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Whole number solutions

ssome help

New member
Apr 7, 2013
3
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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Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
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: $\$$.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Re: $ solutions

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

Amer

Active member
Mar 1, 2012
275
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
[tex] 5 + 3(9) = 32 [/tex]
note that [tex] 1 = 2(5) - 9 [/tex]
so [tex] 33 = 5 + 2(5) + 3(9) - 9 [/tex]
suppose it is true for k>=32 integer there exist a positive integers x,y such that

[tex] 5x + 9y = k [/tex]
for k+1
[tex] k+1 = 5x + 9y +1 [/tex]
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)