Search found 138 matches
- Sun Feb 24, 2013 1:21 pm
- Forum: Number Theory
- Topic: Perfect Square
- Replies: 6
- Views: 5012
Re: Perfect Square
Let $2x^3+9=y^2$ $2x^3=(y+3)(y-3)$ Surely $y$ is odd. Then $(y+3)$ and $(y-3)$ both will be even. Take $(y-3)=2m$. $2x^3=2m(2m+6)$ $x^3=2m(m+3)$ Note that $x$ is even. So both of $m$ and $(m+3)$ are powers of 2. Suppose, $m=2^a$ and $m+3=2^b$ $2^b-2^a=3$ It implies $a$ is 0. So $m=1$. Only solution ...
- Sat Feb 23, 2013 1:23 am
- Forum: Divisional Math Olympiad
- Topic: A Easy One
- Replies: 5
- Views: 4321
Re: A Easy One
(Not from Junior) but I want to say that 14 is not correct.
$240 = 2^4 \times 3 \times 5$
+1 their power and multiply. It will give you the number of factors as well as the answer of actual question.
$(4+1)(1+1)(1+1)=20$
$240 = 2^4 \times 3 \times 5$
+1 their power and multiply. It will give you the number of factors as well as the answer of actual question.
$(4+1)(1+1)(1+1)=20$
- Sat Feb 23, 2013 1:06 am
- Forum: Number Theory
- Topic: Divisibility
- Replies: 3
- Views: 2622
Re: Divisibility
$m^2+m+n^2+n-3$ $=(m+4)(m-3)+(n+4)(n-3)+21$ Take $(m+4)=a$ and $(n+4)=b$. If $a(a-7)+b(b-7)$ is divisible by 7, it will also be divisible by 49. But 49 does not divide 21. So 7|(a,b) are not acceptable here. $a^2+b^2-7(a+b)+21$ $n^2 \equiv 1,2,4$ It implies $(a^2+b^2)$ can't be divisible by 7. So th...
- Thu Feb 21, 2013 10:59 pm
- Forum: Combinatorics
- Topic: Probability :Two gamblers
- Replies: 3
- Views: 3576
Re: Probability :Two gamblers
You're right. I thought it to be a two round game. Even I missed the first round. Then I have to calculate the probability for every possible rounds. It's full of calculations but not difficult. Isn't there any one liner solution ? I saw someone giving the solution in just one line in fb but can't r...
- Thu Feb 21, 2013 2:29 am
- Forum: Combinatorics
- Topic: Probability :Two gamblers
- Replies: 3
- Views: 3576
Re: Probability :Two gamblers
A few months ago I couldn't solve that one. But now I can.
Learning...
Prob of $A = \dfrac {5}{36} = \dfrac {180}{36^2}$
Prob of $B = \dfrac {6}{36} \times \dfrac {31}{36} = \dfrac {186}{36^2}$
So $B$ has better chance.
Learning...
Prob of $A = \dfrac {5}{36} = \dfrac {180}{36^2}$
Prob of $B = \dfrac {6}{36} \times \dfrac {31}{36} = \dfrac {186}{36^2}$
So $B$ has better chance.
- Thu Feb 21, 2013 12:15 am
- Forum: Social Lounge
- Topic: Quadratic Equation
- Replies: 4
- Views: 4696
Re: Quadratic Equation
For the equation $ax^2+bx+c=0$, $x = \frac {-b \pm \sqrt {b^2-4ac}}{2a}$
[$M=b^2-4ac$]
When M>0 ; it has two real roots.
When M<0 ; it has two complex roots.
When M=0 ; its two roots become same.
[$M=b^2-4ac$]
When M>0 ; it has two real roots.
When M<0 ; it has two complex roots.
When M=0 ; its two roots become same.
- Wed Feb 20, 2013 1:30 pm
- Forum: Junior Level
- Topic: Equilateral Triangle
- Replies: 3
- Views: 3780
Re: Equilateral Triangle
I used trigonometry. Draw perpendiculars $PM$ & $PN$ on $AB$ & $CD$ respectively. Let the side of the square $x$. $PM+PN=x$. In $\triangle PBM$ $tan PBM = \frac {PM}{BM}$ $2 - \sqrt {3} = \frac {2PM}{x}$ $2 - \sqrt {3} = \frac {2x-2PN}{x}$ $PN = \frac {\sqrt {3}x}{2}$ In $\triangle PCN$ $tan PCN = \...
- Wed Feb 20, 2013 1:38 am
- Forum: Number Theory
- Topic: IMO Shortlist 2005 N6
- Replies: 8
- Views: 5139
Re: IMO Shortlist 2005 N6
$b^n+n \equiv 0 (mod a^n+n)$ $b^n \equiv -n (mod a^n+n)$ $b^n \equiv a^n (mod a^n+n)$ $b^n-a^n \equiv 0 (mod a^n+n)$ $(b^n-a^n)=(b-a)(....)$ $(b-a)$ will be divisible by $(a^n+n)$ for any value of $n$. Then $(b-a)$ has infinite factors which implies $b-a=0$. Am I correct? I think the 2nd part isn't....
- Sun Feb 17, 2013 8:49 pm
- Forum: Junior Level
- Topic: Welcome to congruence
- Replies: 8
- Views: 5811
Re: Welcome to congruence
$36! \equiv -1 (mod 37)$ $31 \equiv -6 (mod 37) $ $31^2 \equiv -1 (mod 37)$ $(31+1)(31-1)=31^2-1 \equiv -2 (mod 37)$ $(31+2)(31-2)=31^2-4 \equiv -5 (mod 37)$ ......... $(31+5)(31-5)=31^-25 \equiv 11 (mod 37)$ $25!*(-6)*(-2)*(-5)*(-10)*20*11 \equiv -1 (mod 37)$ $25! * 21 \equiv -1 (mod 37)$ $25! * 21...
- Fri Feb 15, 2013 4:46 pm
- Forum: Number Theory
- Topic: 13th Power
- Replies: 4
- Views: 2974
Re: 13th Power
$10^{13} < a^{13} < 100^{13}$. So $a$ is a $2$ digit number. Taking (mod 10) we can observe that $a$'s last digit is $9$. We can write $a=10k-1$ ;where $k$ is an integer & $2 \leq k \leq 10$. $(10k-1)^{13} = (10k)^{13} -.....+ 13*10k*1-1$ We will now work on last $2$ digits. $130k-1 = \_\_ 69$ $k=9$...