I used Simon to get (3a+2)(3b+1) = 128; first factor is 2 mod 3, 2nd factor is 1 mod 3; When checking negative factors, their mod 3 values flip; eg factors (128*1) makes (a,b)=(42,0) a+b=42; then flip to (-1 * -128) makes (a,b)=(-1,-43) a+b= -44. Solution set {-44, -23, -13, -9, 7, 11, 21, 42}
I know everybody has their favourite way of dealing with a rational expression where deg(numerator) = deg(denominator). Like adding 0 to the numerator and splitting. Also used, for example, in integration. Long division is often given a bum rap - claiming longer/more difficult. Not saying that you should change your method, but it is another option that is actually quite easy and short.
@@SyberMath Oh, it's a deeply philosophical japanese video game, where the main character's name is 2B for allegorical reasons, cuz it's about the meaning of existance :)
Your solution is wrong or at least is incomplete. First of all, if you call this equation a Diophantine, then a, b must be > 0. So a = 0 or a = -1 are not solutions. Secondly, you missed solution a=10, b=1 -> a+ b = 11. Basically, there are 2 solutions: 2, 5 (sum = 7) and 10, 1 (sum = 11), if we're talking about Diophantine equation. If we're talking about an ordinal equation with integer solutions, then 0, 21 (21) and -1, -43 (-44) are solutions too.
@@FrancisZerbib Wiki says you're right - it's about integer solutions. Which is a big surprise for me, as I remember from school - Diophantus (like all ancient Greek science) didn't know zero or negative numbers. They only operated natural numbers and fractions.
Feel like a mouse being put into a maze. The question is: how the experience of finding a way out in this maze could help a mouse finding the way out in next one?
I used Simon to get (3a+2)(3b+1) = 128; first factor is 2 mod 3, 2nd factor is 1 mod 3; When checking negative factors, their mod 3 values flip; eg factors (128*1) makes (a,b)=(42,0) a+b=42; then flip to (-1 * -128) makes (a,b)=(-1,-43) a+b= -44. Solution set {-44, -23, -13, -9, 7, 11, 21, 42}
43 not 42 . . . that is what I got
To make it simple, I set b=1 and took it from there; thus making a=10, coming up with a final answer of 11 and verifying it.
I know everybody has their favourite way of dealing with a rational expression where deg(numerator) = deg(denominator). Like adding 0 to the numerator and splitting. Also used, for example, in integration. Long division is often given a bum rap - claiming longer/more difficult. Not saying that you should change your method, but it is another option that is actually quite easy and short.
I agree
Patiently waiting for SyberMath to learn about the existance of Nier Automata 😁
What’s that? 🧐
@@SyberMath Oh, it's a deeply philosophical japanese video game, where the main character's name is 2B for allegorical reasons, cuz it's about the meaning of existance :)
simons favourite method
I got a+b=2+5=7.
Your solution is wrong or at least is incomplete.
First of all, if you call this equation a Diophantine, then a, b must be > 0. So a = 0 or a = -1 are not solutions.
Secondly, you missed solution a=10, b=1 -> a+ b = 11.
Basically, there are 2 solutions: 2, 5 (sum = 7) and 10, 1 (sum = 11), if we're talking about Diophantine equation. If we're talking about an ordinal equation with integer solutions, then 0, 21 (21) and -1, -43 (-44) are solutions too.
Diophantine is for Integer solutions. Including zero and negative numbers. Review your basics
@@FrancisZerbib Wiki says you're right - it's about integer solutions. Which is a big surprise for me, as I remember from school - Diophantus (like all ancient Greek science) didn't know zero or negative numbers. They only operated natural numbers and fractions.
Feel like a mouse being put into a maze. The question is: how the experience of finding a way out in this maze could help a mouse finding the way out in next one?
a = 10, b = 1
Answer: 11
a = 10, b = 1 is the answer