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Weird algebra problem?
- Thread starter thepilot
- Start date
Let a, b, c be integers such that 2005 both ab+9b+81 and bc+9c+81.
Prove that 2005 also divides ca+9a+81.
I've been tinkering with this problem for hours, and can't find a solution.
Any help would be appreciated.
You didn't provide any work. You allege that you have tinkered with it
"for hours."
That looks like a math contest problem from its relative complexity.
And you also didn't provide any shown work from that other problem you posted.
Hello, thepilot!
I think I have a proof . . .
We are told: .\(\displaystyle \begin{Bmatrix}ab + 9b + 81 \;=\; 2005p\;\text{ for some integer }p & [1] \\ bc + 9c + 81 \;=\; 2005q\;\text{ for some integer }q & [2]\end{Bmatrix}\)
\(\displaystyle \begin{array}{ccccc}\text{Multiply [1] by }c: & abc + 9bc + 81c &=& 2005cp & [3] \\
\text{Multiply [2] by }a: & abc + 9ac + 81a &=& 2005aq & [4] \end{array}\)
\(\displaystyle \text{Subtract [4] - [3]: }\;9ac - 9bc + 81a - 81c \:=\:2005aq - 2005 cp\)
. . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle 9ac + 81a \:=\:9bc + 81c + 2005(aq-cp)\)
\(\displaystyle \text{Divide by 9:}\) . . . . . . . . . . . . . . . \(\displaystyle ac + 9a \:=\:bc + 9c + \dfrac{2005(aq-cp)}{9}\)
\(\displaystyle \text{Add 81 to both sides: }\qquad\, ac + 9a + 81 \:=\:\underbrace{bc + 9c + 81}_{\text{This is }2005q} + \dfrac{2005(aq-cp)}{9}\)
\(\displaystyle \text{Hence: }\:ac + 9a + 81 \;=\;2005\left(q + \dfrac{aq-cp}{9}\right)\)
\(\displaystyle \text{If }aq - cp\text{ is divisible by 9, our proof is complete.}\)
\(\displaystyle \text{From [4]: }\;aq \:=\:\dfrac{abc + 9ac + 81a}{2005}.\quad\text{From [3]:}\;cp \:=\:\dfrac{abc + 9bc + 81c}{2005}\)
\(\displaystyle \text{Then: }\:aq - cp \:=\:\dfrac{abc+9ac+81a}{2005} - \dfrac{abc + 9bc + 81c}{2005}\)
. . . . . . . . . . . . \(\displaystyle =\;\dfrac{9ac + 81a - 9bc - 81c}{2005} \;=\;\dfrac{9(ac + 9a - bc = 9c)}{2005}\)
. . . . \(\displaystyle Q.E.D.\)
I think I have a proof . . .
\(\displaystyle \text{Let }a,b,c\text{ be integers such that 2005 divides both }ab + 9b+ 81\text{ and }\) \(\displaystyle bc + 9c + 8a.\)
\(\displaystyle \text{Prove that 2005 also divides }ac + 9a + 81.\)
We are told: .\(\displaystyle \begin{Bmatrix}ab + 9b + 81 \;=\; 2005p\;\text{ for some integer }p & [1] \\ bc + 9c + 81 \;=\; 2005q\;\text{ for some integer }q & [2]\end{Bmatrix}\)
\(\displaystyle \begin{array}{ccccc}\text{Multiply [1] by }c: & abc + 9bc + 81c &=& 2005cp & [3] \\
\text{Multiply [2] by }a: & abc + 9ac + 81a &=& 2005aq & [4] \end{array}\)
\(\displaystyle \text{Subtract [4] - [3]: }\;9ac - 9bc + 81a - 81c \:=\:2005aq - 2005 cp\)
. . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle 9ac + 81a \:=\:9bc + 81c + 2005(aq-cp)\)
\(\displaystyle \text{Divide by 9:}\) . . . . . . . . . . . . . . . \(\displaystyle ac + 9a \:=\:bc + 9c + \dfrac{2005(aq-cp)}{9}\)
\(\displaystyle \text{Add 81 to both sides: }\qquad\, ac + 9a + 81 \:=\:\underbrace{bc + 9c + 81}_{\text{This is }2005q} + \dfrac{2005(aq-cp)}{9}\)
\(\displaystyle \text{Hence: }\:ac + 9a + 81 \;=\;2005\left(q + \dfrac{aq-cp}{9}\right)\)
\(\displaystyle \text{If }aq - cp\text{ is divisible by 9, our proof is complete.}\)
\(\displaystyle \text{From [4]: }\;aq \:=\:\dfrac{abc + 9ac + 81a}{2005}.\quad\text{From [3]:}\;cp \:=\:\dfrac{abc + 9bc + 81c}{2005}\)
\(\displaystyle \text{Then: }\:aq - cp \:=\:\dfrac{abc+9ac+81a}{2005} - \dfrac{abc + 9bc + 81c}{2005}\)
. . . . . . . . . . . . \(\displaystyle =\;\dfrac{9ac + 81a - 9bc - 81c}{2005} \;=\;\dfrac{9(ac + 9a - bc = 9c)}{2005}\)
. . . . \(\displaystyle Q.E.D.\)