## Search found 55 matches

- Mon Apr 25, 2016 11:31 pm
- Forum: Number Theory
- Topic: Integer solution
- Replies:
**5** - Views:
**596**

### Re: Integer solution

SOLUTION 1: From the equation, we can easily see, x & y are of different parity. So, Case 1 : At first,we let, x = 2p and y=2q+1 then, from equation, $4q^2$+4q +1 = $8p^3$ + 7 => $4(q^2 +q)$ = 4($ 2p^3$ +1) +2 .Here, 4 divides LHS but not RHS.Contradiction! Case 2: Here, x=2p+1 ,y=2q as 2nd case. W...

- Sat Apr 23, 2016 4:16 pm
- Forum: Number Theory
- Topic: International Zhautykov Olympiad 2005, Problem 6
- Replies:
**2** - Views:
**592**

### Re: International Zhautykov Olympiad 2005, Problem 6

$pq$ divides $p^2q^2 +8(p^2 +q^2)+64$ so, $pq$ divides $8(p^2 +q^2 +8)$ so, $p$ divides $q^2$ and $q$ divides $p^2$ Ahmed Ittihad ,Your approach is wrong.your start was statisfactory indeed; but in your 3rd line, u made a mistake. look, u told in ur 3rd line that ($p$ divides $q^2$ and $q$ divides ...

- Wed Apr 20, 2016 5:49 pm
- Forum: Combinatorics
- Topic: Binomials
- Replies:
**3** - Views:
**751**

### Re: Binomials

[I assume,u understand case1,as u wanted case2's explaination.].. Case2 mainly tells that,u've to take xyz such that the product is not a multiple of 9, but a multiple of 3(as case2,we find out when the product is neither multiple of 3 nor 9 in case1).So, we r now finding out the quantity of product...

- Wed Apr 06, 2016 4:12 pm
- Forum: Combinatorics
- Topic: Combi-Spanish olympiad_1985
- Replies:
**2** - Views:
**437**

### Re: Combi-Spanish olympiad_1985

It's easy to prove by induction.after base case, let ,2^n divides (n+1).....(2n).(inductive hypothesis). Then,simply prove 2^(n+1) or,(2^n)*2 divides{(n+2).(n+3)....(2n).2(n+1)}. You can see easily,(2^n)*2 divides {2.(n+1).(n+2).......(2n)}[from inductive hypothesis] .So,(n+1)(n+2)...(2n) is divisib...

- Mon Apr 04, 2016 10:57 pm
- Forum: Number Theory
- Topic: Tuymaada 2008,Junior League,D2,P8
- Replies:
**2** - Views:
**526**

### Re: Tuymaada 2008,Junior League,D2,P8

From the ques.,a1+a2+a3+a4-t≡0 (mod 23).so,a1+a2+a3+a4≡t(mod 23). The residue classes of modulo 23 is{0,1,2,.....,22}.so,it can be, t≡0/1/2/3/..../22(mod 23)(***1). Here we see,the integers can't exceed 501.so one residue class among(0,1,2,...,22)must be of the 21 integers not exce...