2023 AMC 12B Problem 16

Below is the professionally curated solution for Problem 16 of the 2023 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 12B solutions, or check the answer key.

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Concepts:Chicken McNugget Theorem

Difficulty rating: 1660

16.

In Coinland, there are three types of coins, each worth $6,$ $10,$ and $15.$ What is the sum of the digits of the maximum amount of money that is impossible to have?

$8$

$10$

$7$

$11$

$9$

Solution:

The amounts $30,31,32,33,34,35$ are all attainable (for instance $30=6\cdot 5,$ $31=6+10+15,$ $32=6\cdot 2+10\cdot 2,$ $33=6\cdot 3+15,$ $34=6\cdot 4+10,$ $35=10+10+15$ ). Adding $6$ 's then reaches every larger amount. Checking below, $29$ is impossible, since $29-6,\ 29-10,\ 29-15$ are all impossible. So the largest impossible amount is $29,$ whose digit sum is $2+9=11.$

Thus, the correct answer is D.

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