LCM

LCM (15, 20, 30) = 60 => They meet at the starting point after every 60 min, i.e., after every 1 hour. Therefore, they will meet at the starting point for the fourth time after 4 hours, i.e., at 12:00:00 pm.

117 = 3 x 3 x 13 As the numbers are co-prime, HCF = 1. So, the numbers have to be 9 and 13. 9² = 81 13² = 169 Therefore, required answer = 250

Product of numbers = LCM x HCF = 11 x 385 = 4235 Let the numbers be of the form 11m and 11n, such that 'm' and 'n' are co-primes. => 11m x 11n = 4235 => m x n = 35 => (m,n) can be either of (1, 35), (35, 1), (5, 7), (7, 5). => The numbers can be (11, 385), (385, 11), (55, 77), (77, 55). But it is given that the numbers cannot differ by more than 50. Hence, the numbers are 55 and 77. Therefore, sum of the two numbers = 55 + 77 = 132

Let 'h' be the HCF of the two numbers. => The numbers are 5h and 7h. We know that Product of Numbers = LCM x HCF => 5h x 7h = 105 x h => h = 3 So, the numbers are 15 and 21. Therefore, difference of their squares = 21² - 15² = 441 - 225 = 216

LCM (8, 16, 24, 32) = 96 7/8 = 84/96 15/16 = 90/96 23/24 = 92/96 31/32 = 93/96 Hence, 31/32 is the largest of all.

1st number = 3x
2nd number =5x
LCM of 3x and 5x is 15x
=> 15x = 75
=> x = 5
sum = 15+25 =40

say, 1st number =3x
2nd number =2x
LCM of numbers = 6x
given LCM = 60
=> x6 = 60
=>x = 10

Let the numbers be 2x, 3x and 4x LCM = 12x 12x=240 ⇒x=20 H.C.F of 40, 60 and 80=20

LCM = 180

LCM of 4,5,6,7 is 420
=>(420k+3) should be divisible by 9
if k =1, 423/9 remainder !=0