Posers and Puzzles
10 Mar 22
- Joined
- 18 Jan 07
- Moves
- 12477
@bigdogg saidI'm going to spoiler this whole answer, except for a few hints.
There is an 11-car train. The sum of passengers for any 3 consecutive cars is 99. There are 381 total passengers.
How many passengers are in car #9?
The first three carriages together must have 99 passengers total. Let's call them A, B and C, so A+B+C=99.
Now
the second, third and fourth carriages must also have 99 passengers. Call the fourth carriage D, we get B+C+D=99=A+B+C, therefore D=A. Continuing this way, the fifth carriage must have B passengers, and so on.
The whole train is ABCABCABCAB
, and carriage 9 is C.
The train as a whole has 4A+4B+3C=381 passengers, but we also know that 3A+3B+3C=3×99=297. Subtracting these, we get A+B=381-297=84.
Subtracting that again from A+B+C=99, we get C
, the 9th carriage, The train as a whole has 4A+4B+3C=381 passengers, but we also know that 3A+3B+3C=3×99=297. Subtracting these, we get A+B=381-297=84.
Subtracting that again from A+B+C=99, we get C
=15, as are the 3rd and 6th carriages.
We cannot determine the numbers for the eight
other carriages, except that they must add up, for each adjoining pair, to 84
, and each pair in the same way
.
10 Mar 22
@shallow-blue saidSOLV'D
I'm going to spoiler this whole answer, except for a few hints.
The first three carriages together must have 99 passengers total. Let's call them A, B and C, so A+B+C=99.
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10 Mar 22
@shallow-blue saidI enjoyed it as well. Takes some thought, but not overly difficult.
It was a nice one.
