Problems For National:1

[protik]

@Mahi, This problem set is quite hard. So may be people are having trouble to solve them. That's why they aren't posting solution.

[Nadim Ul Abrar]

Sol no3:

lemma : if $a^n+1$ be a prime then $a$ be even n will be of form $2^k$ .
proof : let there is an odd fator of $n$ and that is $p$ . if $n=pq$ then $(a^q+1)|a^n+1=(a^q)^p+1$ .
so $n=2^k$ ,

now from the set {1,2,3,4,5....2000} there are $11$ numbers of form $2^k$ .

So that $\mathbb{M}$ can contain at most $11$ primes .

and 1% of $2000$ is $20$ .

[Masum]

I can't understand question no.5
What is the necessity of $n$ here. Is there any bug in this problem ???

I am restating the problem:
Prove that there exist an infinite $n$ such that $n$ does not have any zeros in its base $10$ representation and the sum of digits of $n$ divides $n$.
But why didn't anyone ask this in time? Forum was activated much before the time had been over.
And the exam is over. Discuss over the problems.
I hoped that you will solve at least the first one and the last one. Hint to first one. See $a^2=2b^2$ or something like that happens after factorization. So the rest is clear!!
The last one was actually from me. I proposed this problem to Chamok vai at his last birthday as his gift. And the fact is, it seems to be very interesting that you will take any $2n$ positive integer and their pair-wise differences. And you will find that the product of them is divisible by $2^n(2n-1)$. In fact we can even optimize the exponent of $2$. Hint: pigeonhole principle. $2n-1<2n$ and $n=\lceil\frac{2n}2\rceil$

[Nadim Ul Abrar]

Sol 4

$\angle OBM=\angle LBM=\angle MKL=\angle MAO$
so $AOMB$ is cyclic .

Now $\angle KBO=\angle BAM=\angle BOM$
So $KB||OM $ or $\angle BKM=\angle KMO$

Again$ \angle MBL=\angle MKL=\angle KAO=\angle ABK$
So $\angle ABO=\angle KBM$

and$ \angle BMK=\angle BLK=\angle BOA $
So $\angle BKM=\angle BAO=\angle ABO=\angle KBM=\angle KMO$.

So our proof is complete

[Nadim Ul Abrar]

But why didn't anyone ask this in time? Forum was activated much before the time had been over.

Some of us failed to enter forum in 1st january .
It showed "general Error" .

[Nadim Ul Abrar]

sol 6

$f(x^2-y^2)=(x-y)(f(x)+f(y))$

case 1: $f(x)$ is constant .

if that is then let $f(x)=c; x\to\mathbb{R}$ . then c=(x-y)c or c=0
So $f(x)=0$ is a solution to this equation .

case 2 : $f(x)$ isn't constant
putting $x=y+1$ ye have ,

$f(y+1+y)=f(y+1)+f(y)$. (cauchy's functional equation)

so that $f(x)$ not be constant constant then $f(x)=cx$ ($c$,Constant) .

So all function $f(x)$ such that $f(x)$ satisfy the equation $f(x^2-y^2)=(x-y)(f(x)+f(y))$ are $f(x)=cx$

[Nadim Ul Abrar]

sol 1

$a^4b^4+4a^2b^2=2a^6+2b^6$
$\Leftrightarrow 4a^2b^2-2b^6=2a^6-a^4b^4$
$\Leftrightarrow (a^4-2b^2)(2a^2-b^4)=0$

$a^4-2b^2=0$ then
$\frac{a^2}{b}=\sqrt{2}$
if a,b be rational then $\frac{a^2}{b}$ will also be rational . $\sqrt{2}$ is irrational . so . there doesnt exist such $a,b$ .

[sourav das]

I couldn't understand the change in Problem 5. I thought it was :For all $n \in N$, there is a $m \in N$ consisting $n$ digits such that the sum of digits of $m$ divides $m$
and $m$ does not have any zeros in its representation.

And i also couldn't get the part of new formation problem ""exist an infinite $n$"". Please someone help me.

[*Mahi*]

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 $n$."

[*Mahi*]

case 2 : $f(x)$ isn't constant
putting $x=y+1$ ye have ,

$f(y+1+y)=f(y+1)+f(y)$. (cauchy's functional equation)

so that $f(x)$ not be constant constant then $f(x)=cx$ ($c$,Constant) .

So all function $f(x)$ such that $f(x)$ satisfy the equation $f(x^2-y^2)=(x-y)(f(x)+f(y))$ are $f(x)=cx$

Cauchy equation is valid for $\mathbb Q \rightarrow \mathbb Q$ functions or continuous functions $\mathbb R \rightarrow \mathbb R$

Another flaw: Cauchy equation is only valid when $f(x+y)=f(x)+f(y)$ for all $x,y \in \text{DOM } f$