0's in a row, we realize that these digits are a lot less valuable to us than the 1's through 5's that we'll need when we get a few hundred kilometers out.

Next,

I assumed the answer was close to optimal, so 600-something. I'll need 5, maybe 6 each of digits 1 through 5 just for hundreds places. If I'm climbing by just under 20 at a time, I'll need a bunch of odd digits for tens places, and then at some point I'll run out and need to switch to evens. Every switch is a loss of efficiency, so I'll try to only do it once. The rest of the digits will be used for the ones places. That's the game plan.

Following that plan, here's what I came up with:

20, 40, 60, 80, 100, 120, 133, 141, 159, 179, 199, 219, 239, 259, 279, 299, 317, 337, 357, 373, 388, 408, 428, 448, 468, 488, 507, 527, 547, 567, 586, 606, 626, 646.

I could keep rearranging things, but

given that all I have left after this sequence are two 5's, and we're pretty damn close to the fundamental maximum of 700, I think this is close enough. If this isn't the answer, I'm pretty sure that the real answer is ~670, but not much more than that.