The Six-Loop Paper Chain: Who Is Right About Loop 3?
At Friday enrichment, Maya's group gets a direction card for building a six-link paper chain. The card lists six rules: cut six strips; join each new loop to the one before it; make loop 1 red; never put two loops of the same color side by side; make the last loop yellow; and use exactly two blue loops, exactly one green loop, and red for every other loop. The group has finished loop 1 (red) and loop 2 (blue). Four minutes remain. Jordan says, 'Loop 3 must be red, because the card says red for everything else.' Priya answers, 'No - loop 3 could be green.' The teacher refuses to settle it. She says only that all six rules must be true at the same time, and that the group must be able to show one complete chain that works.
1. Which colors are legal for loop 3, and which are not? Judge Jordan's and Priya's claims separately.
Read the worked answer
Legal: red or green. Illegal: blue and yellow.
- Loop 3 touches loop 2 and loop 4. Loop 2 is already blue, so the side-by-side rule forbids blue at loop 3.
- The yellow loop is pinned to loop 6, so yellow cannot be used at loop 3.
- Red is legal at loop 3 even though loop 1 is red, because loops 1 and 3 are not side by side - loop 2 sits between them. Students who say 'red is used at loop 1, so red is used up' are confusing 'used' with 'touching.'
So Jordan is wrong on the word 'must': nothing forces red. Priya is right but incomplete: green is allowed, and so is red. The honest answer is a set, not a single color, and the group should not let Jordan's confident tone decide it.
2. Build one complete six-loop chain that obeys every rule, then change exactly one loop so the chain breaks only the side-by-side rule and not the color counts.
Read the worked answer
A working chain: red, blue, red, blue, green, yellow.
Check it against the card: two reds (loops 1 and 3), two blues (loops 2 and 4, not touching), one green (loop 5), yellow last. No two neighbors match.
Single-rule break: red, blue, blue, red, green, yellow. The color totals are still correct - two red, two blue, one green, one yellow - and the yellow is still last. The only rule violated is the side-by-side rule, at loops 2 and 3.
That last point is the skill being tested. If you change a loop and the chain breaks three rules at once, you have not tested anything: you cannot tell which rule your change was aimed at. A careful solver changes one thing, then re-checks all six rules against the new chain before announcing an error.
3. The group wants the green loop as late as possible. What is the latest legal position, and what does that force the other loops to be?
Read the worked answer
Latest legal position: loop 5.
Loop 6 is yellow, so green cannot go there. Loop 5 works: red, blue, red, blue, green, yellow satisfies all six rules, so the latest position is at least 5, and 6 is impossible, so it is exactly 5.
Putting green at loop 5 also forces the rest. Loops 3 and 4 must be red and blue in some order, because all that remains after loops 1, 2, 5, 6 is one red and one blue. Loop 3 cannot be blue (loop 2 is blue), so loop 3 is red and loop 4 is blue. One arrangement only: R, B, R, B, G, Y.
Notice what made this solvable: pinning the latest position first, then letting the leftovers fall where they must. Trying to place green first and hoping a chain appears would take far longer than the four minutes the group had.