Search found 592 matches
- Sat Dec 11, 2010 8:08 pm
- Forum: Asian Pacific Math Olympiad (APMO)
- Topic: Easiest problem of Apmo
- Replies: 18
- Views: 12838
Re: Easiest problem of Apmo
মাসুম- সমস্যাকে সহজ/কঠিন বলে বিশেষায়িত না করাই ভাল, কারণ ফোরামে বিভিন্ন দক্ষতার ছাত্র-ছাত্রী থাকবে, তারা এতে নিরুৎসাহিত বোধ করতে পারে। . I meant it easy with respect to APMO,not at all from me. :) Now I am posting solutions using different ideas. First solution(my,using Fermat's Method Of Descent ...
- Fri Dec 10, 2010 11:01 pm
- Forum: Number Theory
- Topic: Largest value of $n$
- Replies: 5
- Views: 4252
Re: Largest value of $n$
I didn't understand the sign $||$ here, because it is used to mean:If $p^a||n$,then $p^{a+1}$ does not divide $n$
However,it was a problem from AIME may be and simialr to my post"find the largest $x$'
$n+10|n^3+10$ and $n^3+100$,then $n+10|990,n_{max}=980$
However,it was a problem from AIME may be and simialr to my post"find the largest $x$'
$n+10|n^3+10$ and $n^3+100$,then $n+10|990,n_{max}=980$
- Fri Dec 10, 2010 10:44 pm
- Forum: Secondary Level
- Topic: Number Theory(congruance)
- Replies: 10
- Views: 8243
Re: Number Theory(congruance)
We use the fact $a^3\equiv 0,\pm 1\ (mod\ 7)$
If none of $x,y,z$ divisible by $7,$then we have:
$x^3+y^3\equiv \pm(1+1)\equiv \pm 2\ (mod\ 7)$ but $z^3$ is not $\equiv \pm 2\ (mod\ 7)$.Contradiction
If none of $x,y,z$ divisible by $7,$then we have:
$x^3+y^3\equiv \pm(1+1)\equiv \pm 2\ (mod\ 7)$ but $z^3$ is not $\equiv \pm 2\ (mod\ 7)$.Contradiction
- Fri Dec 10, 2010 12:01 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2005-4
- Replies: 10
- Views: 9174
Re: Imo 2005-4
Because never is $gcd(a_m,a_n)=1,$ I think there exists neither $n$ nor $a_n$ satisfying the condition.
- Thu Dec 09, 2010 11:57 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2005-4
- Replies: 10
- Views: 9174
Re: Imo 2005-4
But Zzzz,what do you mean by 'determine all positive integers'-$n$ or $a_n$ Probably the problem statement should be: Determine all positive integers such that they are coprime to all the terms of this sequence \[a_n=2^n+6^n+3^n-1^n, \forall n \in N\] আর যে এই সমস্যার সমাধান একবার দেখছে তার ভোলার কথ...
- Thu Dec 09, 2010 8:53 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: Imo 1964-1
- Replies: 2
- Views: 3156
Imo 1964-1
Prove that $n^4+4^n$ is not a prime for $n>1$
- Thu Dec 09, 2010 8:29 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: Imo 1959-1
- Replies: 3
- Views: 3323
Imo 1959-1
Prove that the fraction $\dfrac {21n+4} {14n+3}$ is irreducible.
- Thu Dec 09, 2010 8:24 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: Imo 2010-1
- Replies: 3
- Views: 3573
Imo 2010-1
Find all functions $f:\mathbb R \to \mathbb R$ such that for all $x,y\in \mathbb R$ $f([x]y)=f(x)f([y])$ where $[a]$ denots the greatest integer less than or equal $a$
- Thu Dec 09, 2010 6:38 pm
- Forum: Number Theory
- Topic: On the sum of divisors
- Replies: 2
- Views: 3552
Re: On the sum of divisors
Post your full solution. Here is my solution: Since $p$ prime,ofcourse,$n=q^{p-1}$ for $q$ prime. First solution:The divisors are $1,q,q^2,...,q^{p-2}$ and there sum is $\frac {q^{p-1}-1} {q-1}$ which is less than $q^{p-1}$.So the sum of some divisors can't be equal to the number. Second solution:Le...
- Thu Dec 09, 2010 3:21 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2005-4
- Replies: 10
- Views: 9174
Re: Imo 2005-4
Thanks.Now I am 17 .And this is the easiest problem of Imo(according to me).Just note that $a_n$ is even for all $n$.So $gcd(a_m,a_n)$ is at least $2$,therefore never coprime.