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Savage Road Signs
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Savage Road Signs
Cross-road optimization - what is the proper way to solve this type of puzzle?Ernie and the CalendarErnie and the Alien InvasionDroning On in CirclesBuilding a cheap roadErnie and the Grand TourErnie and the slice of PiErnie and the Disco Unicorn4x4 Sliding Puzzle with a twistFinding unique number properties
$begingroup$
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.
mathematics combinatorics
asked 8 hours ago
Dark ThunderDark Thunder
818114
$endgroup$
add a comment |
$begingroup$
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.
mathematics combinatorics
asked 8 hours ago
Dark ThunderDark Thunder
818114
$endgroup$
$begingroup$
You mention that this isn't a lateral thinking problem; should I take that to mean that I can't use 6's as 9's and vice versa? (ditto for 2's and 5's, though depending on the font, that's a lot more of a stretch).
$endgroup$
– Gilad M
8 hours ago
$begingroup$
@GiladM yeah, pretend it's a font where 6's and 9's look different
$endgroup$
– Dark Thunder
7 hours ago
add a comment |
$begingroup$
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.
mathematics combinatorics
asked 8 hours ago
Dark ThunderDark Thunder
818114
$endgroup$
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.
mathematics combinatorics
mathematics combinatorics
asked 8 hours ago
Dark ThunderDark Thunder
818114
asked 8 hours ago
Dark ThunderDark Thunder
818114
asked 8 hours ago
Dark ThunderDark Thunder
818114
asked 8 hours ago
Dark ThunderDark Thunder
818114
asked 8 hours ago
Dark ThunderDark Thunder
818114
818114
$begingroup$
You mention that this isn't a lateral thinking problem; should I take that to mean that I can't use 6's as 9's and vice versa? (ditto for 2's and 5's, though depending on the font, that's a lot more of a stretch).
$endgroup$
– Gilad M
8 hours ago
$begingroup$
@GiladM yeah, pretend it's a font where 6's and 9's look different
$endgroup$
– Dark Thunder
7 hours ago
add a comment |
$begingroup$
You mention that this isn't a lateral thinking problem; should I take that to mean that I can't use 6's as 9's and vice versa? (ditto for 2's and 5's, though depending on the font, that's a lot more of a stretch).
$endgroup$
– Gilad M
8 hours ago
$begingroup$
@GiladM yeah, pretend it's a font where 6's and 9's look different
$endgroup$
– Dark Thunder
7 hours ago
$begingroup$
You mention that this isn't a lateral thinking problem; should I take that to mean that I can't use 6's as 9's and vice versa? (ditto for 2's and 5's, though depending on the font, that's a lot more of a stretch).
$endgroup$
– Gilad M
8 hours ago
$begingroup$
You mention that this isn't a lateral thinking problem; should I take that to mean that I can't use 6's as 9's and vice versa? (ditto for 2's and 5's, though depending on the font, that's a lot more of a stretch).
$endgroup$
– Gilad M
8 hours ago
$begingroup$
@GiladM yeah, pretend it's a font where 6's and 9's look different
$endgroup$
– Dark Thunder
7 hours ago
$begingroup$
@GiladM yeah, pretend it's a font where 6's and 9's look different
$endgroup$
– Dark Thunder
7 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered 6 hours ago
Weather VaneWeather Vane
3,8121118
$endgroup$
$begingroup$
It seems to be beneficial to preserve some of the 0s until later on. Even though Gilad M has made an error, with a slight correction, they can still get easily over 600.
$endgroup$
– hexomino
6 hours ago
$begingroup$
@hexomino yes I noticed that: deleting the last two terms beats this. I decided to post something after the false start from another answer (too many zeros) although I feel this could be better.
$endgroup$
– Weather Vane
6 hours ago
$begingroup$
Yeah, sorry, fixed now, I think. Listen, counting is hard. :P
$endgroup$
– Gilad M
6 hours ago
$begingroup$
@hexomino I found a way to improve the answer.
$endgroup$
– Weather Vane
5 hours ago
$begingroup$
The new solution is neat! It uses all 100 digits exactly!
