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I was hoping someone could help me express this problem as a system of congruences: 3 trucks carrying deliveries for the same destination leave the warehouse at 5pm. Each truck travels between 5am and 5pm each day, and the drivers rest for the rest of the time. The first one to arrive is truck A, a few days later truck B arrives, and a few days later still truck C arrives. Truck A traveled 30km a day and reached its destination at 5pm, truck B traveled 24km a day (24/12 km/hour) and reached its destination at 2pm. Finally, truck C traveled 18km a day (18/12km hour) and reached its destination at 9am. What is the minimum distance between the warehouse and the trucks' destination?

I was confident that the system would be: mod(d,30)=0; mod(d,24)=9; mod(d,18)=4;

but I assume this can't be the case because the problem is presented as being solvable by the Chinese Remainder Theorem, and since 30,24, and 18 aren't coprime, and the condition for applying the theorem for non-comprime divisors isn't fulfilled, it doesn't seem to me that this system can be solved in that way.

Please do forgive me if I missed something obvious, and I appreciate any help you can provide. Thanks!

Bill Dubuque
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Alexandre Ramos
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  • The Chinese Remainder Theorem can be used when the moduli are not relatively prime, you simply need to confirm that the congruences are compatible. You can't have $n\equiv 0 \pmod 4$ and $n\equiv 1 \pmod 2$, for instance. – lulu May 31 '23 at 17:29
  • Should note that the problem seems odd. $24$ km a day? Granted, it's a math problem, but it seems odd to introduce plausible sounding context but then use unreasonable numbers. Also you say that they travel between $5$ am and $5$ pm, but that they head out at $5$ pm. So I guess they can travel after $5$ pm? – lulu May 31 '23 at 17:30
  • @lulu Thanks you for replying! Are you aware of any way to check if the congruences are compatible? About the context, as you could probably tell I'm not a native english speaker, and the problem wasn't originally in english, so I changed the context into one I could more easily translate, I apolagize for any confusion I may have caused! – Alexandre Ramos May 31 '23 at 17:40
  • A good general approach is to split each congruence into prime power components. Thus $a\equiv 5\pmod {24}$ can be rewritten as $a\equiv 5\pmod 8$ AND $a\equiv 5\equiv 2\pmod 3$. Then you can look at any instances where powers of the same prime appear and check that the relevant congruences can be met simultaneously. – lulu May 31 '23 at 17:43
  • See e.g. here for CRT with noncoprime moduli. – Bill Dubuque May 31 '23 at 18:18

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