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

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The discussion focuses on proving that any amount \( n \ge 32 \) can be expressed as \( 5x + 9y \) where \( x \) and \( y \) are non-negative integers. An initial case is provided with \( n = 32 \) showing \( x = 1 \) and \( y = 3 \) as a valid solution. The proof employs mathematical induction, assuming the statement holds for some integer \( k \ge 32 \) and demonstrating it for \( k + 1 \). The discussion also explores specific cases for \( k + 1 \) based on the values of \( y \), leading to valid constructions for \( n \). The conclusion affirms that all integers \( n \) from 32 onward can indeed be represented in this form.
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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 $n \ge 32$.
 
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Re: $ solutions

ssome help said:
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: $\$$.
 
Re: $ solutions

Alternatively you can type:
Code:
\$
which comes out as \$.
 
Re: $ solutions

ssome help said:
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)
 
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