$endgroup$
– Dr Xorile
5 hours ago
|
show 1 more comment
$begingroup$
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.
answered 7 hours ago
Gilad MGilad M
1837
$endgroup$
$begingroup$
There are only 10 of each digit available.
$endgroup$
– Weather Vane
6 hours ago
$begingroup$
...I only used 10 of each digit, I'm pretty sure. Did I miss something? If this is about the first sequence that used an extra 0, I fixed that in my final answer.
$endgroup$
– Gilad M
6 hours ago
$begingroup$
Counting them??
$endgroup$
– Weather Vane
6 hours ago
1
$begingroup$
5 is used 12 times: 159, 259, 357, 507, 527, 547, 566, 586, 605, 625, 645, 665.
$endgroup$
– hexomino
6 hours ago
2
$begingroup$
...which is a shame, because I really liked the "one kilometer short of hell" line :P
$endgroup$
– Gilad M
6 hours ago
|
show 2 more comments
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 6 hours ago
Weather VaneWeather Vane
3,8121118
$endgroup$
$begingroup$
It seems to be beneficial to preserve some of the 0s until later on. Even though Gilad M has made an error, with a slight correction, they can still get easily over 600.
$endgroup$
– hexomino
6 hours ago
$begingroup$
@hexomino yes I noticed that: deleting the last two terms beats this. I decided to post something after the false start from another answer (too many zeros) although I feel this could be better.
$endgroup$
– Weather Vane
6 hours ago
$begingroup$
Yeah, sorry, fixed now, I think. Listen, counting is hard. :P
$endgroup$
– Gilad M
6 hours ago
$begingroup$
@hexomino I found a way to improve the answer.
$endgroup$
– Weather Vane
5 hours ago
$begingroup$
The new solution is neat! It uses all 100 digits exactly!
$endgroup$
– Dr Xorile
5 hours ago
|
show 1 more comment
$begingroup$
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.
answered 6 hours ago
Weather VaneWeather Vane
3,8121118
$endgroup$
$begingroup$
It seems to be beneficial to preserve some of the 0s until later on. Even though Gilad M has made an error, with a slight correction, they can still get easily over 600.
$endgroup$
– hexomino
6 hours ago
$begingroup$
@hexomino yes I noticed that: deleting the last two terms beats this. I decided to post something after the false start from another answer (too many zeros) although I feel this could be better.
$endgroup$
– Weather Vane
6 hours ago
$begingroup$
Yeah, sorry, fixed now, I think. Listen, counting is hard. :P
$endgroup$
– Gilad M
6 hours ago
$begingroup$
@hexomino I found a way to improve the answer.
$endgroup$
– Weather Vane
5 hours ago
$begingroup$
The new solution is neat! It uses all 100 digits exactly!
$endgroup$
– Dr Xorile
5 hours ago
|
show 1 more comment
$begingroup$
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.
answered 6 hours ago
Weather VaneWeather Vane
3,8121118
$endgroup$
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.
answered 6 hours ago
Weather VaneWeather Vane
3,8121118
edited 5 hours ago
answered 6 hours ago
Weather VaneWeather Vane
3,8121118
answered 6 hours ago
Weather VaneWeather Vane
3,8121118
answered 6 hours ago
Weather VaneWeather Vane
3,8121118
3,8121118
$begingroup$
It seems to be beneficial to preserve some of the 0s until later on. Even though Gilad M has made an error, with a slight correction, they can still get easily over 600.
$endgroup$
– hexomino
6 hours ago
$begingroup$
@hexomino yes I noticed that: deleting the last two terms beats this. I decided to post something after the false start from another answer (too many zeros) although I feel this could be better.
$endgroup$
– Weather Vane
6 hours ago
$begingroup$
Yeah, sorry, fixed now, I think. Listen, counting is hard. :P
$endgroup$
– Gilad M
6 hours ago
$begingroup$
@hexomino I found a way to improve the answer.
$endgroup$
– Weather Vane
5 hours ago
$begingroup$
The new solution is neat! It uses all 100 digits exactly!
