Question

find the smallest number which when divided by 24,36 and 64 leaves 4 as remainder in each case.​

Answer 1

24 = 2³ × 3

36 = 2² × 3²

64 = 2^6

LCM = product of prime factors with highest power

☛ LCM = 2^6 × 3²

☛ LCM = 64 × 9

☛ LCM = 576

Required number which when divided by these numbers gives remainder 4 in each case :

===> LCM - 4

===> 576 - 4

===> 572

Hence, the required number is 572