@albiesriram said:
(9x²Sin²x + 4)/(xsinx) = 9xSinx + 4/(xSinx)
Here significance of 0 9xSinx and 4/(xSinx) are positive
This means we can use AM ≥ GM, as product of 9xSinx and 4/(xSinx) is 36 (a constant)
Using this we will get the least value as 12
@albiesriram said:One more geometry ,
This question went unsolved yesterday... @scrabbler @chillfactor
two more counting problems..
1)
@albiesriram said:two more counting problems..1)
1st 115?
Either I pick one each from 3 different sides (which can be done in 1*2*3 + 1*2*4 + 1*3*4 + 2*3*4 = 50 ways) or else 2 from 1 side and one from any other (2C2 * 8 + 3C2 * 7 + 4C2 * 6 = 65 ways) total 50 + 65 = 115.
Previous problem ka kuch idea nahin hai 😃 Will try it again after a while....
regards
scrabbler
Either I pick one each from 3 different sides (which can be done in 1*2*3 + 1*2*4 + 1*3*4 + 2*3*4 = 50 ways) or else 2 from 1 side and one from any other (2C2 * 8 + 3C2 * 7 + 4C2 * 6 = 65 ways) total 50 + 65 = 115.
Previous problem ka kuch idea nahin hai 😃 Will try it again after a while....
regards
scrabbler
@albiesriram I think we can not add all the numbers in that digit by that summation formula of 'n' natural numbers. Like the addition of (1,2,3,4,.....,9,1,0,1,1,1,2,1,3,.... ) can not be determined by that addition formula.
"In triangle ABC, the angle subtended by side BC at the orthocenter is 100 degrees, find the angle subtended by BC at the incenter."
Can I have a diagram too, please.
@albiesriram said:two more counting problems.. 1)
Total number of ways = Taking two points from each side + Taking one point from each side = {8+3C2*7+4C2*6} +{2*3*(4+1) + 1*4*(2+3)}= 115
Team BV - Vineet
@veertamizhan said:"In triangle ABC, the angle subtended by side BC at the orthocenter is 100 degrees, find the angle subtended by BC at the incenter."Can I have a diagram too, please.
Do you have an OA? I am getting 100 only...fig thoda time lagega...in office :(
Edit: getting 130!
regards
scrabbler
Edit: getting 130!
regards
scrabbler
@Gsathe89 said:find the remainder when 123456789101112.....40 is divided by 36???/
4a=9b+1=> 0=(b+1)(mod4) (since a is an integer)=>min(b)=3. Hence the remainder =9*3+1=28
Team BV - Vineet
Q: 107^109^101 div 111 R?
@albiesriram said:two more counting problems..1)
10C3-4C3-3C3=115
with 2 and 1 dot triangle is not possible hence not req to consider
with 2 and 1 dot triangle is not possible hence not req to consider
@soham701 said:@albiesriram I think we can not add all the numbers in that digit by that summation formula of 'n' natural numbers. Like the addition of (1,2,3,4,.....,9,1,0,1,1,1,2,1,3,.... ) can not be determined by that addition formula.
We are going to split 12345678910 like
1|2|3|4|5|6|7|8|9|10|11|12| .. and not by single digits.
divisibility rule for 9 allows such partition. coz (10^n-1) is divisible by 9.
we can partition a number according to tour need in case of 9.
@viewpt said:Q: 107^109^101 div 111 R?
107 ??
111 = 3*37
with 3 rem is 2
and with 37 it is 33
3k+2 = 37m + 33
gives 107
111 = 3*37
with 3 rem is 2
and with 37 it is 33
3k+2 = 37m + 33
gives 107
@veertamizhan said:@scrabbler no I don't have OA(?)the answer is 125 saar.
Getting 130 actually. See attached. Miscalculated last time. 125 to nahin aa raha hai kisi bhi angle se :(
regards
scrabbler
regards
scrabbler
@amresh_maverick said:107 ??111 = 3*37with 3 rem is 2and with 37 it is 333k+2 = 37m + 33gives 107
hw did u get 33 as rem wid 37?
@viewpt said:Q: 107^109^101 div 111 R?
E(111)=36
109^101 mod 36
=(108+1)^101 mod 36
=1 mod 36
=> 107^109^101 mod 111
=107^1 mod 111