## Chicken Nuggets

Chicken nuggets come in packets of $6,9$, or $20$. What is the largest number of nuggets that you cannot buy when combining various packets?
Source: NCTM Mathematics Teacher 2006

SOLUTION
Consider the numbers $36,38$, and $40$
$36=9+9+9+9$
$38=9+9+20$
$40=20+20$
All even numbers greater than or equal to $42$ can be formed by adding $6$s to these $3$ numbers. For example,
$42=36+6$
$44=38+6$
$46=40+6$
$48=36+6+6$
$50=38+6+6$
$52=40+6+6$
$54=36+6+6+6$
$56=38+6+6+6$
$58=40+6+6+6$
$\cdots$
Similarly, all odd numbers greater than or equal to $45$ can be formed by adding $9$s. For example,
$45=36+9$
$47=38+9$
$49=40+9$
$51=42+9$
$53=44+9$
$55=46+9$
$57=48+9$
$59=50+9$
$\cdots$
Some of the numbers that cannot be formed from $6,9$, or $20$ because they are either not multiple of $3$ or $20$ subtracted from them are not multiple of $3$
$10,11,13,14,16,17,19,22,23,25,28,31,34,37$, and $43$
$43$ is not a multiple of $3$
$43-20=23$ is not a multiple of $3$
$23-20=3$ is too small
$43$ is the largest number of nuggets that you cannot buy when combining various packets.

Answer: $43$

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## About mvtrinh

Retired high school math teacher.
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