Problem
Parity bounds the fewest three-coin flips
Five coins lie picture side up. One turn means flipping exactly three of the five together. All five must end number side up. Find the fewest turns needed.
Reasoning · Grade 6-2 Swapping and Flipping
Your answer
Watch out Two turns seems plausible, but each turn flips the parity — it must be odd.
How to solve
Strategy Look for a Pattern — Instead of chasing which particular coins get flipped, watch a single number: how many coins are currently showing a number. Listing the only four things a turn can do to that count reveals a pattern that never breaks -- the count always jumps by an odd amount, so it flips between even and odd on every single turn. That settles half the problem instantly, because 0 is even and 5 is odd, so the number of turns has to be odd. Then only the small candidates 1 and 3 are left to guess and check, and five real coins (or five coins drawn on paper) are the quickest way to test them.
1STEP 1
Track one number: how many coins show a number
The count must go from 0 to 5.
0 ⟶ 5
Laying five real coins out and just counting the number sides after each turn turns a five-coin puzzle into a single number to watch, and a fourth grader can keep one number in their head far more reliably than five separate faces.
4.OA.C.5Create A Physical Representation2STEP 2
List the only four things one turn can do to that count
One turn moves the count by 3 or by 1.
(3-t)-t = 3-2t → +3, +1, -1, -3
There are only four ways three chosen coins can be split between number sides and picture sides, so a complete list is four short lines -- and a complete list is what makes the next step airtight.
3.OA.D.8Make A Systematic List3STEP 3
Notice the pattern: every turn switches even and odd
So each turn flips odd and even.
even → odd → even → odd → …
Even-and-odd is exactly the kind of thing you can decide by pairing objects up and seeing whether one is left over, and the fact that odd steps always break the pairing is what makes the switching unavoidable.
2.OA.C.3Look For A PatternEvery turn switches the count of number-side coins between even and odd, so the parity is fixed by how many turns have been made.
Why?
Each turn changes that count by an odd amount, and adding an odd amount always swaps even with odd.
🧱Even and odd behave predictablyTwice any whole number is even and adding an odd amount always swaps even with odd, so a total's parity is settled by its parts before anything is computed.
Why?
There are only four ways one turn can fall, and every one of them changes the count by an odd amount, so no case escapes the rule.
🧱Disjoint cases addSplit the possibilities so no case overlaps another and none is left out, and the case counts simply add.
4STEP 4
Read off which turn counts are even possible
Going 0 to 5 needs an odd number of turns.
This is the same even-and-odd reasoning as knowing you can never make an odd handful out of an even one by adding pairs; here every turn adds an odd amount, so the parity flips like a light switch.
2.OA.C.3Look For A Pattern5STEP 5
Guess and check the smallest odd number, 1
One turn reaches only 3, so it falls short.
0 + 3 = 3 ≠ 5
There is only one odd number below 3, so checking it is a single quick test rather than a search.
3.OA.D.8Guess And Check6STEP 6
Show that 3 turns really work
Three turns give 0, 3, 2, 5.
0 → 3 → 2 → 5
Writing the whole row out after each turn makes every claim checkable coin by coin, and the counts 0, 3, 2, 5 alternate even, odd, even, odd exactly as the pattern predicted.
4.OA.C.5Make A Systematic List7STEP 7
Put the two halves together
So the fewest is 3 turns.
Neither half alone is enough: the parity argument alone would leave open whether 3 is reachable, and the example alone would leave open whether 1 or 2 might sneak through.
2.OA.C.3Look For A PatternAnswer
3 turns
0 → 3 → 2 → 5
The answer is a count of turns, so a whole number is right, and 3 is comfortably small for a five-coin puzzle. Every count along the way -- 0, 3, 2, 5 -- stays between 0 and 5, as it must with only five coins, and the counts alternate even, odd, even, odd, matching the parity rule found before any coins were flipped. A useful sanity check on the parity claim: try to reach 5 in 2 turns. From all pictures the first turn must land on exactly 3 numbers, and from 3 numbers (with only 2 pictures left to choose from) the second turn can only land on 4, 2 or 0 -- all even, never 5. Notice also that the answer would change if there were 4 coins instead of 5: the target 4 is even, so an even number of turns would be needed. The parity of the target really is what drives the answer.
Takeaway
Flipping three coins always changes the number of number-sides by an odd amount, so even and odd take turns -- and that alone tells you the answer must be an odd number of flips.
- Track one number: how many coins show a number
- List the only four things one turn can do to that count
- Notice the pattern: every turn switches even and odd
- Read off which turn counts are even possible
- Guess and check the smallest odd number, 1
- Show that 3 turns really work
- Put the two halves together