Someone is hired to watch the owners dog. When will he watch it again?
asked Feb 17 in least common multiple by anonymous

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Let's assume that they both go on business trips the first time on the first day. That establishes a reference point. Charlie will go again in 6 months' time, then 12, then 18. Dasha will go again in 9 months' time, then 18. So every 18 months they will need someone to look after the dog.

But if Dasha goes on business 1 month after Charlie, their schedules will never clash. Using Charlie as the reference point 0, he goes next at month 9, 18, 27, 36 etc, while Dasha goes at month 1, 7, 13, 19, 25, 31 etc. The series are 9C and 6D+1 for Charlie and Dasha where C and D are integers. They next go on business at the same time when 9C=6D+1. C=(6D+1)/9=(9D-3D+1)/9=D+(1-3D)/9. Since C must be an integer and (1-3D) has values 1, -2, -5, -8, -11 (=-9-2), -14 (=-9-5), -17 (=-9-8), ... 9 will never divide into 1-3D. So the business trips won't ever clash. A similar argument applies if there's 2 months initially between trips.

Note that 18 is the LCM of 6 and 9. 

If the time between trips is initially 3 months then the cycle of 18 months applies, because (6D+3)/9=(2D+1)/3 and D=1, 4, 7 satisfies the requirement, gaps of 3 periods of 6 months, since 3*6=18.

answered Feb 17 by Rod Top Rated User (478,380 points)
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