Problem 13 worked answer

Lakitu’s Hidden Boost Setting

Designed for Grade 3 · Uses Middle Grade 3 math · About 13–30 minutes

Earn one jewelMaster-Cut ZirconLevel 5

At a glance

Part 1: One Mushroom moves Mario 7 checkpoints.
Part 2: Checkpoint 17 is unreachable, while every checkpoint 18 or greater is reachable.
Part 3: The equations and number-line jumps below count every turbo and Mushroom used in the proof.

1. Prove the Mushroom setting

The hidden whole-number setting must be 5, 6, 7, 8, 9, or 10.

It cannot be 6, 8, or 10. Those moves and the 4-checkpoint turbo are all even. Sums of even moves are always even, so infinitely many odd checkpoints after 17 would remain unreachable. Then 17 could not be the greatest unreachable checkpoint.
It cannot be 5 because $5 + 3 \times 4 = 17$ . The jumps are $0 \to 5 \to 9 \to 13 \to 17$ , so 17 would be reachable.
It cannot be 9 because $9 + 2 \times 4 = 17$ . The jumps are $0 \to 9 \to 13 \to 17$ , so 17 would again be reachable.

The only allowed setting left is 7 checkpoints.

2. Prove 17 is the last unreachable checkpoint

First show that 17 cannot be made from moves of 4 and 7. Only zero, one, or two Mushrooms need to be checked before passing 17:

With no Mushrooms, multiples of 4 go from 16 to 20 without hitting 17.
With one Mushroom, $7 + 2 \times 4 = 15$ and $7 + 3 \times 4 = 19$ .
With two Mushrooms, $2 \times 7 = 14$ and $2 \times 7 + 4 = 18$ .
Three Mushrooms already give $3 \times 7 = 21$ .

These cases cover every possible number of Mushrooms before reaching or passing 17, so 17 is unreachable.

Now show four consecutive reachable checkpoints:

$18 = 2 \times 7 + 4$
$19 = 7 + 3 \times 4$
$20 = 5 \times 4$
$21 = 3 \times 7$

Adding one more 4-checkpoint turbo reaches 22, 23, 24, and 25. Repeating that addition covers every later block of four checkpoints. Therefore every checkpoint greater than 17 is reachable.

3. Show the jumps and count the boosts

18: $0 \to 7 \to 14 \to 18$ , counted by $2 \times 7 + 4 = 18$ .
19: $0 \to 7 \to 11 \to 15 \to 19$ , counted by $7 + 3 \times 4 = 19$ .
20: $0 \to 4 \to 8 \to 12 \to 16 \to 20$ , counted by $5 \times 4 = 20$ .
21: $0 \to 7 \to 14 \to 21$ , counted by $3 \times 7 = 21$ .

The order of 4s and 7s may change, but the landing checkpoint depends only on their sum.

Check

Settings 5 through 10 have all been considered, and only 7 survives. The proof separately establishes both requirements in Lakitu’s statement: 17 itself is unreachable, and every larger checkpoint is reachable.

Technical fit and rating

MJ 3.3 · C5 · W3 Olympiad challenge · Heavy workload

Number and operations

How the rating works →

mathjewels.com/answers/problems/13/