@saurav205 said:@bodhi_vriksha was pondering over this question for the last 10 mins...good that you pointed it out..P.S. guys please see that you post correct questions..it wastes a lot of time trying questions that are not even correct....
@Faruq said:Find Remainder when 888222888222. . . . . up to 9235 digits divided by 53
@saurav205 said:3 aayega ..cycle check kiya...3^0/13 r=13^1/13 r=33^2/13 r =-43^3/13 r =1..the repeats so on...hence 1+3+4 + 1+3+4 +1+3+4 +......till 201/3 terms8*67/1316/13 , so R=3OA kya hai?
@saurav205 said:3 aayega ..cycle check kiya...3^0/13 r=13^1/13 r=33^2/13 r =43^3/13 r =1..the repeats so on...hence 1+3+4 + 1+3+4 +1+3+4 +......till 201/3 terms8*67/1316/13 , so R=3OA kya hai?
@grkkrg said:Question is correct. It is just out of our reach.
@Koushik98 said:In how many ways can 1000 be written as a sum of 'n' consecutive natural numbers, where 'n' isgreater than 1?(a) 0 (b) 1 (c) 2 (d) 3
@bodhi_vriksha said:Ok let me explain you from scratch. Here we go...As i already explained earlier that there could be two cases: odd number of terms and even number of terms.I dont think you have any issues understanding with odd number of factors. Still there is an important point to make here. The odd factors of 1000 are 1,5,25, and 125. We cannot take n as 125 because then the consecutive numbers will not be natural numbers and will involve some negative integers as well. Also, we are supposed to consider odd factors greater than 1. So only valid values of n are 5 and 25.When there are even number of terms, since these numbers are consecutives and since the middle term must be the average of two middle terms, which are consecutive terms, hence the middle term is a rational number of the form X.5The least divisor, i.e. n, of 1000, such that we get the quotient in the form X.5 is 16. The we could also have other values of n as 80 and 400 such that the quotient are 12.5 and 2.5 respectively. But the problem with the latter values of n =80 and 400 is that we will not be able to consecutive natural numbers as the nubers will extend towards negative side on the number line. So, the only valid value of n is 16.So there is not trial and error, but pure logic. Hope you get a transparent picture now
@amresh_maverick said:OA = 676f(x/(2x+3)) =3x, find f(1)
@Koushik98 said:There are 140 students in a school. The number of students who play Cricket, Football and Hockeyare 50, 80 and 70 respectively. The ratio of the number of students who play more than one of thethree sports to the number of students who play all the three sports is 3 : 2. If each student of theschool plays at least one of the three sports, then how many students play exactly two of the threesports?(a) 12 (b) 14 (c) 16 (d) 20
@Koushik98 said:There are 140 students in a school. The number of students who play Cricket, Football and Hockeyare 50, 80 and 70 respectively. The ratio of the number of students who play more than one of thethree sports to the number of students who play all the three sports is 3 : 2. If each student of theschool plays at least one of the three sports, then how many students play exactly two of the threesports?(a) 12 (b) 14 (c) 16 (d) 20
@Koushik98 said:There are 140 students in a school. The number of students who play Cricket, Football and Hockeyare 50, 80 and 70 respectively. The ratio of the number of students who play more than one of thethree sports to the number of students who play all the three sports is 3 : 2. If each student of theschool plays at least one of the three sports, then how many students play exactly two of the threesports?(a) 12 (b) 14 (c) 16 (d) 20
@Koushik98 said:There are 140 students in a school. The number of students who play Cricket, Football and Hockeyare 50, 80 and 70 respectively. The ratio of the number of students who play more than one of thethree sports to the number of students who play all the three sports is 3 : 2. If each student of theschool plays at least one of the three sports, then how many students play exactly two of the threesports?(a) 12 (b) 14 (c) 16 (d) 20
@Koushik98 said:There are 140 students in a school. The number of students who play Cricket, Football and Hockeyare 50, 80 and 70 respectively. The ratio of the number of students who play more than one of thethree sports to the number of students who play all the three sports is 3 : 2. If each student of theschool plays at least one of the three sports, then how many students play exactly two of the threesports?(a) 12 (b) 14 (c) 16 (d) 20
Question :
@Koushik98 said:There are 140 students in a school. The number of students who play Cricket, Football and Hockeyare 50, 80 and 70 respectively. The ratio of the number of students who play more than one of thethree sports to the number of students who play all the three sports is 3 : 2. If each student of theschool plays at least one of the three sports, then how many students play exactly two of the threesports?(a) 12 (b) 14 (c) 16 (d) 20
