$Exactly$ $one$ or $at$ $least$ $one$?
Search found 73 matches
- Fri Feb 02, 2018 10:43 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2017 National Round Secondary 7
- Replies: 17
- Views: 19777
Re: BdMO 2017 National Round Secondary 7
- Thu Feb 01, 2018 7:55 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO 2017 National round Secondary 6
- Replies: 9
- Views: 8238
Re: BDMO 2017 National round Secondary 6
OK. A very big misunderstanding. I didn't see it was either a octagon or a octahedron!
- Fri Dec 01, 2017 2:07 pm
- Forum: Junior Level
- Topic: BDMO 2016: National Junior/2
- Replies: 5
- Views: 8908
Re: BDMO 2016: National Junior/2
Capture.PNG Let $CF'G'$ be the reflection of $C'FG$. And as $CDE$ is the reflection of $C'DE$, we can say, $F'G'||DE||AB$ So, $(CF'G')=(C'FG)=80$. (ABC)=$\frac{1}{2}AB.CP=500$ or, $10CP=500$ or,$CP=50$ We know, $\triangle ABC~\triangle CF'G'$ $\frac {AB}{F'G'}=\frac {CP}{CQ}=k$ So, $F'G'=\frac{AB}{...
- Thu Nov 30, 2017 11:26 pm
- Forum: Secondary Level
- Topic: coxbazar divisional 2014 Q.no.-1
- Replies: 1
- Views: 2832
Re: coxbazar divisional 2014 Q.no.-1
It can be solved in the way of Physics! The velocity of Kamrul $a$ m/s The velocity of Tusher $4a$ m/s The velocity of Avik $8a$ m/s After a limited time $t$, Kamrul, Tusher and Avik passed $at$ m, $4at$ m and $8at$ m respectively. Then, $8at-at=150$ $7at=105$ $13at=\frac{105*13}{7}$ As the total pa...
- Mon Nov 27, 2017 12:04 am
- Forum: Secondary Level
- Topic: Sylhet - 2014
- Replies: 2
- Views: 2970
Re: Sylhet - 2014
Let's find out the sum of some consecutive numbers greater than 976 with the lowest difference. That is:
$1+2+3+...+44=(41*44)/2=990$
990-976=14 ; the sum of the missing page number.
Look, $14=2+12=2+3+9=2+3+4+5$ ; that we want.
So, there were 4 pages missing.
[Maybe it is the easiest way ]
$1+2+3+...+44=(41*44)/2=990$
990-976=14 ; the sum of the missing page number.
Look, $14=2+12=2+3+9=2+3+4+5$ ; that we want.
So, there were 4 pages missing.
[Maybe it is the easiest way ]
- Thu Jan 26, 2017 10:08 pm
- Forum: Divisional Math Olympiad
- Topic: BDMO Divisional_2014
- Replies: 2
- Views: 2695
Re: BDMO Divisional_2014
You can't say EF||BC only to see that EF=BC/2. Because E and F mayn't be the midpoints. Thanks
- Thu Jan 26, 2017 8:47 pm
- Forum: Junior Level
- Topic: COMBINATORICS!!!
- Replies: 4
- Views: 3746
Re: COMBINATORICS!!!
We assume that a number of 7 digits where at least 3 digits are same. As for example, if we take such a number, n = 111222 and the last digit will be 1 or 2. So, total combination: $\frac{7!}{3!4!}$ for the last digit 1 and same for 2. So, like (1,2), there are {(1,3),(1,4),(1,5),...,(1,9)} and 70 c...
- Tue Jan 24, 2017 10:47 pm
- Forum: Junior Level
- Topic: COMBINATORICS!!!
- Replies: 4
- Views: 3746
Re: COMBINATORICS!!!
Not understand. In the second case, I see 7 has appeared once. So, where each digiit is appearing thrice (at least)?
- Tue Jan 24, 2017 10:42 pm
- Forum: Divisional Math Olympiad
- Topic: BDMO Divisional_2014
- Replies: 2
- Views: 2695
BDMO Divisional_2014
The area of ABC and OBC triangle is 120 and 24 respectively. BC=16, EF=8. Find out the area of OEAF Quadrilateral.
- Sun Jan 22, 2017 8:11 pm
- Forum: Divisional Math Olympiad
- Topic: Chittagong Regional 2014
- Replies: 1
- Views: 2426
Re: Chittagong Regional 2014
We assume that the 6-digit integer is: abcdef 11|abd and 11|cef where, c is not equal to d. Number of such 3-digit integer divisible by 11 is: (90-10+1)=81 If abd={110, 220, 330, 440, 550, 660, 770, 880, 990} where d=0, c must be different from d each time. For each of this abd, there are 81 choices...