Search found 327 matches

by sm.joty
Sat Dec 29, 2012 1:55 pm
Forum: International Olympiad in Informatics (IOI)
Topic: DP-নিয়ে সমসসা-১
Replies: 2
Views: 1336

DP-নিয়ে সমসসা-১

দেখে বুঝা যায় এইটা একটা DP প্রব্লেম। কিন্তু এখনও তো রিকারসনই বানাইতে পারলাম না। :( কেউ একটু সাহায্য করেন।

http://www.lightoj.com/volume_showprobl ... oblem=1032
by sm.joty
Fri Dec 14, 2012 8:48 pm
Forum: National Math Olympiad (BdMO)
Topic: ২০১৩ সালের জন্য সমস্যা!!(১)
Replies: 15
Views: 3080

Re: ২০১৩ সালের জন্য সমস্যা!!(১)

মাসুম ভাই,
১ নাম্বারে
"Consider a regular convex polygon marked with n− vertices dividing into n"

কি হবে ?
$(n-1)$ ??????
by sm.joty
Thu Dec 06, 2012 6:57 pm
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 126
Views: 23660

Re: Secondary and Higher Secondary Marathon

New problem please . :D
by sm.joty
Wed Dec 05, 2012 6:31 pm
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 126
Views: 23660

Re: Secondary and Higher Secondary Marathon

No One post any problem !!!!!!
Ok ,so I think, I can post one. :)
Problem - $\boxed {14}$
Let $n$ is a natural number.How many solution are there in ordered positive integers pairs $(x,y)$ to the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{n} ?$
Source : Putnam 1960
by sm.joty
Tue Dec 04, 2012 1:05 pm
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 126
Views: 23660

Re: Secondary and Higher Secondary Marathon

Solution - $13$ (may be I'm correct :? ) At first proof a nice lemma, the product of all divisors of a number $N$ is "$N^{k/2}$" where $k$ is the number of divisors. if we consider only the proper divisors, then according to the condition we get, $\frac{N^{k/2}}{N}=N$ so $k=4$ Now $k=4=(3+1)=(1+1)(1...
by sm.joty
Mon Dec 03, 2012 1:55 pm
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 126
Views: 23660

Re: Secondary and Higher Secondary Marathon

Nadim Ul Abrar wrote:If the Ans is $90$ . Then I'm ready to post my solution .. :?
joty ভাই , confirmation দেন ।
Good job, Nadim. :D
Ans is $90$
Now post your solution and a new problem :)
And Shahrier, you also can post your solution
by sm.joty
Thu Nov 29, 2012 1:43 pm
Forum: Algebra
Topic: Floors and roots
Replies: 4
Views: 958

Re: Floors and roots

you can check one by one. but the another way is that to check only squire number. Now notethat, $x=4$ implies $\lfloor \sqrt {x +15} \rfloor - \lfloor \sqrt x \rfloor ═1$ because $16<19<25$ so $ 4 <\sqrt { 4+15} <5$ now checking one by one squire number you will see that for $x=49$ you can calculat...
by sm.joty
Wed Nov 28, 2012 8:09 pm
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 126
Views: 23660

Re: Secondary and Higher Secondary Marathon

Problem $\boxed {12}$
How many different $4 \times 4$ arrays whose entries are all $1$'s and $-1$'s have the property that the sum of entries in each row is $0$ and sum of entries in each column is $0$

Source: AIME 1997
by sm.joty
Wed Nov 28, 2012 8:00 pm
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 126
Views: 23660

Re: Secondary and Higher Secondary Marathon

we know $1+2+3+...+n = \frac{n(n+1)}{2}$ Let, $S=1^k+2^k+3^k+...+n^k$ We know that for any odd $n$ $a+b|a^n+b^n$ so $1+(n-1)|1^k+(n-1)^{k}$ $2+(n-2)|2^k+(n-2)^{k}$ . . . so all terms of this sequence is divisible by $n$ Now, $1+(n+1)-1|1^k+n^k$ $2+(n+1)-2|2^k+(n-1)^k$ . . . so $S$ is also divisible ...
by sm.joty
Tue Sep 18, 2012 3:24 pm
Forum: Combinatorics
Topic: Putnam, 1962: Summing Binomials
Replies: 9
Views: 2313

Re: Putnam, 1962: Summing Binomials

using second derivative can give a nice proof. :)