Search found 57 matches
- Mon Jun 05, 2017 2:31 pm
- Forum: Divisional Math Olympiad
- Topic: 2017 Regional no.9 Dhaka
- Replies: 11
- Views: 17556
Re: 2017 Regional no.9 Dhaka
Check the parity of the equation and the rest should be clear.
- Mon Jun 05, 2017 2:09 pm
- Forum: News / Announcements
- Topic: MPMS Problem Solving Marathon
- Replies: 11
- Views: 27051
Re: MPMS Problem Solving Marathon
Simplier solution to problem $1$,I used this after getting to know what Wilson's Theorem is from dshasan's post.Let, $a$=$1$ . $2$...... $2k$.Telescope $a^2$ as $1$ . $2$.... $2k$.($-2k$)....($-2$)($-1$).From the telescoping,it follows $a^2$ $\equiv$ $1$.$2$....$2k$($p-2k$)....($p-2$)($p-1$) (mod $p...
- Mon Jun 05, 2017 1:47 pm
- Forum: News / Announcements
- Topic: MPMS Problem Solving Marathon
- Replies: 11
- Views: 27051
Re: MPMS Problem Solving Marathon
$Problem 4$: If $7^m$+$11^n$ is a perfect square,then $m$ must be even and $n$ must be odd. (By using mod $3$ and mod $4$);let $m$=$2q$ and $n$=$2k$+$1$ and $49^q$+$121^k.11$=$a^2$ $\Rightarrow$ ($x$+$7^q$)($x$-$7^q$)=$121^k$.$11$.Now make a total of $2$ cases along with $6$ subcases each.$1$ case w...
- Mon Jun 05, 2017 1:17 pm
- Forum: News / Announcements
- Topic: MPMS Problem Solving Marathon
- Replies: 11
- Views: 27051
Re: MPMS Problem Solving Marathon
$Alternative Solution to 3$:It is obvious that if $4n^2$-$6n$+$45$ is a perfect square,then $4n^2$-$6n$+$45$ $\equiv$ $1$ (mod $2$) $\Rightarrow$ $4n^2$-$6n$+$45$ $\equiv$ $1$ (mod $4$) $\Rightarrow$ $1$-$6n$ $\equiv$ $1$ (mod $4$) $\Rightarrow$ $n$ $\equiv$ $0$ (mod $2$);so $n$ cannot be odd.
- Mon May 29, 2017 3:53 pm
- Forum: News / Announcements
- Topic: MPMS Problem Solving Marathon
- Replies: 11
- Views: 27051
Re: MPMS Problem Solving Marathon
For problem 1,if we substitute $a^2$=$4k$ where $k$ is an integer,we can easily find that $p$ |$a^2$+$1$.
- Mon May 29, 2017 12:34 am
- Forum: Asian Pacific Math Olympiad (APMO)
- Topic: APMO 2017 P2
- Replies: 2
- Views: 8539
Re: APMO 2017 P2
Let the reflection of $DM$ over $M$ be $D'M$.Thus we concurr that $AD'BD$ is a parallelogram as $M$ is designed to be the midpoint of $AB$.Let the midpoint of $AC$ be $X$.Thus we have $AX$=$CX$; $ZX$ as a common side of triangles $ZXA$ and $ZXC$;$\angle ZXA$=$\angle ZXC$.So,by $SAS$,we have $\bigtri...
- Tue Apr 25, 2017 4:10 pm
- Forum: Number Theory
- Topic: JBMO:NT
- Replies: 1
- Views: 2430
JBMO:NT
Consider the equation:$2^x$.$3^y$-$5^w$.$7^z$=$1$;$x$, $y$, $z$, $w$ are non negative integers.
- Sat Apr 22, 2017 12:55 pm
- Forum: Number Theory
- Topic: USAJMO/USAMO 2017 P1
- Replies: 3
- Views: 4186
Re: USAJMO/USAMO 2017 P1
When $a,b$ € $N$ and it follows that $a^x$=$b^y$,there exists $t$ € $N$ such that $a$=$t^k$,$b$=$t^q$.This lemma can be quite handy in this problem.
- Wed Apr 19, 2017 6:05 pm
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies: 68
- Views: 45635
Re: Beginner's Marathon
Problem $17$ : Let $\bigtriangleup$ $ABC$ be an acute triangle.$D$ is the foot of perpendicular drawn from $C$ on $AB$.Let the bisector of $\angle$ $ABC$ intersect $CD$ at $E$ and $\bigcirc$ $ADE$ at $F$.If $\angle$ $ADF$ =45˚,prove that $CF$ is tangent to $\bigcirc$ $ADE$.
- Tue Apr 18, 2017 9:26 pm
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies: 68
- Views: 45635
Re: Beginner's Marathon
The proof is trivial for some $n$ which has the form $2k$ where $k$ is an integer.Now,to say,this problem is equivalent to proving that the number of reducible fractions is even when $n$ is odd because the aforementioned sequence has precisely 'n-1' fractions(I.g:even numbered fractions).Now,we take...