### Instruction: Dave's sister baked $3$ dozen pies of which a third contained chocolate, a quarter contained marshmallows, a sixth contained cayenne, and one twelfth contained salted soy nuts.  What is the smallest possible number of pies that had none of these ingredients?### Response: I need to find the fraction of pies that had none of these ingredients, and then multiply it by $3$ dozen, or $36$.
To find the fraction of pies that had none of these ingredients, I can subtract the fractions of pies that had each ingredient from $1$.
[*, *] That is, $1 - \frac{1}{3} - \frac{1}{4} - \frac{1}{6} - \frac{1}{12}$.
I notice that the denominators of these fractions are all multiples of $12$, so I can find a common denominator by using $12$.
I can rewrite each fraction as an equivalent fraction with a denominator of $12$.
That is, $1 - \frac{4}{12} - \frac{3}{12} - \frac{2}{12} - \frac{1}{12}$.
Now I can subtract the numerators and keep the denominator of $12$.
That is, $\frac{12}{12} - \frac{4}{12} - \frac{3}{12} - \frac{2}{12} - \frac{1}{12}$.
That simplifies to $\frac{2}{12}$.
I can reduce this fraction by dividing both the numerator and the denominator by $2$.
That is, $\frac{1}{6}$.
This means that one sixth of the pies had none of these ingredients.
To find the number of pies that had none of these ingredients, I can multiply this fraction by $36$.
That is, $\frac{1}{6} \times 36$.
That simplifies to $6$.
Therefore, the smallest possible number of pies that had none of these ingredients is $6$.