Reasoning · Grade 5-2 Parity

Problem

Flip count parity pins down hidden faces

Three coins start with all three heads up. One coin at a time, any coin, is turned over. After 11 turns coins 1 and 3 show heads; in a second run, after 20 turns coin 1 shows tails. Work out which face the hidden coins show.
LIBERTY LIBERTY LIBERTY (1) LIBERTY LIBERTY LIBERTY LIBERTY LIBERTY (2) LIBERTY LIBERTY LIBERTY QUARTER
Your answer
How to solve
Strategy Look for a Pattern — There are far too many possible turn orders to list — 11 turns on 3 coins already gives thousands of them — so listing cases is hopeless. Instead I look for the one thing that never changes no matter how the turns are shuffled: what a single coin shows depends only on whether IT was turned an odd or an even number of times, never on when. That turns the whole puzzle into a statement about 3 whole numbers that add up to 11 or to 20, and the odd/even pattern of a sum settles both parts. Flipping a real coin a few times is enough to see the pattern, and trying it first with 1 coin instead of 3 makes the rule obvious.
1STEP 1

Turn one coin over and over and watch it

Turning one coin makes its face alternate.

0 → H, 1 → T, 2 → H, 3 → T, 4 → H, …
2STEP 2

Record only the counts, not the order

The per-coin counts add to the total.

(turns of coin 1) + (turns of coin 2) + (turns of coin 3) = (total turns)
3STEP 3

State the invariant that does all the work

Even turns leave heads, odd turns leave tails.

count even ⇔ heads, count odd ⇔ tails
4STEP 4

Part (1): 11 turns, coins 1 and 3 are heads

In (1) the hidden coin's count is odd.

(even)_coin 1 + (coin 2) + (even)_coin 3 = 11 → (coin 2) = 11 - (even) = odd
5STEP 5

Part (1): sanity-check with one real sequence

So the second coin shows tails.

2 + 7 + 2 = 11
6STEP 6

Part (2): 20 turns, coin 1 is tails

In (2) the other two counts add to an odd number.

(coin 2) + (coin 3) = 20 - (odd)_coin 1 = odd
7STEP 7

Part (2): read that back as faces

So those two show one of each face.

1 + 18 + 1 = 20 (coin 2 H, coin 3 T), 1 + 19 + 0 = 20 (coin 2 T, coin 3 H)
Answer
tails / one of each
11 − even = odd
Both answers are faces, not numbers, which is what was asked, and each one is forced rather than guessed. The check that matters most is that the answers survive every legal turn sequence: for part (1) the run 2 + 7 + 2 = 11 fits the picture and gives tails, and no run with coin 2 even can total the odd number 11 while coins 1 and 3 stay even. For part (2) the runs 1 + 18 + 1 = 20 and 1 + 19 + 0 = 20 both fit the picture, and in each one the 2 hidden coins disagree — as they must, since coins 2 and 3 have counts adding to an odd number. It is also worth noticing that the two parts have to behave differently: 11 is odd and 20 is even, so an odd number of coins must have changed in (1) and an even number in (2), and the answers match that.
Takeaway

A coin only remembers how many times it was turned, not when — so the odd-or-even of the total number of turns tells you about the coins you cannot even see.

  • Turn one coin over and over and watch it
  • Record only the counts, not the order
  • State the invariant that does all the work
  • Part (1): 11 turns, coins 1 and 3 are heads
  • Part (1): sanity-check with one real sequence
  • Part (2): 20 turns, coin 1 is tails
  • Part (2): read that back as faces