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)