$endgroup$
– Dr Xorile
5 hours ago
|
show 1 more comment
$begingroup$
It seems to be beneficial to preserve some of the 0s until later on. Even though Gilad M has made an error, with a slight correction, they can still get easily over 600.
$endgroup$
– hexomino
6 hours ago
$begingroup$
@hexomino yes I noticed that: deleting the last two terms beats this. I decided to post something after the false start from another answer (too many zeros) although I feel this could be better.
$endgroup$
– Weather Vane
6 hours ago
$begingroup$
Yeah, sorry, fixed now, I think. Listen, counting is hard. :P
$endgroup$
– Gilad M
6 hours ago
$begingroup$
@hexomino I found a way to improve the answer.
$endgroup$
– Weather Vane
5 hours ago
$begingroup$
The new solution is neat! It uses all 100 digits exactly!
$endgroup$
– Dr Xorile
5 hours ago
$begingroup$
It seems to be beneficial to preserve some of the 0s until later on. Even though Gilad M has made an error, with a slight correction, they can still get easily over 600.
$endgroup$
– hexomino
6 hours ago
$begingroup$
It seems to be beneficial to preserve some of the 0s until later on. Even though Gilad M has made an error, with a slight correction, they can still get easily over 600.
$endgroup$
– hexomino
6 hours ago
$begingroup$
@hexomino yes I noticed that: deleting the last two terms beats this. I decided to post something after the false start from another answer (too many zeros) although I feel this could be better.
$endgroup$
– Weather Vane
6 hours ago
$begingroup$
@hexomino yes I noticed that: deleting the last two terms beats this. I decided to post something after the false start from another answer (too many zeros) although I feel this could be better.
$endgroup$
– Weather Vane
6 hours ago
$begingroup$
Yeah, sorry, fixed now, I think. Listen, counting is hard. :P
$endgroup$
– Gilad M
6 hours ago
$begingroup$
Yeah, sorry, fixed now, I think. Listen, counting is hard. :P
$endgroup$
– Gilad M
6 hours ago
$begingroup$
@hexomino I found a way to improve the answer.
$endgroup$
– Weather Vane
5 hours ago
$begingroup$
@hexomino I found a way to improve the answer.
$endgroup$
– Weather Vane
5 hours ago
$begingroup$
The new solution is neat! It uses all 100 digits exactly!
$endgroup$
– Dr Xorile
5 hours ago
$begingroup$
The new solution is neat! It uses all 100 digits exactly!
$endgroup$
– Dr Xorile
5 hours ago
|
show 1 more comment
$begingroup$
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.
answered 7 hours ago
Gilad MGilad M
1837
$endgroup$
$begingroup$
There are only 10 of each digit available.
$endgroup$
– Weather Vane
6 hours ago
$begingroup$
...I only used 10 of each digit, I'm pretty sure. Did I miss something? If this is about the first sequence that used an extra 0, I fixed that in my final answer.
$endgroup$
– Gilad M
6 hours ago
$begingroup$
Counting them??
$endgroup$
– Weather Vane
6 hours ago
1
$begingroup$
5 is used 12 times: 159, 259, 357, 507, 527, 547, 566, 586, 605, 625, 645, 665.
$endgroup$
– hexomino
6 hours ago
2
$begingroup$
...which is a shame, because I really liked the "one kilometer short of hell" line :P
$endgroup$
– Gilad M
6 hours ago
|
show 2 more comments
$begingroup$
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.
answered 7 hours ago
Gilad MGilad M
1837
$endgroup$
$begingroup$
There are only 10 of each digit available.
$endgroup$
– Weather Vane
6 hours ago
$begingroup$
...I only used 10 of each digit, I'm pretty sure. Did I miss something? If this is about the first sequence that used an extra 0, I fixed that in my final answer.
$endgroup$
– Gilad M
6 hours ago
$begingroup$
Counting them??
$endgroup$
– Weather Vane
6 hours ago
1
$begingroup$
5 is used 12 times: 159, 259, 357, 507, 527, 547, 566, 586, 605, 625, 645, 665.
