New Math Challenge (may or may not be simple)

[Gleb Gamow]

http://www.census.gov/popclock/







At what time will ALL four (4) bars be solid gray?

[vlajce]

If I solve this can I get a tap on a back? Just kidding.. here is the answer.

Its not possible to get them all gray but they can collide at their maximum/minimum. Also animation doesnt spends 1s on gray so... min/max

Eevery 8s goes a birth and 12s death

We need a real/whole number to get max min. Lets focus on death bar. From begining it needs 3*12 to get full bar with birth and that means that when death bar goes 3 times full animation it will collide with max bar with birth.

Now we 3x death and match with One international migrant.

3*12/33 = 36/33 = 12/11... again we 11*12/11 and we get that 3*11*12 is the time when death birth and One international migrant collide..

Net gain of one person every

.. same thing and we add 13*(formla above)

13*396=5148s that all bars reach (max/min) or the GRAY area

edit: This is example when it oscilates (amplitude) but knowing the exact time i need to count exact time for every of their pos. so i can tell you when it will accure. If you really want it tell me.

[Gleb Gamow]

If I solve this can I get a tap on a back? Just kidding.. here is the answer.

Its not possible to get them all gray but they can collide at their maximum/minimum. Also animation doesnt spends 1s on gray so... min/max

Eevery 8s goes a birth and 12s death

We need a real/whole number to get max min. Lets focus on death bar. From begining it needs 3*12 to get full bar with birth and that means that when death bar goes 3 times full animation it will collide with max bar with birth.

Now we 3x death and match with One international migrant.

3*12/33 = 36/33 = 12/11... again we 11*12/11 and we get that 3*11*12 is the time when death birth and One international migrant collide..

Net gain of one person every

.. same thing and we add 13*(formla above)

13*396=5148s that all bars reach (max/min) or the GRAY area

edit: This is example when it oscilates (amplitude) but knowing the exact time i need to count exact time for every of their pos. so i can tell you when it will accure. If you really want it tell me.

I thought that the answer would be every 8 X 12 X 13 X 33 = 41,184 seconds (686.4 minutes/11.44 hours).

It's true that when first started, three bars would be gray, with the death bar being 100% blue. Come to think of it, this may be a tad more difficult than I originally envisioned, but surely there's a unique solution.

Thanks for the input, bud. Let's tackle it some more.

[High Plains Drifter]

Oh gosh, I just about had it until I thought of the Asian thread.  I'll be back later 

[jeffthebaker]

If I solve this can I get a tap on a back? Just kidding.. here is the answer.

Its not possible to get them all gray but they can collide at their maximum/minimum. Also animation doesnt spends 1s on gray so... min/max

Eevery 8s goes a birth and 12s death

We need a real/whole number to get max min. Lets focus on death bar. From begining it needs 3*12 to get full bar with birth and that means that when death bar goes 3 times full animation it will collide with max bar with birth.

Now we 3x death and match with One international migrant.

3*12/33 = 36/33 = 12/11... again we 11*12/11 and we get that 3*11*12 is the time when death birth and One international migrant collide..

Net gain of one person every

.. same thing and we add 13*(formla above)

13*396=5148s that all bars reach (max/min) or the GRAY area

edit: This is example when it oscilates (amplitude) but knowing the exact time i need to count exact time for every of their pos. so i can tell you when it will accure. If you really want it tell me.

I thought that the answer would be every 8 X 12 X 13 X 33 = 41,184 seconds (686.4 minutes/11.44 hours).

It's true that when first started, three bars would be gray, with the death bar being 100% blue. Come to think of it, this may be a tad more difficult than I originally envisioned, but surely there's a unique solution.

Thanks for the input, bud. Let's tackle it some more.

It's actually quite simple, just look for the least common multiple of the four numbers. In other words, the smallest number that is divisible by 8, 12, 13, and 33. I'm not sure what the correct formula would be for this, but just by a bit of multiplication I found that all bars will turn gray every 3432 seconds, however it is possible that they could all become gray at a more frequent rate.

[Gleb Gamow]

If I solve this can I get a tap on a back? Just kidding.. here is the answer.

Its not possible to get them all gray but they can collide at their maximum/minimum. Also animation doesnt spends 1s on gray so... min/max

Eevery 8s goes a birth and 12s death

We need a real/whole number to get max min. Lets focus on death bar. From begining it needs 3*12 to get full bar with birth and that means that when death bar goes 3 times full animation it will collide with max bar with birth.

Now we 3x death and match with One international migrant.

3*12/33 = 36/33 = 12/11... again we 11*12/11 and we get that 3*11*12 is the time when death birth and One international migrant collide..

Net gain of one person every

.. same thing and we add 13*(formla above)

13*396=5148s that all bars reach (max/min) or the GRAY area

edit: This is example when it oscilates (amplitude) but knowing the exact time i need to count exact time for every of their pos. so i can tell you when it will accure. If you really want it tell me.

I thought that the answer would be every 8 X 12 X 13 X 33 = 41,184 seconds (686.4 minutes/11.44 hours).

It's true that when first started, three bars would be gray, with the death bar being 100% blue. Come to think of it, this may be a tad more difficult than I originally envisioned, but surely there's a unique solution.

Thanks for the input, bud. Let's tackle it some more.

It's actually quite simple, just look for the least common multiple of the four numbers. In other words, the smallest number that is divisible by 8, 12, 13, and 33. I'm not sure what the correct formula would be for this, but just by a bit of multiplication I found that all bars will turn gray every 3432 seconds, however it is possible that they could all become gray at a more frequent rate.

https://en.wikipedia.org/wiki/Least_common_multiple#A_method_using_a_table

Looks like 3,432 [secs] is the LCM: 2*2*2*3*11*13

Now all we need to do is to calculate the time for the next (or any) time all four bars are gray (or all Blue).

[vlajce]

however it is possible that they could all become gray at a more frequent rate.

You mean on long time period or in that one oscillation period?

[Gleb Gamow]

however it is possible that they could all become gray at a more frequent rate.

You mean on long time period or in that one oscillation period?

I assumed he meant the former, with the oscillation period being fixed and now established (I believe).

[Gleb Gamow]

I guess this problem is a lot harder than it seems on the surface. This challenge may be a good candidate for one of them YouTube math gurus to tackle since none of us (mostly yous) math whizzes have yet to crack it.

If there's no solution, then a proof should be presented depicting such, else a Q.E.D. could be had.

To date, all we have is the oscillation period for when three bars are solid gray and one bar is solid blue. We've yet to determine the oscillation period for when ALL bars are gray (or blue).

[catena5260]

at the moment someone hacks the site just for fun and make all bars full

[Gleb Gamow]

at the moment someone hacks the site just for fun and make all bars full

http://www.census.gov/popclock/

WTF! They're gone! It's probably that Kugelis Fest goer.

[Gleb Gamow]

Apologies for the bump, but still seeking a solution. Thanks in advance, all.