Question #738f7

2 Answers
Dec 6, 2017

#24 " minutes"#

Explanation:

#lcm(6,8) = 24#
(least common multiple)
#lcm(a,b) = (a*b)/gcd(a,b)#
#=> lcm(8,6) = (8*6)/gcd(8,6)#
#and gcd(8,6) = gcd(6,2) = 2 #
# "because" gcd(a,b) = gcd(a-b, b) if a > b #
(greatest common divisor)
#=> lcm(8,6) = 48/2 = 24#

Dec 6, 2017

We need to calculate LCM of the two given timings: #6and8# min.

factors of #6->ul2xx3#
factors of #8->ul2xx2xx2#
LCM#=2xx3xx2xx2# (#ul2# is included once)
LCM#=24# min