Dhaka regional higher secondary/8

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Ragib Farhat Hasan
Posts: 20
Joined: Sun Mar 30, 2014 10:40 pm

Dhaka regional higher secondary/8

Unread post by Ragib Farhat Hasan » Thu Jan 14, 2016 7:45 pm

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) =?

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asif e elahi
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Location: Sylhet,Bangladesh

Re: Dhaka regional higher secondary/8

Unread post by asif e elahi » Mon Jan 25, 2016 9:33 pm

Prove that if $5 \nmid x$, then $f(x)=0$ using the first 2 conditions.

Ragib Farhat Hasan
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Joined: Sun Mar 30, 2014 10:40 pm

Re: Dhaka regional higher secondary/8

Unread post by Ragib Farhat Hasan » Thu Jan 28, 2016 7:01 pm

But f(2015)=?

mdhasib
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Re: Dhaka regional higher secondary/8

Unread post by mdhasib » Mon Jan 09, 2017 1:15 pm

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

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