Problems For National:1
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@Nadim
Pro-1: Nice! But a=b=0 is a solution.
Pro-3: You forgot to include 10+1; So, there can be at most 11 primes.
Pro-4: Nice!
Pro-6:First of all, Cauchy's Functional equation is not a direct formula. You need specific conditions to use it.
i)$f(b+1+b) =f(b+1)+f(b)$ satisfy only for (a,b)=(b,b+1); But not for all (a,b)
ii)It is not given (or you didn't prove) that $f$ is continuous or, monotone or, for$x\geq 0$ $f(x)\geq 0$ or, $f$ is continuous at $0$. (I think other conditions vary with situations.)
Pro-1: Nice! But a=b=0 is a solution.
Pro-3: You forgot to include 10+1; So, there can be at most 11 primes.
Pro-4: Nice!
Pro-6:First of all, Cauchy's Functional equation is not a direct formula. You need specific conditions to use it.
i)$f(b+1+b) =f(b+1)+f(b)$ satisfy only for (a,b)=(b,b+1); But not for all (a,b)
ii)It is not given (or you didn't prove) that $f$ is continuous or, monotone or, for$x\geq 0$ $f(x)\geq 0$ or, $f$ is continuous at $0$. (I think other conditions vary with situations.)
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
- Nadim Ul Abrar
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Re: Problems For National:1
Sourav da
1.0 j rational jantam na .
3.yes , $2^0$ count korte vule gechi .
4.thanku .
6.Ami kichu PDF theke f.equation shekhar cheshta korechi, onek kichui bujhini but vebechilam bujhe gesi , Keu amar vul gula dhoraye dey ni too .
Thanku
1.0 j rational jantam na .
3.yes , $2^0$ count korte vule gechi .
4.thanku .
6.Ami kichu PDF theke f.equation shekhar cheshta korechi, onek kichui bujhini but vebechilam bujhe gesi , Keu amar vul gula dhoraye dey ni too .
Thanku
$\frac{1}{0}$
Re: Problems For National:1
There is only $m$ now. Prove that there exists infinite $m$ such that $m$ is divisible by the sum of its digits and has no zeros. Nothing more than that.*Mahi* wrote:The problem statement should be, "Prove, for all $m \in \mathbb N$, $\exists n$ with $m$ digits such that $n$ does not have any zeros in its base $10$ representation and the sum of digits of $n$ divides $n$."
One one thing is neutral in the universe, that is $0$.
Re: Problems For National:1
You may discuss about the problems and solutions for some more time before the next problems come.
Here is the official solution for $5$.
Problem: Prove that there exist infinite $m$ such that $m$ has no $0$ in its base $10$ representation and is divisible by the sum of its digits.
Solution: Let $n=33..33$ where $n$ have $3^k$ $3's $ for $k\in\mathbb N$. Then sum of digits of $n$ is $3^{k+1}$. So we need to prove that \[3^{k+1}|n=3\cdot11...11\]
But \[11...11=\frac{10^{3^k}-1}{9}\]
Or we need to prove that \[3^{k+2}|10^{3^k}-1\]
You may prove it easily using induction or invoke Lifting The Exponent Lemma(if you know it). Ask me for any clarification.
Here is the official solution for $5$.
Problem: Prove that there exist infinite $m$ such that $m$ has no $0$ in its base $10$ representation and is divisible by the sum of its digits.
Solution: Let $n=33..33$ where $n$ have $3^k$ $3's $ for $k\in\mathbb N$. Then sum of digits of $n$ is $3^{k+1}$. So we need to prove that \[3^{k+1}|n=3\cdot11...11\]
But \[11...11=\frac{10^{3^k}-1}{9}\]
Or we need to prove that \[3^{k+2}|10^{3^k}-1\]
You may prove it easily using induction or invoke Lifting The Exponent Lemma(if you know it). Ask me for any clarification.
One one thing is neutral in the universe, that is $0$.
Re: Problems For National:1
$n=333......33$ এর জন্য হিসাব করা হল কেন ?? $n$ অন্যান্য মানের জন্য কি এমন কোন সংখ্যা নাই যারা তাদের digit এর যোগফল দ্বারা বিভাজ্য ??? বুঝি নাই
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
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Re: Problems For National:1
Problem 1:
Problem 2:
Problem 3:
Problem 4:
Problem 5:
Problem 6:
Problem 7:
Nice problem set. Hope you'll (Masum bhai) give us another soon.
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
- FahimFerdous
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Re: Problems For National:1
I solved problem 1 quite easily. Tell me if there's a bug.
The equation can be written as:
${ab(ab+2)}^2=2(a^3+b^3)^2$
as $p^2=2q^2$ has no rational solution, so it doesn't have one.
The equation can be written as:
${ab(ab+2)}^2=2(a^3+b^3)^2$
as $p^2=2q^2$ has no rational solution, so it doesn't have one.
Your hot head might dominate your good heart!
- FahimFerdous
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Re: Problems For National:1
My solution to problem 3, 4, 7 are as same as Sourav. But I didn't understand problem 6. Any help?
Your hot head might dominate your good heart!
Re: Problems For National:1
Can you optimize $7$ more? I mean the exponent of $2$?
One one thing is neutral in the universe, that is $0$.
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Re: Problems For National:1
I think I'm the only one who make the worst look of problem 1.
@Fahim, Where is the problem? In understanding problem or solution? Which part?
I think you made a typo. $ab(ab+2)^2$.
@Fahim, Where is the problem? In understanding problem or solution? Which part?
I think you made a typo. $ab(ab+2)^2$.
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )