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

  • Thread starter Thread starter ssome help
  • Start date Start date
AI Thread Summary
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.
ssome help
Messages
3
Reaction score
0
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:
Physics news on Phys.org
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)
 
Hello, I'm joining this forum to ask two questions which have nagged me for some time. They both are presumed obvious, yet don't make sense to me. Nobody will explain their positions, which is...uh...aka science. I also have a thread for the other question. But this one involves probability, known as the Monty Hall Problem. Please see any number of YouTube videos on this for an explanation, I'll leave it to them to explain it. I question the predicate of all those who answer this...
I'm taking a look at intuitionistic propositional logic (IPL). Basically it exclude Double Negation Elimination (DNE) from the set of axiom schemas replacing it with Ex falso quodlibet: ⊥ → p for any proposition p (including both atomic and composite propositions). In IPL, for instance, the Law of Excluded Middle (LEM) p ∨ ¬p is no longer a theorem. My question: aside from the logic formal perspective, is IPL supposed to model/address some specific "kind of world" ? Thanks.
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
Back
Top