Advance P-2(BOMC-2)

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sourav das
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Advance P-2(BOMC-2)

Unread post by sourav das » Sat Mar 31, 2012 7:30 pm

Let $S(x)$ be the sum of the digits of the positive integer $x$ in its decimal
representation.
(a) Prove that for every positive integer $x$,
$\frac{S(x)}{S(2x)}\leq 5$ Can this bound be
improved?
(b) Prove that $\frac{S(x)}{S(3x)}$is not bounded.
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*Mahi*
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Re: Advance P-2(BOMC-2)

Unread post by *Mahi* » Sat Mar 31, 2012 11:03 pm

Posting only the hints again. Do not open them all at once. Do it serially .This is because it is easier to get the later hints when you know the earlier ones.
Warning: My solution is too much abstract :S I think I should see the official solution as well.
First and basic:
Counter example is enough (in contest, not for learning) for part (a)-2 and part (b).
$\frac ab \leq \frac{a+c}{b+d} \leq \frac cd$
Then:
Study the numbers $2n$ and $3n$ for one digit $n$'s.
Desperate:
$S(n)=S(n+9)$ for $2$ digit $n$
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nafistiham
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Re: Advance P-2(BOMC-2)

Unread post by nafistiham » Sun Apr 01, 2012 1:43 am

what is meant by can this bound be improved ?
does it mean whether it can be shown or not that there can be a greater value than $5$ ? :? :?
if it is then
as it is the decimal system it cant be greater than $5$.so, the relation cant be improved
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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*Mahi*
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Re: Advance P-2(BOMC-2)

Unread post by *Mahi* » Sun Apr 01, 2012 12:32 pm

nafistiham wrote:what is meant by can this bound be improved ?
does it mean whether it can be shown or not that there can be a greater value than $5$ ? :? :?
if it is then
as it is the decimal system it cant be greater than $5$.so, the relation cant be improved
It is actually "a smaller value than $5$".
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nafistiham
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Re: Advance P-2(BOMC-2)

Unread post by nafistiham » Sun Apr 01, 2012 8:18 pm

but, $\frac {5}{1}=5$ where, $x=5$ and $2x=10$ so it cant be improved, right ?
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Advance P-2(BOMC-2)

Unread post by *Mahi* » Sun Apr 01, 2012 8:55 pm

Yes, that's right.
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