Hey i think the answer would be 3^17 only as u have extracted the powers of 3 so x and y can not be 3 or a multiple of 3 so in that case they wont be divisible by 3 hence 3^17
I would also like to ask divyaakanksha to post the solution or answer to the above problem in case she has it other wise we can ask quant devils like implex or made_for_iims people who can confirm the asnwers but i think it is fine
lets c what others have to say
There is a possibility of 3^10*x-y being divisible by 3 '=For eg: 5-2=3 has 3 as a factor or 101-2 =99 has 3^2 as factors
There is a possibility of 3^10*x-y being divisible by 3 '=For eg: 5-2=3 has 3 as a factor or 101-2 =99 has 3^2 as factors
see i think chance of 3^10*x-y being divisible by 3 is zero as (3^10*x-y) mod 3 have (0-y mod 3) as the first term is already divisible by 3 hance until and unless the value of y is not a multiple of 3 it cannot divide the complete term.
juz c this approch, from fact58 - fact38 lets take fact 38 common which have 3's power as17 and in remaining factor, some multiple of three -1 will be the number so ans is 17.
gives remainder and {.} denotes the fractrional part of that. The fractional part is of the form (0.bx). The value of x could be
A) 2 B) 4 C) 6 D) 8 E) None of these
ans is 2 3^32 is of the form 4n.hence the unit's digit would b 1.on dividing 3^32 by 50 u'll get d last digit of remainder always as 1.hence wen v divide 1 by 50 takin decimal the quotient would always b 2. so whatever d value of 'b' ,' x' would always b 2.
Q1 denotes the greatest integer value just below x and {x} its fractional value.the sum of ^3 and {x}^2 is -7.91.find x?
Q2 find the 28383rd term of the series:123456789101112....... ???
Q3 some birds settled on the branches of a tree.first,they sat one to a branch and there was one bird too many.next they sat two to a branch and there was one branch too many.how many branches were there??
Q1 denotes the greatest integer value just below x and {x} its fractional value.the sum of ^3 and {x}^2 is -7.91.find x?
Q2 find the 28383rd term of the series:123456789101112....... ???
Q3 some birds settled on the branches of a tree.first,they sat one to a branch and there was one bird too many.next they sat two to a branch and there was one branch too many.how many branches were there??
Q1) well the answer should have been x=-1.03(which is not in the options) -7.91 = -8 + 0.09(since we have to write it as sum of a cube & a square) -8 = (-2)^3 and 0.09=(0.3)^2 which means x= -1.03 (since =-2 & {-1.03}=0.03) Q3) Let Br - branches , Bi - birds From 1st statement we get Bi - 1 = Br. ----> 1 From 2nd statement we get Br= (Bi/2) +1 ------>2
I dont want to be posting questions who's solutions have already being figured out but everything here is presented in a haphazard way. Does any body have all the chapterwise solutions & explanations in a file format?
Neways i had a doubt from Number System. M posting the question:
32/9 will give remainder as 5. nw, 5^32^32/9 => 25^16^32/9 25/9 gives remainder as 7. nw 7^16^32/9 => 49^8^32/9 49/7 gives remainder as 4. 4^256/9 will give final remainder as 4. hence ans is 4. i hope its clear..though its pretty lenghty..
arun,bikas and chetakar have a total of 80 coins among them.arun triples the number of coins with the others by giving them some coins from his own collection. next,bikas repeats the same process.after this bikas now has 20 coins.find the number of coins he had at the beginning. a) 11 b) 10 c) 9 d) 12
arun,bikas and chetakar have a total of 80 coins among them.arun triples the number of coins with the others by giving them some coins from his own collection. next,bikas repeats the same process.after this bikas now has 20 coins.find the number of coins he had at the beginning. a) 11 b) 10 c) 9 d) 12
Hey can you please recheck the question. See this is how I approached the problem and I am getting the answer as 20. Check out my method and point out any mistakes that you find.
Let us consider that arun(A) ,bikas(B) and chetakar(C) each have a, b, c respectively initially
after A gives the coin to B and C. B and C will each have 3 times what they had initially and A that much less. A: a - 2b - 2c B: 3b C: 3c
After B gives the coins, A: 3(a - 2b - 2c ) B: 3b - 2*(3c) - 2*(a - 2b - 2c ) -------(1) C: 3*(3c)