## Search found 327 matches

Sat Dec 29, 2012 1:55 pm
Forum: International Olympiad in Informatics (IOI)
Topic: DP-নিয়ে সমসসা-১
Replies: 2
Views: 3821

### 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
Fri Dec 14, 2012 8:48 pm
Topic: ২০১৩ সালের জন্য সমস্যা!!(১)
Replies: 15
Views: 6250

### Re: ২০১৩ সালের জন্য সমস্যা!!(১)

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

কি হবে ?
$(n-1)$ ??????
Thu Dec 06, 2012 6:57 pm
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 127
Views: 57558

### Re: Secondary and Higher Secondary Marathon

Wed Dec 05, 2012 6:31 pm
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 127
Views: 57558

### 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 $\frac{1}{x}+\frac{1}{y}=\frac{1}{n} ?$
Tue Dec 04, 2012 1:05 pm
Forum: Higher Secondary Level
Topic: Secondary and Higher Secondary Marathon
Replies: 127
Views: 57558

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... Mon Dec 03, 2012 1:55 pm Forum: Higher Secondary Level Topic: Secondary and Higher Secondary Marathon Replies: 127 Views: 57558 ### 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. Ans is$90$Now post your solution and a new problem And Shahrier, you also can post your solution Thu Nov 29, 2012 1:43 pm Forum: Algebra Topic: Floors and roots Replies: 4 Views: 2061 ### 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$you can calculat... Wed Nov 28, 2012 8:09 pm Forum: Higher Secondary Level Topic: Secondary and Higher Secondary Marathon Replies: 127 Views: 57558 ### Re: Secondary and Higher Secondary Marathon Problem$\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 Wed Nov 28, 2012 8:00 pm Forum: Higher Secondary Level Topic: Secondary and Higher Secondary Marathon Replies: 127 Views: 57558 ### 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$na+b|a^n+b^n$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,$1+(n+1)-1|1^k+n^k2+(n+1)-2|2^k+(n-1)^k$. . . so$S\$ is also divisible ...
Tue Sep 18, 2012 3:24 pm
Forum: Combinatorics
Topic: Putnam, 1962: Summing Binomials
Replies: 9
Views: 4333

### Re: Putnam, 1962: Summing Binomials

using second derivative can give a nice proof.