Problem
Repeated shuffle returns in a cycle
Seven cards numbered 1 to 7 start in order. A shuffle slips the right three into the gaps between the left four. The same shuffle repeats. Find the second card from the left after 10 shuffles.
Reasoning · Grade 6-2 Swapping and Flipping
Your answer
Watch out Doing all ten looks necessary, but the row returns every three shuffles.
How to solve
Strategy Look for a Pattern — Doing ten shuffles by hand is ten chances to make a copying mistake, so instead I do the easier related problem -- a few shuffles -- and watch for the row to repeat. Since there are only finitely many ways seven cards can lie, and the shuffle is always the same move, the row is bound to come back to where it started sooner or later. Once I know how many shuffles that takes, the tenth shuffle is found by dividing 10 by that length and keeping the remainder, which is the whole point of a repeating pattern.
1STEP 1
Read the shuffle rule off the picture
One shuffle gives 1, 5, 2, 6, 3, 7, 4.
1,2,3,4,5,6,7 ⟶ 1,5,2,6,3,7,4
Stating the move as 'left cards to the odd places, right cards to the even places' turns a picture full of arrows into one sentence you can apply to any row without looking back at the figure.
4.OA.C.5Draw A Diagram2STEP 2
Do the second shuffle
Two shuffles give 1, 3, 5, 7, 2, 4, 6.
1,5,2,6,3,7,4 ⟶ 1,3,5,7,2,4,6
The rule is applied to the row as it is now, not to the original row, and writing out the seven places one at a time is the safest way to keep that straight.
4.OA.C.5Make A Systematic List3STEP 3
Do the third shuffle and spot the repeat
Three shuffles return the start.
1,3,5,7,2,4,6 ⟶ 1,2,3,4,5,6,7
Once a repeating move brings a position back to the start, it must loop forever, because from the same row the same move always produces the same next row.
4.OA.C.5Look For A PatternThe third shuffle brings back an arrangement already seen, so the shuffles must repeat in a loop from there on.
Why?
There are only finitely many arrangements the cards can take, so an arrangement is bound to come round a second time.
🧱More items than boxes forces a shared boxn boxes can hold at most n items one apiece, so item number n plus one has nowhere new to go and must share.
Why?
Once an arrangement repeats, the same shuffle produces the same next one, so everything after it repeats too.
🧱A cycle returns after a full periodA fixed repeating cycle of length p comes back to the same position every p steps.
4STEP 4
Divide 10 by the length of the loop
10 divided by 3 leaves 1.
10 ÷ 3 = 3 remainder 1
This is exactly the same trick as working out a day of the week: peel off whole loops and only the remainder matters.
4.NBT.B.6Look For A Pattern5STEP 5
Read off the second card
The second card after one shuffle is 5.
1, 5, 2, 6, 3, 7, 4
Second from the left means one card in from the end, so the count is short enough to do with a finger and check twice.
4.OA.C.5Make A Systematic ListAnswer
5
10 ÷ 3 = 3 R 1
The answer has to be one of the seven card numbers, and 5 is. It cannot be 1: the leftmost card never moves, so the 1 stays in first place forever and can never be the second card. The remainder step is worth checking against the loop directly -- after 3, 6 and 9 shuffles the row is back to 1, 2, 3, 4, 5, 6, 7, so the tenth shuffle acts on the starting row and must give 1, 5, 2, 6, 3, 7, 4, which is the picture's own bottom row. A classic slip here is to use 10 ÷ 3 = 3 remainder 1 but then read the row after 3 shuffles instead of after 1; the check above rules that out, since after 3 shuffles the second card would be 2, and 10 is plainly not a multiple of 3.
Takeaway
Shuffle a few times until the row comes back to the start, then divide 10 by the length of that loop -- only the remainder 1 decides the answer.
- Read the shuffle rule off the picture
- Do the second shuffle
- Do the third shuffle and spot the repeat
- Divide 10 by the length of the loop
- Read off the second card