i understand. in this method if (2n+1) is a prime then i guess d=(2n+1) am i right ?
thanks.
Search found 56 matches
- Sun May 01, 2011 1:32 am
- Forum: Number Theory
- Topic: Secondary Special Camp 2011: NT P 1
- Replies: 20
- Views: 11343
- Sun May 01, 2011 1:21 am
- Forum: Asian Pacific Math Olympiad (APMO)
- Topic: APMO 1999-2
- Replies: 6
- Views: 4873
Re: APMO 1999-2
the thing i understand is find a largest integer suppose $n^3$ which divisible by all integer less than n means all 1~n numbers are factor of $n^3$. then n=420 or $n^3$=420 not correct value. what u mean ?
- Sat Apr 30, 2011 10:27 pm
- Forum: Higher Secondary Level
- Topic: factorials
- Replies: 4
- Views: 3723
Re: factorials
i try to prove last non-zero digit of n! not periodic suppose 1~n, last non-zero digit of n!is periodic and last non-zero digits are L1,L2,...Ln. suppose A=n!,B=(n+1)!,C=(n+1) and last non-zero digit of A=Ln,B=L1,C=c1. so B=A.C and L1=last digit of (Ln.c1). every time when Ln come as last digit at ...
- Sat Apr 30, 2011 3:16 am
- Forum: Higher Secondary Level
- Topic: find 90 degree Angle in a Clock
- Replies: 13
- Views: 8653
Re: find 90 degree Angle in a Clock & Equation
my answer also $22$ let make an equation start from 12'o clock. difference in d between h and m prick is 0 degree. after 1hour 330 degree in clockwise. it will work as mod. after difference 360 degree it will round again. difference d increase with time same relation as gravity increase with mass. l...
- Sat Apr 30, 2011 2:40 am
- Forum: Higher Secondary Level
- Topic: factorials
- Replies: 4
- Views: 3723
Re: factorials
Answer=6. actually after some factorial last digit of number multiple by last digit of next number. example suppose $x!=A$ and $a$ is last non-zero digit of $A$ and $(x+1)!=B$ and $b$ is last nonzero digit of $B$ and $c$ is last digit of $(x+1)$. finally i say $b$ is last digit of $a.c$ i try it and...
- Fri Apr 29, 2011 5:05 pm
- Forum: Asian Pacific Math Olympiad (APMO)
- Topic: APMO 1999-2
- Replies: 6
- Views: 4873
Re: APMO 1999-2
Answer=5. $n!$ divisor of $n^3$. sterlimg's formula\[n!~\sqrt{2\left ( pi \right )n}\left ( n/e \right )^{n}\] then \[n>\left ( n/e \right )^{n}\]
\[n>\left ( n/e \right )^{\left ( n-2 \right )}\]
\[e^{n-2/n-3}>n\]
\[n>\left ( n/e \right )^{\left ( n-2 \right )}\]
\[e^{n-2/n-3}>n\]
- Fri Apr 29, 2011 1:19 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Higher Secondary 2011/1
- Replies: 7
- Views: 5723
Re: BdMO National Higher Secondary 2011/1
first x+(x+3)=(x+1)+(x+2)=2x+3 is true for every x.
then all 4n numbers can divide into two group of equal sum.
n+(n+3)=(n+1)+(n+2).....(n+m)+(n+m+3)=(n+m+1)+(n+m+2).....(4n-3)+4n=(4n-2)+(4n-1)
then all 4n numbers can divide into two group of equal sum.
n+(n+3)=(n+1)+(n+2).....(n+m)+(n+m+3)=(n+m+1)+(n+m+2).....(4n-3)+4n=(4n-2)+(4n-1)
- Fri Apr 29, 2011 12:56 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Primary 2011/4
- Replies: 4
- Views: 4440
Re: BdMO National Primary 2011/4
a=16, minimum value of a is 16.
- Fri Apr 29, 2011 12:51 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National Primary 2011/2
- Replies: 2
- Views: 3226
Re: BdMO National Primary 2011/2
2011 is smallest common prime factor of both given number. $gcd=2011$.
- Fri Apr 29, 2011 12:41 am
- Forum: Secondary: Solved
- Topic: Rangpur Secondary 2011/2
- Replies: 5
- Views: 11063
Re: Rangpur Secondary 2011/2
if f=3 and g=-9 then -27<fg<-3