DILR guide, section 11

Process sets and cubes

Process sets reward simulation: write the state after every step in a table and watch for a repeat. Cube sets reward the counting formulas. Both usually go on the skip list in the first pass, but their simple versions are quick marks.

Worked set 13, medium. Attempt it in 15 minutes first.

Process set, a token game

Four friends, A, B, C and D, start with 9, 7, 5 and 3 tokens. In each round, the person with the most tokens gives one token to each of the other three. If two people tie for the most, the one who comes first alphabetically gives.

Questions

  1. Who has the most tokens after round 3?
  2. (TITA) How many tokens does A have after round 5?
  3. After which round does a distribution first repeat an earlier one?
  4. (TITA) How many tokens does D have after round 10?
Show solution

Simulate in a table; the total stays at 24, a free check on every row.

After roundABCDWho gave
Start9753—
16864A
27575B
34686A (tied with C)
45757C
56468B (tied with D)
67575D

Answers. (1) C, with 8. (2) 6. (3) Round 6 repeats round 2. (4) 5: from round 2 the pattern repeats every 4 rounds, so round 10 matches rounds 6 and 2.

Takeaway. Once any state repeats, everything after it cycles. Find the cycle length and you can answer “after round 50” in seconds.

8 minutes, timed

Practice set 13: process

Three jars, X, Y and Z, hold 12, 6 and 2 marbles. In each step, the jar with the most marbles gives 2 marbles to the jar with the fewest. Ties for most or for fewest go to the earlier letter.

Questions

  1. (TITA) How many marbles are in Y after step 3?
  2. After step 2, which two jars hold equal numbers?
  3. (TITA) How many marbles are in X after step 20?
Show answers

(1) 8. (2) Y and Z, with 6 each. (3) 8. The states run: step 1 (10, 6, 4), step 2 (8, 6, 6), step 3 (6, 8, 6), step 4 (8, 6, 6), step 5 (6, 8, 6). From step 3 the jars alternate, so every even step from 4 onwards gives X 8.

Worked set 14, easy to medium. Attempt it in 15 minutes first.

Painted cube

A 5 × 5 × 5 cube is painted red on all six faces and cut into 125 unit cubes.

Questions

  1. (TITA) How many unit cubes have exactly two red faces?
  2. (TITA) How many have exactly one red face?
  3. (TITA) How many have no red face?
  4. (TITA) The top layer of 25 cubes is removed, and the remaining 5 × 5 × 4 block is painted blue on all its outer faces. How many of its 100 cubes have no paint at all?
Show solution

With n = 5: three red faces 8 (the corners); two faces 12 × 3 = 36 (edges without corners); one face 6 × 3² = 54; none 3³ = 27. Check: 8 + 36 + 54 + 27 = 125. For question 4, the unpainted cubes are the new block’s interior: 3 along the length, 3 along the width and 2 in height (the block is only 4 layers tall), so 3 × 3 × 2. All of them were inside the original cube, so they never had red either.

Answers. (1) 36. (2) 54. (3) 27. (4) 18.

Takeaway. For any a × b × c block painted on the outside: 8 corner cubes have three painted faces and (a − 2)(b − 2)(c − 2) have none. Learn those two and get the rest by subtraction.

6 minutes, timed

Practice set 14: cubes

Questions

  1. (TITA) A 4 × 4 × 4 cube painted on all faces is cut into 64 unit cubes. How many have exactly two painted faces?
  2. (TITA) In the same cube, how many have exactly one painted face?
  3. (TITA) A 3 × 4 × 5 cuboid is painted on all faces and cut into 60 unit cubes. How many have no paint?
  4. (TITA) In that cuboid, how many unit cubes have exactly three painted faces?
Show answers

(1) 24: 12 × 2. (2) 24: 6 × 2². (3) 6: 1 × 2 × 3. (4) 8: the corners.