Prove $5x+9y=n$ Solutions for $n \ge 32$, $\mathbb{Z}_0^+$

[ssome help]

Prove that $n \ge \$32$ can be paid in \$5 and \$9 dollar bills ie the equation $5x+9y=n$ has solutions $x$ and $y$ element $\mathbb{Z}_0^+$ for $n$ element of $\mathbb{Z}^+$ and $n \ge 32$.

 

[Evgeny.Makarov]

Re: $ solutions

Prove that $n >=(greater than or equal to) 32 can be paid in $5 and $9 dollar bills ie the equation 5x+9y=n has solutions x and y element (Z(sub0)^+) for n element of Z^+ and n >=32.

See a similar problem here.

On this forum, the dollar sign starts a "mathematical mode" where one can use special commands to produce symbols like $\pi$ and $\int$. If you want to write a dollar sign, you can type dollar, backslash and two dollars, like this: $\$$.

 

[I like Serena]

Re: $ solutions

Alternatively you can type:

\$

which comes out as \$.

 

[Amer]

Re: $ solutions

Prove that $n >=(greater than or equal to) 32 can be paid in $5 and $9 dollar bills ie the equation 5x+9y=n has solutions x and y element (Z(sub0)^+) for n element of Z^+ and n >=32.

induction at n for n=32 x=1,y=3
$$5 + 3(9) = 32$$
note that $$1 = 2(5) - 9$$
so $$33 = 5 + 2(5) + 3(9) - 9$$
suppose it is true for k>=32 integer there exist a positive integers x,y such that

$$5x + 9y = k$$
for k+1
$$k+1 = 5x + 9y +1$$
choose 1 = 2(5) - 9, if y>=1
if y = 0
then k multiple of 5 which is 35 or larger, x>=7 so choose
1= -7(5) + 4(9)

 

[blue_raver22]

I would approach this problem by first defining all the terms and variables involved. In this case, we are dealing with the equation $5x+9y=n$, where $x$ and $y$ are both positive integers (denoted by $\mathbb{Z}_0^+$) and $n$ is a positive integer (denoted by $\mathbb{Z}^+$).

Next, I would state the given condition, which is that $n \ge 32$. This means that we are looking for solutions to the equation $5x+9y=n$ where $n$ is at least 32. In other words, we are looking for combinations of $x$ and $y$ that add up to a number that is at least 32.

To prove that such solutions exist, we can use a method called induction. We start by showing that the equation holds for the base case, which in this case is $n = 32$. This means we need to find values for $x$ and $y$ that satisfy $5x+9y=32$. One possible solution is $x=3$ and $y=1$, since $5(3)+9(1)=32$. This shows that the equation holds for $n=32$.

Next, we assume that the equation holds for some arbitrary value of $n$, and we want to show that it also holds for $n+1$. In other words, we want to show that if $5x+9y=n$, then $5x'+9y'=n+1$ for some $x'$ and $y'$. We can rewrite this as $5x'+9y'=n+5+9$, which is equivalent to $5x'+9y'=n+14$. This means that we need to find values for $x'$ and $y'$ that satisfy $5x'+9y'=n+14$. One possible solution is $x'=3$ and $y'=2$, since $5(3)+9(2)=32+14=46$. This shows that if the equation holds for $n$, then it also holds for $n+1$.

Therefore, by the principle of mathematical induction, we have shown that the equation $5x+9y=n$ has solutions for all values of $n$ that are at least 32. In other words, $n \