# Whole number solutions

#### 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$n \ge 32$. Last edited by a moderator: #### Evgeny.Makarov ##### Well-known member MHB Math Scholar 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
Re: $solutions Alternatively you can type: Code: \$
which comes out as \$. #### Amer ##### Active member Re:$ 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)