Some problems of last year divisionals, I need help for

[SANZEED]

@Tiham vaia,the larger triangle $ABC$ is obtuse, and that means its circumcenter is outside. But we are not working on the circumcenter of $ABC$,rather on the circumcenter of $\triangle DEF$.

[sowmitra]

I think there is a bug in the problem involving $"ababab"$...

[SANZEED]

I think there is a bug in the problem involving "$ababab$"...

I suppose you're right.

[nafistiham]

@Tiham vaia,the larger triangle $ABC$ is obtuse, and that means its circumcenter is outside. But we are not working on the circumcenter of $ABC$,rather on the circumcenter of $\triangle DEF$.

oops... you are right

and in the ababab problem, I have just copied from english. The bengali version says to find $a/b$

[sowmitra]

Here's what I think is wrong with the problem:
Let, $N=ababab$. Then,
\[\displaystyle N=ab\times10101=ab\times3\times7\times13\times37\]
So, $N$ has at least $4$ distinct primes in its factorization.

Again, $60$ cannot be expressed as a product of more than $4$ numbers (namely, $2\times2\times3\times5$). This means that there can't be more than $4$ primes in the factorization of $N$. Because, if there are, then, $60$ would be a product of $5$ or more numbers, each $\geq2$, which is impossible.

Therefore, $N$ has exactly $4$ primes in its factorization, which are : $3,7,13,37$.
Now, let,
\[\displaystyle N=p_1^{\alpha_1}\times p_2^{\alpha_2}\times p_3^{\alpha_3}\times p_4^{\alpha_4}, \text{where}, p_i\in \{3,7,13,37\}\]
So, no. of divisors of $N$, $\tau(n)=(\alpha_1+1)(\alpha_2+1)(\alpha_3+1)(\alpha_4+1)=2\times2\times3\times5$.
Equating, we get, $(\alpha_1,\alpha_2,\alpha_3,\alpha_4)=\{(2-1),(2-1),(3-1),(5-1)\}=\{1,1,2,4\}$
WLOG, assume that, $\alpha_1=1, \alpha_2=1, \alpha_3=2, \alpha_4=4$.

Therefore, $N=p_1^1\times p_1^1\times p_3^2\times p_4^4
\Rightarrow ab\times(3\cdot7\cdot13\cdot37)=(p_3\cdot p_4^3)\times(p_1\cdot p_2\cdot p_3\cdot p_4)$
Since,$p_1\times p_1\times p_3\times p_4= 3\times7\times13\times37$. So, we get, $ab=p_3\times p_4^3$.

Now, the least value of $ab$ is $3^3\times7=189$, which is not a two-digit number, and thus, a contradiction.

[SANZEED]

If $ababab$ is indicating a $6$ digit number, then it's surely wrong. If you multiply 60 different integers, you are not going to get such small number.

You are not multiplying $60$ integers here. Just $60$ divisors. And even $442368$has $60$ divisors.
And yes...Sowmitra vaia is right.That is the problem here I found.

[Fahim Shahriar]

@SANZEED. I could understand that just after submitting. That's why I removed that instantly in few seconds. But I wonder how you quoted that.

There was a bug in that question and also one still walking in my head.

[nafistiham]

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

Dhaka Secondary 2012

$5.$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?

[This problem has been stated in another post in this forum. Still, I'm posting it. As far as I remember in বিনোদন গনিত by রমজান আলী সরকার, a general solution has been stated. Unfortunately, the book is not with me, now. I believe, you'll be able to solve it.]


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

The figure may not be as per scale

90'=90'.png (3.97KiB)Viewed 4935 times

$9.$ 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 $f_{2012}$ .


Solve these, and there are more.....

[protik]

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?

[protik]

Assuming That $F$ is the midpoint. If We draw the circumcircle of $\triangle ABC$, $AC$ and $DE$ will be the Diameter of the circle. As $FE$ is perpendicular to $BC$, $\widehat {BE} = \widehat {EC} = \frac {\widehat {BC}}{2}$ . so, $\angle BAC=2 \cdot \angle BDE$.
$\frac {(AB \sin \angle BAC)}{(BD \sin \angle BDE)}=\sqrt {3}$. and $\frac {AB}{BD} = \frac {\cos \angle BAC}{\cos \angle BDE}$ .
If I didn't made any mistake from here we will get $\angle BDE=15 ^ {\circ}$.

Sorry for any typo and Latex(which I can't write well.) Please some one edit my latex if possible.