Some problems of last year divisionals, I need help for

[nafistiham]

More problems have arrived............

Dhaka Secondary 2012

$10.$ In the given diagram, both $ABC$ and $DBE$ are right triangles, $B$ being the right angle for both. They have hypotenuses of same length. $DE$ is perpendicular on $BC$. Area of $ABC$ is $\sqrt 3$ times of $DEF$. Find $\angle BDE$.

Isn't F is the midpoint of AC and DE?

Yes. It is actually. (But, only for the bangla medium )

[protik]

Hehehe..

[Tahmid Hasan]

5. There are some small stones in a box. Weight of each of the stones is an integer (in units of gram), weight of two different stones cannot be same. If you use them on one side of a scale, you can weigh any integer numbered weight upto $2012$. What is the weight of the heaviest stone in that box?

If you want the value of the heaviest stone with minimum numbers of stones needed: Use induction to prove with stones weighing $2^0,2^1,2^2 \cdots 2^n$ units it is possible to measure all the weights upto $2^{n+1}-1$ with minimum numbers of stones needed.
And if you want the minimum value of the highest weighing stone then $\sum i_{i=1}^{n} \ge 2012$.
There are multiple answers depending on different requirements as the requirements in the original problem don't ensure a unique solution. The question seems quite confusing.

[sowmitra]

$9.$ Consider a family of functions $f_{m}:N\cup \left \{ 0 \right \}\rightarrow N\cup \left \{ 0 \right \}$ that follows the relations:
$f_{m}(a+b)=f_{m}((f_{m}(a)+f_{m}(b)))$
$f_{m}(km)=0$
Here, $m$ is any positive integer apart from $1$. Find the number of elements in the range of the function $f_{2012}$ .

Well, the given function looks a lot like $"mod"$ function.
Suppose, $g_m(x)=g_m(c)$, where, $c\equiv x(\text{mod}m)$

So, if $a\equiv c_1(\text{mod} m)$, and, $b\equiv c_2(\text{mod}m)$, then, $a+b\equiv c_1+c_2(\text{mod}m)
\Rightarrow a+b\equiv g_m(a)+g_m(b)(\text{mod}m)$.
$\Rightarrow g_m(a+b)=g_m(g_m(a)+g_m(b))$ (This matches with the 1st property.)

And, $km\equiv 0(\text{mod}m)$. So, $g_m(km)=g_m(0)$.(This matches with the second property.)
So, $g_m $ and $f_m$ treat $x$ similarly. This means that they should have the same no. elements in their range.

Now, no. of elements in the range of $g_m$ is equal to the no. of elements in the complete residue system of $m$. Therefore, no. of elements in the range of $f_{2012}$ is $\boxed{2012}$.

I am sorry that my reasoning might be a little hazy.... , but, you get the idea.

[Tahmid Hasan]

So, if $a\equiv c_1(\text{mod} m)$, and, $b\equiv c_2(\text{mod}m)$, then, $a+b\equiv c_1+c_2(\text{mod}m)
\Rightarrow a+b\equiv g_m(a)+g_m(b)(\text{mod}m)$.
$\Rightarrow g_m(a+b)=g_m(g_m(a)+g_m(b))$ (This matches with the 1st property.)

I don't get the second line

[sowmitra]

I think that I have also become confused....
Here, since,
$a\equiv c_1(\text{mod}m)$, and, $b\equiv c_2(\text{mod}m)$
So, $g_m(a)=g_m(c_1)$, and, $g_m(b)=g_m(c_2)$.
Now, $a+b\equiv c_1+c_2(\text{mod}m)\Rightarrow g_m(a+b)=g_m(c_1+c_2)$

[Tahmid Hasan]

$a+b\equiv c_1+c_2(\text{mod}m)\Rightarrow g_m(a+b)=g_m(c_1+c_2)$

How can you get $g_m(a)+g_m(b)=g_m(g_m(a)+g_m(b))$ from there? It is still not clear to me.

[sowmitra]

@Tahmid: Sorry for making the solution so complex. Here's a different approach:

We need to find all the different values that $f_{2012}(x)$ can have (which is the range of $f_{2012}$).

For starters, we need to show that, $\displaystyle f_m(x)=f_m(f_m(x))$.
This is quite straight-forward, for, if we assume,
\[P(a,b)\Rightarrow f_m(a+b)=f_m(f_m(a)+f_m(b))\]
\[\text{Then, } P(x,0)\Rightarrow f_m(x+0)=f_m(f_m(x)+f_m(0))\Rightarrow f_m(x)=f_m(f_m(x))\]
Now, firstly, $f_{2012}(0)=0$, and, for $1\leq k \leq 2011$, $f_{2012}(k)=f_{2012}(k)$.
Again, $f_{2012}(2012)=f_{2012}(2012\times 1)=0$ ;

And, $f_{2012}(2013)=f_{2012}(1+2012)$
$=f_{2012}(f_{2012}(1)+f_{2012}(2012))=f_{2012}(f_{2012}(1))=f_{2012}(1)$.

Similarly,$f_{2012}(2014)=f_{2012}(2)$, $f_{2012}(2015)=f_{2012}(3)$,......, and thus, the function repeats periodicaly.
Therefore, the elements in the range of $f_{2012}$ are: $rangef_{2012}=\{0,f_{2012}(1), f_{2012}(2)......f_{2012}(2011)\}$.
So,\[\displaystyle \mid rangef_{2012}\mid=\boxed{2012}.\]

[Tahmid Hasan]

Yes, that's quite close to my solution .
My approach was after proving $f_m(f_m(x))=f_m(x) \forall x \in \mathbb{N}_0$
We show $f_m(km+c)=f_m(c)[k \in \mathbb{N},0 \le c<m]$
which implies $f_m$ is a periodic function with period $m$. So the range of $f_m$ is $m$ 'at most'.
[$f_m$ can be a constant fuction too ].

[kfoozminus]

Suppose, "Probability of opened door $A$"= $P_{A}$
then the probability will be $P_A \cdot (P_B+P_C+P_D \cdot P_E)$

first, i tried it same way too, but it gives 1.2 as answer!
but, i'm not sure about sourav's solution too...