That means the required number is 18 more than the L.C.M.nquired number = 360 +18= 378Exercise 4.5(v) 35 and 1091. Using prime factorisation, find the LCM of:20,35 ( 42.86 (l) 12,30,81(v) 220, 440,(iv) 120,502. Find the LCM of the following by inspection :4 and 16 (1) 7 and 28 (ii) 24 and 72 (iv) 36 and 72 (v) 35 an3. Find the LCM of the following by inspection :2x3 and 3 x 2 ) 2x 3 x 5 and 2 x 34. Using the common division method, find the LCM. of:(iii) 22 x 7 x 5' and( 54,81(iii) 224, 280, 336*5 and 2 x 7'() 117, 221Find the greatest -digit number which is eactly divisible by 12, 16, 20 and 24.Find the least digit number which on dividing by 6, 9, 12 and 18 leaves a remau&emainder 31. Find the smallest number acuy divisible by 12,20,24 and 32Find the smallest number wuch divided by 15, 20 and 51and 25 leaves a remainder 5 inein each orProperties and Relationship of H.C.F. and L.C.M.rally observed the following properties while doing prevtibet is to greater than any of the numbers.numbers is not less than any of the numbers.doing previous exerciseYou would have maturally observed the following1. The HC.F. of given number is tot greater tha2 The LCM of given numbers is not less than3. HCF. of two co-prime taumbers is 1.LCM of two or more come trumbersli a number is a factor the HCHCF of two numbers is always a fa7. The product of HC andThe HCFXLCMThe LCM and HCF.then find the second thumbco primetimes is the productof aand bis d and their LCM isis always a factor of thew.CMand two wambers is equal to the produAxleb er Second rumberHCE. Tambers are 180 and 6 respecthuice product of the nunnumbersFirst number SecondSecondorespectively. If first52NN-XLC6180