The Romankachi Clock Problem !!!
Math brain teasers require computations to solve.
I always remember the days when I was at high school, whiling away my time with my friends. Those were the days when I'd spent the maximum time with my grandfather Rudolf the worldfamous Romankachi. He'd come over to our house at one of the weekends leaving behind his Romankachi estate in the village in capable hands. One day when I returned home from school, he put this clock problem in front of me which I was required to solve in the alloted time frame.
Considering time represented in twelve hour format H:M1M2 where H is a singledigit or a twodigit number indicating the hours and M1 and M2 the digits indicating the minutes (M1 representing the tens digit of the minutes and M2 the units digit of the minutes), he wanted me to find out that in a day how such instances would occur so that H is exactly divisible by M1 and gives M2 as quotient?
He further added that if H:M1M2 was the time represented in 24hour format instead of a 12hour format, then how many instances were there in a day such that H is exactly divisible by M1 giving M2 as the quotient? In the 24 hour format, 12 o' clock midnight is represented as 00:00.
Answer
Clearing the scope of the problem, we have to divide H by M1 getting M2 as the quotient with remainder as 0.
First let us consider the time format to be a 12hour format.
Here H can take values from 1 to 12.
For M1 = 0, no value exists since divisor itself is 0.
For M1 = 1, M2 takes values from 1 to 9 i.e., 9 values, for H = 1 to 9.
For M1 = 2, M2 takes values from 1 to 6 i.e., 6 values, for H = 2, 4, 6, 8, 10 and 12.
For M1 = 3, M2 takes values from 1 to 4 i.e., 4 values, for H = 3, 6, 9, and 12.
For M1 = 4, M2 takes values from 1 to 3 i.e., 3 values, for H = 4, 8, and 12.
For M1 = 5, M2 takes values from 1 to 2 i.e., 2 values, for H = 5 and 10.
Any further values of M1 cannot be considered since M1M2 will then exceed 60 minutes which is not possible.
Hence the total = 9+6+4+3+2 = 24
This is for 12 hours.
In a day the number of instances where H is exactly divisible by M1 giving M2 as quotient is =
24 * 2 = 48 instances in a 12hour format.
Now let us consider the time format to be a 24hour format.
Here H can take values from 0 to 23.
For M1 = 0, no value exists since divisor itself is 0.
For M1 = 1, M2 takes values from 0 to 9 i.e., 10 values, for H = 0 to 9.
For M1 = 2, M2 takes values from 0 to 9 i.e., 10 values, for H = 0, 2, 4, 6, 8, 10, 12, 14, 16, and 18.
For M1 = 3, M2 takes values from 0 to 7 i.e., 8 values, for H = 0, 3, 6, 9, 12, 15, 18 and 21.
For M1 = 4, M2 takes values from 0 to 5 i.e., 6 values, for H = 0, 4, 8, 12, 16, and 20.
For M1 = 5, M2 takes values from 0 to 4 i.e., 5 values, for H = 0, 5, 10, 15 and 20.
Any further values of M1 cannot be considered since M1M2 will then exceed 60 minutes which is not possible.
Hence the total = 10+10+8+6+5 = 39
This is for 24 hours.
In a day the number of instances where H is exactly divisible by M1 giving M2 as quotient is =
39 instances in a 24hour format.
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