How did you find $a \equiv 11 (mod 13)$ ?
Did you use divisibility rule ? or what?
Search found 138 matches
- Fri Feb 15, 2013 4:27 pm
- Forum: Number Theory
- Topic: 13th Power
- Replies: 4
- Views: 2982
- Fri Feb 15, 2013 12:39 pm
- Forum: Introductions
- Topic: PERSONAL ANNOUNCEMENT
- Replies: 3
- Views: 5389
Re: PERSONAL ANNOUNCEMENT
Hello, :P I'm Shakkhor. No one here knows me personally. From the very beginning I had interest in Math. But I started reading other Math books except text book 3 years ago when I participated in Math Olympiad for the first time. Number Theory is also my favorite but weak in Geo. @Adib. I liked how ...
- Fri Feb 15, 2013 12:02 pm
- Forum: Number Theory
- Topic: 13th Power
- Replies: 4
- Views: 2982
13th Power
$21982145917308330487013369$ is the 13^{th} power of which positive integer ?
Na..Na. Don't use calculator.
Na..Na. Don't use calculator.
Re: a+b=1
$(a)$
$\frac {a+b}{2} \geq \sqrt {ab}$
$\frac {1}{4} \geq ab$
$\frac {a^8+b^8}{2} \geq \sqrt {(ab)^8}$
$\frac {a^8+b^8}{2} \geq (ab)^4 $
$\frac {a^8+b^8}{2} \geq \frac {1}{256}$
$7(a^8+b^8) \geq \frac {7}{128}$
Similarly, $8(a^7+b^7)= \frac{1}{8}$
Now we have to substract.
$\frac {a+b}{2} \geq \sqrt {ab}$
$\frac {1}{4} \geq ab$
$\frac {a^8+b^8}{2} \geq \sqrt {(ab)^8}$
$\frac {a^8+b^8}{2} \geq (ab)^4 $
$\frac {a^8+b^8}{2} \geq \frac {1}{256}$
$7(a^8+b^8) \geq \frac {7}{128}$
Similarly, $8(a^7+b^7)= \frac{1}{8}$
Now we have to substract.
- Tue Feb 12, 2013 7:27 pm
- Forum: Secondary Level
- Topic: Number Theory Proof please
- Replies: 1
- Views: 2525
Re: Number Theory Proof please
$a=be$ and $c=de$
$\frac {a-c}{b-d} = \frac {be-de}{b-d} = \frac {e(b-d)}{b-d} = e$
$\frac {a-c}{b-d} = \frac {be-de}{b-d} = \frac {e(b-d)}{b-d} = e$
- Wed Feb 06, 2013 11:09 pm
- Forum: Divisional Math Olympiad
- Topic: Some problems of last year divisionals, I need help for
- Replies: 36
- Views: 20529
Re: Some problems of last year divisionals, I need help for
$cos ABC = cos 60° = \frac {BC}{AB}$. $BC=0.5$
$(90° - \angle ACD) + (60° - \angle DBE) = 90°$
$\angle DBE = 60° = \angle BCE$.
$\triangle EBC$ is equilateral. So $AE=BE=BC=0.5$
$(90° - \angle ACD) + (60° - \angle DBE) = 90°$
$\angle DBE = 60° = \angle BCE$.
$\triangle EBC$ is equilateral. So $AE=BE=BC=0.5$
- Wed Feb 06, 2013 10:35 pm
- Forum: Secondary Level
- Topic: A Ge0metry
- Replies: 2
- Views: 2601
Re: A Ge0metry
Similar Problem:
http://www.matholympiad.org.bd/forum/vi ... 833#p11833
http://www.matholympiad.org.bd/forum/vi ... 833#p11833
- Wed Feb 06, 2013 10:19 pm
- Forum: Secondary Level
- Topic: A Ge0metry
- Replies: 2
- Views: 2601
Re: A Ge0metry
There is something wrong. I think $D$ point is on $AC$ and $E$ point is on $AB$.
- Mon Feb 04, 2013 12:41 pm
- Forum: Secondary Level
- Topic: Combinatorics
- Replies: 5
- Views: 7123
Re: Combinatorics
Have a look.
6 Heads, No Tail
5 Heads, 1 Tail
4 Heads, 2 Tails
.........
No Head , 6 Tails
It's $7$. For $n$ indistinguishable fair coins, $(n+1)$ outcomes.
6 Heads, No Tail
5 Heads, 1 Tail
4 Heads, 2 Tails
.........
No Head , 6 Tails
It's $7$. For $n$ indistinguishable fair coins, $(n+1)$ outcomes.
- Sat Feb 02, 2013 2:01 am
- Forum: National Math Olympiad (BdMO)
- Topic: Let us help one another preparing for BdMO national 2013
- Replies: 12
- Views: 13265
Re: Let us help one another preparing for BdMO national 2013
I only calculated on my mind- When I expand it, $1+ ^{2n} C_2 * 2^2 + ^{2n} C_2 * 2^4 +..$ Here every $2$ has even number as power. Which means it is either $(1or-1) \equiv (mod 5)$ Then I had a look on Pascal's triangle. The sum of the coefficients is $2^{n-1}$. Here my problem starts. While doing ...