$endgroup$
– hexomino
6 hours ago
2
$begingroup$
...which is a shame, because I really liked the "one kilometer short of hell" line :P
$endgroup$
– Gilad M
6 hours ago
|
show 2 more comments
$begingroup$
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.
answered 7 hours ago
Gilad MGilad M
1837
$endgroup$
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.
answered 7 hours ago
Gilad MGilad M
1837
edited 6 hours ago
answered 7 hours ago
Gilad MGilad M
1837
answered 7 hours ago
Gilad MGilad M
1837
answered 7 hours ago
Gilad MGilad M
1837
1837
$begingroup$
There are only 10 of each digit available.
$endgroup$
– Weather Vane
6 hours ago
$begingroup$
...I only used 10 of each digit, I'm pretty sure. Did I miss something? If this is about the first sequence that used an extra 0, I fixed that in my final answer.
$endgroup$
– Gilad M
6 hours ago
$begingroup$
Counting them??
$endgroup$
– Weather Vane
6 hours ago
1
$begingroup$
5 is used 12 times: 159, 259, 357, 507, 527, 547, 566, 586, 605, 625, 645, 665.
$endgroup$
– hexomino
6 hours ago
2
$begingroup$
...which is a shame, because I really liked the "one kilometer short of hell" line :P
$endgroup$
– Gilad M
6 hours ago
|
show 2 more comments
$begingroup$
There are only 10 of each digit available.
$endgroup$
– Weather Vane
6 hours ago
$begingroup$
...I only used 10 of each digit, I'm pretty sure. Did I miss something? If this is about the first sequence that used an extra 0, I fixed that in my final answer.
$endgroup$
– Gilad M
6 hours ago
$begingroup$
Counting them??
$endgroup$
– Weather Vane
6 hours ago
1
$begingroup$
5 is used 12 times: 159, 259, 357, 507, 527, 547, 566, 586, 605, 625, 645, 665.
$endgroup$
– hexomino
6 hours ago
2
$begingroup$
...which is a shame, because I really liked the "one kilometer short of hell" line :P
$endgroup$
– Gilad M
6 hours ago
$begingroup$
There are only 10 of each digit available.
$endgroup$
– Weather Vane
6 hours ago
$begingroup$
There are only 10 of each digit available.
$endgroup$
– Weather Vane
6 hours ago
$begingroup$
...I only used 10 of each digit, I'm pretty sure. Did I miss something? If this is about the first sequence that used an extra 0, I fixed that in my final answer.
$endgroup$
– Gilad M
6 hours ago
$begingroup$
...I only used 10 of each digit, I'm pretty sure. Did I miss something? If this is about the first sequence that used an extra 0, I fixed that in my final answer.
$endgroup$
– Gilad M
6 hours ago
$begingroup$
Counting them??
$endgroup$
– Weather Vane
6 hours ago
$begingroup$
Counting them??
$endgroup$
– Weather Vane
6 hours ago
1
1
$begingroup$
5 is used 12 times: 159, 259, 357, 507, 527, 547, 566, 586, 605, 625, 645, 665.
$endgroup$
– hexomino
6 hours ago
$begingroup$
5 is used 12 times: 159, 259, 357, 507, 527, 547, 566, 586, 605, 625, 645, 665.
$endgroup$
– hexomino
6 hours ago
2
2
$begingroup$
...which is a shame, because I really liked the "one kilometer short of hell" line :P
$endgroup$
– Gilad M
6 hours ago
$begingroup$
...which is a shame, because I really liked the "one kilometer short of hell" line :P
$endgroup$
– Gilad M
6 hours ago
|
show 2 more comments
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$begingroup$
You mention that this isn't a lateral thinking problem; should I take that to mean that I can't use 6's as 9's and vice versa? (ditto for 2's and 5's, though depending on the font, that's a lot more of a stretch).
$endgroup$
– Gilad M
8 hours ago
$begingroup$
@GiladM yeah, pretend it's a font where 6's and 9's look different
$endgroup$
– Dark Thunder
7 hours ago