## The Wizard's Three clocks - Russian Math Olympiad.

A typical challenging question once asked in one of the Russian Math Olympiads, meant for Grade-5 and Grade-6. Give it a try !!
After you are done ,take the “Reader’s Challenge

The Problem :
Once there was a wizard named Symonds. He owns three ancient clocks of perpetual motion. Everyday , Clock #1 runs 20 minutes fast; Clock #2 runs 30 minutes fast, and Clock  #3 runs 40 minutes fast. He sets all three clocks to the correct time and leaves to visit his uncle. After many days, he returns home and, to his surprise , finds all 3 clocks showing the correct time again !
What is the fewest possible number of days Symonds could have spent away from home ?

The Solution :
At first glance , the problem may seem too difficult, specially for a student of grade-5 or 6 , but solving this problem requires the  logical application of the concept of LCM . A regular practice of such tricky mathematics only can prepare the potential little champs for the “REAL MATH OLYMPIADS”, winning where, gives you the real PRIDE.
Coming to the problem,
Considering the fact that the ancient clocks do not show AM or PM.
Minutes in 12 hours = 12 hrs × 60 mins per hr = 720 mins
Clock #1 gains 20 minutes in a day , so it will show correct time again after = (720/20) days = 36 days
Similarly,  Clock #2 gains 30 minutes in a day, so it will show correct time after = 720/30 = 24 days
Finally, Clock #3 gains 40 minutes in a day, so it will show correct time after = 720/40 = 18 days. Now, we can say that all the three clocks will show same time and that’s too correct when all the clocks’ coincide latest by the LCM of 36, 24 and 18  days.
Working out the LCM of 36, 24 and 18 we get it’s 72.
So, the fewest possible number of days after which all the three clocks show correct time is 72.
Huh !! I hope I have been able to explain the solution. If you have any easier approach to the problem , please let me know in the comment section. I will definitely add that to this blog.