Problems on H.C.F and L.C.M
In a school, the number of students in Sections A , B and C of 10th class are 70,98 and 126 respectively. Due to overload of student in each class , the administration wants to increase the number roo
The number students in each room = HCF of 70,98 and 126 =14 In each room , maximum 14 students can be seated. Total number students = 70 + 98 +126 =294 Number of rooms required = 294/14 =21
The Largest number of four digits is 9999. Required number must be divisible by L.C.M. of 12,15,18,27 i.e. 540. On dividing 9999 by 540,we get 279 as remainder . Required number = (9999-279) = 9720.
The least number which when divided by 4,5,6,8 and 10 is the LCM of these numbers. But each time to get 3 as remainder, we have to add the remainder 3 to the obtained LCM.LCM of (4, 5, 6, 8, 10)+ 3=120+3=123
Let the numbers be 37a and 37b.Then, 37a x 37b = 4107 ab = 3.Now, co-primes with product 3 are (1, 3).So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111). Greater number = 111.
EXAMIANS