428
Nice
(lemmy.fmhy.ml)
His head hurts because of the unnecessary brackets, right?
Please excuse my dumb ass self.
Huh, that's true of any number that ends in 9.
XY + X + Y = 10*X + Y
Y's cancel,
XY = 9X => Y = 9 for any non-zero finite value of X.
so for 69? X = 6, Y=9
(6*9) + 6 + 9 = 10*6 + 9
54 + 15 = 69
69 = 69 (nice!)
429? X = 42 Y = 9
(42*9) + 42 + 9 =10*42 + 9
(378) + 51 = 429
429 = 429
Even if 10X+Y doesn't equal something that ends in 9 it works
X=3.14 Y=9
(3.14*9) + 3.14 + 9 = 10*3.14 + 9
28.26 + 12.14 = 40.4
40.4 = 40.4
Doesn't work if Y =\= 9:
68? X = 6 Y = 8
(6*8) + 6 + 8 ?= 10*6 + 8
(48) + 14 ?= 68
62 =\= 68
I wanted to try to properly prove that it didn't work for y!=9, but I think you covered the edge cases - X=0 or unbounded. Well done!
I'm an engineering major, we learn all of the edge cases as "well technically this isn't always true, but we'll just pretend it is because the results are close enough"
Every value of Y works for X=0, the equation simplifies to Y=Y, so X=0 is just like Y=9.
In the limit as X->infinity, you get Y = 9 again.
X(1+Y) + Y = 10*X + Y lim X->inf Assuming Y is finite, you drop the non-X terms
X(1+Y) = 10*X lim x->inf
Here, because X is non-zero and equal to itself, you can cancel them (I assume, IANA Mathematician) 1 + Y = 10 Y = 9
Big if true
Large if accurate
Gargantuan if verifiable
...Nice.
Nice
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this post was submitted on 13 Jul 2023
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