Problems For National:1

[sourav das]

@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.)

[Nadim Ul Abrar]

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

[Masum]

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 $n$."

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.

[Masum]

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.

[sm.joty]

$n=333......33$ এর জন্য হিসাব করা হল কেন ?? $n$ অন্যান্য মানের জন্য কি এমন কোন সংখ্যা নাই যারা তাদের digit এর যোগফল দ্বারা বিভাজ্য ??? বুঝি নাই

[sourav das]

Problem 1:

If any of $a,b$ is equal to $0$ then both will be $0$. So consider non-zero rational numbers.
Setting $a=\frac{m}{n}$ $b=\frac{p}{q}$ g.c.d($m,n$)=$1$; ($p,q$)=$1$....(i); and our equation becomes:
$m^2n^2p^2q^2(m^2p^2+2^2n^2q^2)= 2(m^6q^6+n^6p^6)$

Now consider power of 2 for both side. $m^2n^2p^2q^2$ must have a even power of 2. $m^2p^2+2^2n^2q^2$ is either odd or also have a even power of 2. So, $m^6q^6+n^6p^6$ must have a odd power of 2.
It is possible only when both $m^6q^6$ and $n^6p^6$ will have same power of 2. [Other wise R.H.S. will have odd power of 2]

So let $2^c || mq$ then $2^c||pn$ with $c\geq 1$ . Using (i)
$2^c||mn$ ; $2^c||pq$ . Now R.H.S. has a power of 2 equal to $6c+2$ (as sum of two odd squares is divisible by 2 but not by 4 using (mod 4) ..(ii)).

But then L.H.S can have a power of 2 at most $4c$ (When $2^n||mp$ and $2^n||nq$; it contradicts $2^c || mq$ and $2^c||pn$ by (i))

A contradiction

Problem 2:

We know that for a triangle with sides $a,b,c$ and median $l_a$ to side $a$ ;
$4l_a^2=2b^2+2c^2-a^2$....(i).
Using this we'll get $4(AP^2 + BQ^2 + CR^2 + DS^2)=4(AB^2 + BC^2 + CD^2 +DA^2)$
$+AC^2+BD^2$ So now we have to prove that $AB^2 + BC^2 + CD^2 +DA^2$
$ \geq AC^2+BD^2$
Let $S,T$ be the midpoints of $AC,BD$. Join $(S,T)$, $(A,T)$ $(C,T)$
Using (i) we'll get,
$AB^2 + BC^2 + CD^2 +DA^2=AC^2+BD^2+4ST^2\geq AC^2+BD^2$ Equality holds for all parallelograms for which $ST=0$

Problem 3:

For having any odd prime factor of $n$; $10^n+1$ is not a prime as $a^n+b^n$ is divisible by $a+b$ for $n$ odd. So only elements in form of $10^{2^{i}}+1$ can be prime. And $M$ has 11 elements like this (from i=0 to 10; as for 11 it is greater than 2000).So surely 2000-11=1989 composite numbers and $\frac{1989}{2000}$ is greater than $99$%

Problem 4:

Using given conditions:
$\angle BOA=\angle BLK= \angle BMK= \angle BMA$
So, B,O,A,M are cyclic. So,
$\angle OMK=\angle OMA= \angle OBA=\angle OBK+\angle KBA=$
$ \angle LBK+\angle KAO= $
$\angle MAO+\angle LBK=\angle LBK+\angle OBM=$
$ \angle LBK+\angle LBM=\angle KBM$

But that's why $OM$ is tangent to circumcircle of $KBL$

Problem 5:

I couldn't crack it within my own limited like 4:30 hours like national time limit. So i'm not posting it.(I was trying the previous formation of this problem.Is the previous formation really holds?) Although now i would like to get a little help from all of you . Besides what could be the key technique of solving this kind of representation-divisibility problem? I really want some help.

Problem 6:

Setting $x=y$, $f(0)=0$ So let's find $f$ for non-zero reals.
Setting $(y,x)$ instead of $(x,y)$ and adding with main equation we'll get $f(x^2-y^2)=-f(y^2-x^2)$. Any real number $a$ can be written in form of $x^2-y^2$.So $f(a)=-f(-a)$ So setting (x,-y) in main equation and comparing the new equation with main equation we'll get:
$yf(x)=xf(y)$. So for any non-zero $x_0$ if $f(x_0)=0$; then for all $x$; $f(x)=0$
Considering for all $f(x)$ not equal to 0, $\frac{f(x)}{x}=\frac{f(y)}{y}=k$ where $k$ is a non zero-constant in this case.
So general solution is $f(x)=kx$ where k is a constant.

It satisfy clearly as $f(x^2-y^2)=k(x^2-y^2)=(x-y)(f(x)+f(y))$

Problem 7:

For $n=1$ it is not always true (Take an odd and an even number).
For $n=2$; 2 odd, 2 even; or greater or equal then 3 are same parity. So $2^2$ clearly divide the differences' product as at least 2 differences are even.
For$n\geq 3$. We'll get at least $n$ numbers of same parity. So,at least $\binom{n}{2}\geq n$
factors are even. So, $2^n$ divides the product for all $n\geq 2$

And for $(2n-1)$ by pigeon-hole theorem from $2n$ numbers at least 2 numbers will give same remainder (mod 2n-1) as there are 2n-1 different remainder (mod 2n-1). That's why at least one of pair difference is divisible by $2n-1$

Nice problem set. Hope you'll (Masum bhai) give us another soon.

[FahimFerdous]

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.

[FahimFerdous]

My solution to problem 3, 4, 7 are as same as Sourav. But I didn't understand problem 6. Any help?

[Masum]

Can you optimize $7$ more? I mean the exponent of $2$?

[sourav das]

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. $ab(ab+2)^2$.