Brainsteem Mathematics Challenges: Divisibility by 65
This question is slightly more challenging, hence the red logo. It requires some algebra and number theory but is certainly doable by a secondary level student considering a national pre-Olympiad mathematics competition.
Question
Let f(n) = 5n13 + 13n5 + 9kn
Find the least positive integer k such that 65 divides f(n) for every integer n.
The question is taken from the IrMO 2000 paper, that also gives permission to distribute questions so that interested people may openly discuss solutions. It was also featured on my blog GiftedMathematics.com but received no valid answers.
Let's see if Steemians can do better!
As complicated algebra is horrible to write online, I'll accept a logical sequence of key steps in the solution.
The first correct answer and further interesting comments will be rewarded with an upvote.
Enjoy!
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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
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