## Search found 86 matches

Mon Jan 23, 2017 8:18 am
Topic: Divisional MO 2015
Replies: 4
Views: 886

### Re: Divisional MO 2015

See, $9800 = 2^3 \times 5^2 \times 7^2$ Let the two numbers will be $p = 2^a \times 5^b \times 7^c$ and $q = 2^x \times 5^y \times 7^z$. Now consider the pairs of $(a,x)$ which are possible: $(0,3), (1,3),(2,3),(3,3), (3,0), (3,1), (3,2),(1,2),(2,1)$ , numbers of pairs is $9$. Again, similarly the p...
Thu Jan 19, 2017 6:54 pm
Forum: Secondary Level
Topic: Dhaka Regional '16 P8
Replies: 3
Views: 960

### Re: Dhaka Regional '16 P8

I have to solve it like this!?
Thu Jan 19, 2017 2:55 am
Forum: Secondary Level
Topic: Dhaka Regional '16 P8
Replies: 3
Views: 960

### Dhaka Regional '16 P8

How many eight digit number can be formed by using the digits $1, 2, 3, 4, 5, 6, 7, 8$ so that each number has $6$ digits in such place where that digit is less than the next digit? Example: In number $2314$; $2, 1$ are two digits such that each of them is less than the next digit. Translated into B...
Thu Jan 19, 2017 2:13 am
Topic: National BDMO 2016 : Junior 8
Replies: 4
Views: 1265

### Re: National BDMO 2016 : Junior 8

Solution: Take $X,Y$ reflections of $B$ about $P,Q$ respectively and $Z$ reflection of $C$ about $P$.Now see that $\triangle ZYC$ is equilateral triangle, so $ZY=ZC=ZD$, $Z$ is circumcenter of $\triangle CXY$, thus $\angle CXY=30^\circ$, but $PQ$ is midline of $\triangle CXY$, so $PQ\parallel XY$, t...
Thu Jan 19, 2017 2:00 am
Forum: Site Support
Topic: How to prepare and study for BdMO in Secondary Group
Replies: 3
Views: 1849

### Re: How to prepare and study for BdMO in Secondary Group

Sun Dec 25, 2016 11:35 am
Forum: Number Theory
Topic: USAMO 2015 P5
Replies: 2
Views: 736

### USAMO 2015 P5

Let $a, b, c, d$ be distinct positive integers such that $a^4 + b^4 = c^4 + d^4 = e^5$. Show that $ac + bd$ is a composite number.

Can't solve it properly . Any help will be appreciated!

Thanks You!
Fri Dec 23, 2016 9:18 pm
Forum: Algebra
Topic: Blended with equations
Replies: 1
Views: 436

Given that, $ab+bc =130 ... ... ... ... ...$ (i) $bc+ca =168 ... ... ... ... ...$ (ii) $ca+ab =228 ... ... ... ... ...$ (iii) Sum this three equation and you will get, $2 (ab + bc + ca ) = 586$ $\Rightarrow ab + bc + ca = 263 ... ... ... ... ...$ (iv) From (i) and (iv) we get, $130 + ca = 263$ $... Tue Dec 20, 2016 6:19 pm Forum: Site Support Topic: Problem! Help anybody! Replies: 4 Views: 1252 ### Re: Problem! Help anybody! Yes, it is working now. THANK YOU , Vaia. Tue Dec 20, 2016 2:47 pm Forum: Combinatorics Topic: Boy and Girls! Replies: 0 Views: 639 ### Boy and Girls! There are$n$boys and$n+1$girls standing in a straight line in some order (from left to right). A group of girls is a sequence of at least two neighboring girls with no girl immediately to the left and no girl immediately to the right of it. In a move, we take the leftmost group of girls and make... Tue Dec 20, 2016 11:21 am Forum: Number Theory Topic: Pretty Diophantine Equation Replies: 2 Views: 909 ### Pretty Diophantine Equation Find all pairs of integers$ (x,y)\$, such that

$x^2 - 2009y + 2y^2 = 0$