Search found 85 matches
- Fri Nov 09, 2012 11:14 pm
- Forum: News / Announcements
- Topic: Active users for marathon
- Replies: 23
- Views: 17502
Re: Active users for marathon
ok i'm in to...probably in nt section.
- Wed Oct 24, 2012 10:43 pm
- Forum: Number Theory
- Topic: Smart divisor
- Replies: 5
- Views: 4252
Re: Smart divisor
i wanted to say
if a divides n and a+b-1 divides n then a divides a+b-1 as a+b-1>or=a
working on equation editor.
if a divides n and a+b-1 divides n then a divides a+b-1 as a+b-1>or=a
working on equation editor.
- Wed Oct 24, 2012 9:45 pm
- Forum: Number Theory
- Topic: Smart divisor
- Replies: 5
- Views: 4252
Re: Smart divisor
i'm not being able to make my eq editor work.
if $n \equiv 0 \bmod a,0 \bmod b,0 \bmod a+b-1$ then $a+b-1 \equiv 0 \bmod a$ and $a+b-1 \equiv 0 \bmod b$, $\gcd(a,b)=1$.so $a$ divides $b-1$ and $b$ divides $a-1$.a contradiction unless $a=b=1$.works for all $n$.
if $n \equiv 0 \bmod a,0 \bmod b,0 \bmod a+b-1$ then $a+b-1 \equiv 0 \bmod a$ and $a+b-1 \equiv 0 \bmod b$, $\gcd(a,b)=1$.so $a$ divides $b-1$ and $b$ divides $a-1$.a contradiction unless $a=b=1$.works for all $n$.
- Tue Oct 23, 2012 8:05 pm
- Forum: Number Theory
- Topic: USAMO 2003
- Replies: 1
- Views: 2137
USAMO 2003
prove that for every positive integer n there exists an n digit number divisible by 5^n all of whose digits are odd.
- Sat Jun 23, 2012 2:52 pm
- Forum: Number Theory
- Topic: APMO-2001
- Replies: 2
- Views: 2613
Re: APMO-2001
i have a confusion. :? :? let \[n=10a_{1}+a_{0},s(n)=a_{1}+a_{0},9T(n)=9a_{1}\] if stump means the integer which can be found by removing digits from r.h.s. of d.r. then why is only the left coefficient,multiplied by 9 gives the ans?why not the base?does stump mean the sum of left coefficients only?...
- Sat Jun 16, 2012 3:20 pm
- Forum: Number Theory
- Topic: $x^5-x=k^2$
- Replies: 4
- Views: 3511
Re: $x^5-x=k^2$
for even,both x^2+1 and x^2-1 have to be perfect square,which is a contradiction again.
- Thu Jun 14, 2012 12:35 am
- Forum: Combinatorics
- Topic: sum
- Replies: 8
- Views: 6502
Re: sum
of course the notation denotes ans is 16.bt the provlem was not to find that.bt it was given in the book.
- Thu Jun 14, 2012 12:33 am
- Forum: Combinatorics
- Topic: sum
- Replies: 8
- Views: 6502
Re: sum
actualyy i think i should've explained it.the first problem was to find s.the ans is 64.
the second one was to find the sum of all integers that can be formed from the set.i tried it doing with nt which did work.the ans is 117856.how can i do it with combi?
the second one was to find the sum of all integers that can be formed from the set.i tried it doing with nt which did work.the ans is 117856.how can i do it with combi?
- Tue Jun 12, 2012 9:21 pm
- Forum: Number Theory
- Topic: $x^5-x=k^2$
- Replies: 4
- Views: 3511
Re: $x^5-x=k^2$
let,\[x(x^4-1)=y^2\]
\[x\equiv 1(modk),x^4-1\equiv -1(modk),gcd(x^4-1,x)=1\]
\[(x^2-1)(x^2+1)=a^2,x=b^2\]
\[(x^2-1)/2.(x^2+1)/2=(a/2)^2\]
\[gcd(x^2+1)/2,(x^2-1)/2=1\]
both of the forms can't form square integers.so there are no such integers but 1.
\[x\equiv 1(modk),x^4-1\equiv -1(modk),gcd(x^4-1,x)=1\]
\[(x^2-1)(x^2+1)=a^2,x=b^2\]
\[(x^2-1)/2.(x^2+1)/2=(a/2)^2\]
\[gcd(x^2+1)/2,(x^2-1)/2=1\]
both of the forms can't form square integers.so there are no such integers but 1.
- Tue Jun 12, 2012 8:28 pm
- Forum: Combinatorics
- Topic: sum
- Replies: 8
- Views: 6502
sum
let s={1,3,5,7}
find\[\sum_{n \in s}n\]
the sol is given in P.t.c. but i'm unable to understand it. can anyone give a brief sol please?
find\[\sum_{n \in s}n\]
the sol is given in P.t.c. but i'm unable to understand it. can anyone give a brief sol please?