Some problems of last year divisionals, I need help for
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Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
@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 $\triangle DEF$.
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Re: Some problems of last year divisionals, I need hel for
I think there is a bug in the problem involving $"ababab"$...
Re: Some problems of last year divisionals, I need hel for
I suppose you're right.sowmitra wrote:I think there is a bug in the problem involving "$ababab$"...
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Re: Some problems of last year divisionals, I need hel for
oops... you are rightSANZEED wrote:@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 $\triangle DEF$.
and in the ababab problem, I have just copied from english. The bengali version says to find $a/b$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Some problems of last year divisionals, I need hel for
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.
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.
Re: Some problems of last year divisionals, I need hel for
You are not multiplying $60$ integers here. Just $60$ divisors. And even $442368$has $60$ divisors.Fahim Shahriar wrote: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.
And yes...Sowmitra vaia is right.That is the problem here I found.
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Re: Some problems of last year divisionals, I need hel for
@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.
There was a bug in that question and also one still walking in my head.
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College
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Re: Some problems of last year divisionals, I need hel for
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 $\angle BDE$.
$9.$ 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.....
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 $\angle BDE$.
$9.$ 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.....
Last edited by sourav das on Thu Jan 17, 2013 11:48 am, edited 2 times in total.
Reason: Edited a typo
Reason: Edited a typo
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
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Re: Some problems of last year divisionals, I need hel for
Isn't F is the midpoint of AC and DE?nafistiham wrote: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 $\angle BDE$.
Re: Some problems of last year divisionals, I need hel for
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.
$\frac {(AB \sin \angle BAC)}{(BD \sin \angle BDE)}=\sqrt {3}$. 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.
Last edited by sourav das on Thu Jan 17, 2013 12:12 pm, edited 1 time in total.
Reason: $LAT^{E}X_{E}D$
Reason: $LAT^{E}X_{E}D$