How many birds of each kind did I buy?

[Math100]

Homework Statement

I bought sparrows at $3$ for a penny, turtle doves at $2$ for a penny, and doves at $2$ pence each. If I spent $30$ pence buying $30$ birds and bought at least one of each kind of bird, how many birds of each kind did I buy?

Relevant Equations

Diophantine equation: $ax+by=c$ where $a, b, c$ are integers.

Let $x$ denote the number of sparrows, $y$ denote the number of turtle doves and $z$ denote the number of doves.
Then we have $\frac{1}{3}x+\frac{1}{2}y+2z=30$ such that $x+y+z=30$.
Observe that
\begin{align*}
&\frac{1}{3}x+\frac{1}{2}y+2(30-x-y)=30\\
&\frac{1}{3}x+\frac{1}{2}y+60-2x-2y=30\\
&-\frac{5}{3}x-\frac{3}{2}y=-30\\
&-10x-9y=-180\\
&10x+9y=180.\\
\end{align*}
Consider the Diophantine equation $10x+9y=180$.
By the Euclidean Algorithm, we have that $gcd(10, 9)=1$.
Since $1\mid 180$, it follows that the Diophantine equation $10x+9y=180$ can be solved.
From $1=10-1(9)$, we get $180=180[10-1(9)]=180(10)-180(9)$.
Thus $x_{0}=-180, y_{0}=180$.
All solutions in the integers are determined by:
$x=-180+(\frac{10}{1})t=-180+10t$ for some $t\in\mathbb{Z}$,
$y=180-(\frac{9}{1})t=180-9t$ for some $t\in\mathbb{Z}$.
Thus, $x=-180+10t$ and $y=180-9t$.
To find all solutions in the positive integers of the Diophantine equation $10x+9y=180$,
we solve the following inequalities for $t$:
$-180+10t\geq 0$ and $180-9t\geq 0$.
Now we have $18\leq t\leq 20$, or $t=19$.
This implies $x=10, y=9$, and $z=30-10-9=11$.
Therefore, ten sparrows, nine turtle doves and eleven doves were purchased.

 

[fresh_42]

Homework Statement:: I bought sparrows at $3$ for a penny, turtle doves at $2$ for a penny, and doves at $2$ pence each. If I spent $30$ pence buying $30$ birds and bought at least one of each kind of bird, how many birds of each kind did I buy?
Relevant Equations:: Diophantine equation: $ax+by=c$ where $a, b, c$ are integers.

Let $x$ denote the number of sparrows, $y$ denote the number of turtle doves and $z$ denote the number of doves.
Then we have $\frac{1}{3}x+\frac{1}{2}y+2z=30$ such that $x+y+z=30$.
Observe that
\begin{align*}
&\frac{1}{3}x+\frac{1}{2}y+2(30-x-y)=30\\
&\frac{1}{3}x+\frac{1}{2}y+60-2x-2y=30\\
&-\frac{5}{3}x-\frac{3}{2}y=-30\\
&-10x-9y=-180\\
&10x+9y=180.\\
\end{align*}
Consider the Diophantine equation $10x+9y=180$.
By the Euclidean Algorithm, we have that $gcd(10, 9)=1$.
Since $1\mid 180$, it follows that the Diophantine equation $10x+9y=180$ can be solved.
From $1=10-1(9)$, we get $180=180[10-1(9)]=180(10)-180(9)$.

This notation is new to me: $1=10-1(9)$ I have no idea what it means.

Thus $x_{0}=-180, y_{0}=180$.
All solutions in the integers are determined by:
$x=-180+(\frac{10}{1})t=-180+10t$ for some $t\in\mathbb{Z}$,
$y=180-(\frac{9}{1})t=180-9t$ for some $t\in\mathbb{Z}$.
Thus, $x=-180+10t$ and $y=180-9t$.
To find all solutions in the positive integers of the Diophantine equation $10x+9y=180$,
we solve the following inequalities for $t$:
$-180+10t\geq 0$ and $180-9t\geq 0$.
Now we have $18\leq t\leq 20$, or $t=19$.

You should note that $t\in \{18,20\}$ is excluded by $x,y>0.$

This implies $x=10, y=9$, and $z=30-10-9=11$.
Therefore, ten sparrows, nine turtle doves and eleven doves were purchased.

Do you have a metal saw? Otherwise, you bought nine sparrows, ten turtle doves, and eleven doves.

 

[Math100]

This notation is new to me: $1=10-1(9)$ I have no idea what it means.

You should note that $t\in \{18,20\}$ is excluded by $x,y>0.$

Do you have a metal saw? Otherwise, you bought nine sparrows, ten turtle doves, and eleven doves.

I found my mistakes.

 

[WWGD]

I'm not sure if I understand. You're using both pennies and pence. How do they relate to each other?

 

[pasmith]

I'm not sure if I understand. You're using both pennies and pence. How do they relate to each other?

'Penny' is either a currency unit worth 0.01 GBP, in which case the plural is 'pence', or a coin with a face value of one penny, in which case the plural is 'pennies'.

 

[Steve4Physics]

@Math100, FWIW, you can simplify the working by utilising the fact that sparrows are bought in threes and turtle-doves are bought in twos.

S = number of sparrow triplets
T = number of turtle-dove doublets
D= number of doves

S+T+2D =30 (since total cost 30pence)
3S+2T+D =30 (since total no. of birds = 30)

A couple of lines of algebra gives:
5S+3T=30
which (in view of the small values) can easily be solved by inspection or by constructing a small table. (Though admittedly not a general/rigorous approach.)