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RE: Brainsteem Mathematics Challenges: Divisibility by 65
f(1)=18+9k=9 * (2+k)
f(1)=9 * 65 * x
18+9k=585x
let x=1 => 18+9k=585
k=63
f(n)=5n13 + 13n5 + 567n
f(1)=18+9k=9 * (2+k)
f(1)=9 * 65 * x
18+9k=585x
let x=1 => 18+9k=585
k=63
f(n)=5n13 + 13n5 + 567n
That seems to be correct to me. I had the same aproach but since you wrote it down I won't write pretty much exactly the same thing a little diffrent.
How do we then know it works for all n?
Just the idea! The algebra can be long.
It's easy)))
f(1) divided by 65
let f (n) is divided by 65
divide by 65 this f(n+1)-f(n)
f(n+1)-f(n)=65n12+390n11+1430n10+3575n9+6435n8+8580n7+8580n6+6435n5+3575n4+1430n3+390n2+65n+65n4+130n3+130n2+65n+585
all coefficients are divided by 65
therefore, the f(n) is divided into 65 for all n