For all positive integers x,y; f(x) ≥ 0 and f(xy) = f(x) + f(y). If the digit at
the one’s of x is 6, then f(x) = 0. If f(1920) = 420 then f(2015) =?
Dhaka regional higher secondary/8
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Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
- asif e elahi
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Re: Dhaka regional higher secondary/8
Prove that if $5 \nmid x$, then $f(x)=0$ using the first 2 conditions.
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Re: Dhaka regional higher secondary/8
But f(2015)=?
Re: Dhaka regional higher secondary/8
I think F(2015)=420. Because F(5*384)=F(1920)=420=F(5)+F(384)=F(5) and F(384)=0 because 5 does not divide 384 . So F(2015)=F(5*13*31)=F(5)+F(13)+F(31)=F(5)=420