Love to watch cartoon.. ?

.

Today is 2 2 March, 2012 2012 . My friend asked me when was the last show of Dragon Ball Z telecast. He wanted to watch the next show.

I replied that it was telecast 800 800 hours 480 480 minutes and 5400 5400 seconds ago and the next episode is telecast after 1010 hours of the previous one.

It was 4:00 PM when I told him the telecast time.

What will be the time and date when he should turn on the television to watch the next show ?

12:30 AM of 10 March,2012 10:30 AM of 29 January,2012 10:30 AM of 28 January,2012 10:30 AM of 11 March,2012 12:30 AM of 11 March,2012

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1 solution

Rishabh Tripathi
Apr 18, 2015

Initial date: 2 March,2012.

Let us first check if it is a leap year.

We know, if [ Y E A R ] 0 ( m o d 4 ) [YEAR] \equiv 0 \pmod 4 , then it is a leap year. So, 2012 is a leap year.

Number of hours to be deducted = 800 = 800

800 8 ( m o d 24 ) 800 \equiv 8 \pmod {24}

OR

800 = 24 × 33 + 8 800 = 24 \times 33 + 8

So, Number of days to be deducted = 33 = 33

Number of hours to be further deducted = 800 24 × 33 = 8 =800 - 24 \times 33 = 8

Number of minutes to be deducted = 480 =480

480 m i n s = 480 60 h o u r s = 8 h o u r s 480 mins = \frac{480}{60} hours = 8 hours

Number of seconds to be deducted = 5400 =5400

5400 s e c = 5400 3600 h o u r s = 1.5 h o u r s 5400 sec = \frac{5400}{3600} hours = 1.5 hours

So, total number of hours to be further deducted = 8 + 8 + 1.5 = 17.5 = 8+8+1.5 = 17.5

Now, 33 = 2 + 29 + 31 28 33=2+29+31-28

Therefore, after deducting 33 33 days the date and time would be 4 : 00 P M 4:00 PM of 28 January,2012 (Since the time remains same if we deduct 24 hours)

Further deducting 17.5 h o u r s 17.5 hours we get 10 : 30 P M 10:30 PM of 28 January,2012

Now we have to add 1010 h o u r s 1010 hours

Following the similar pattern, 1010 2 ( m o d 24 ) 1010 \equiv 2 \pmod {24}

Number of days to be added = 42 =42

Number of hours to be added = 1010 24 × 42 = 2 =1010-24 \times 42 = 2

So, the required date and time is 12 : 30 A M o f 11 M a r c h , 2012 \boxed{12:30 AM of 11 March,2012}

An equivalent approach is to convert everything in terms of hours first before we begin calculations of time.

800 hrs 480 min 5400 sec = 809.5 hrs ( 6.5 ) hrs ( m o d 24 ) 800\textrm{ hrs }480\textrm{ min }5400\textrm{ sec }= 809.5\textrm{ hrs }\equiv (-6.5)\textrm{ hrs}\pmod{24}

As we're working backward in time, we add the absolute value of the residue obtained to our current time and then add the 1010 hrs m o d 24 1010\textrm{ hrs}\bmod 24 to obtain time of telecast as follows:

4 : 00 PM + 6.5 hrs 10 : 30 PM + ( 1010 hrs ) m o d 24 12 : 30 AM 4:00\textrm{ PM}\stackrel{+6.5\textrm{ hrs}}\longrightarrow 10:30\textrm{ PM}\stackrel{+(1010\textrm{ hrs})~\bmod~24}\longrightarrow 12:30\textrm{ AM}

Now, comes the date. We first have 809.5 = 816 6.5 809.5=816-6.5 . Since 4 : 00 4:00 PM and 10 : 30 10:30 PM comes in a single day after 6.5 6.5 hrs, we have the day of last episode as 816 24 = 34 \frac{816}{24}=34 days before current day. Working from current day, we have,

34 days = ( 2 index on March 1 ) + ( 29 index on Feb 1 ) + ( 31 index on Jan 1 ) ( 28 current index to final day ) 34\textrm{ days }=\left(\underset{\textrm{index on March }1}{2}\right)+\left(\underset{\textrm{index on Feb }1}{29}\right)+\left(\underset{\textrm{index on Jan }1}{31}\right)-\left(\underset{\textrm{current index to final day}}{28}\right)

Hence, the last telecast was on 2 8 th 28^{\textrm{th}} Jan 2015 ^\prime2015 at 10 : 30 PM 10:30\textrm{ PM} . Now, we calculate day-hour increments after 1010 hrs 1010\textrm{ hrs} as follows:

1010 hrs = 1008 24 days + 2 hrs = 42 days + 2 hrs 1010\textrm{ hrs }=\frac{1008}{24}\textrm{ days }+2\textrm{ hrs }=42\textrm{ days }+2\textrm{ hrs}

Working from day of last telecast, we have,

42 days = ( 3 going to Feb 1 ) + ( 29 going to March 1 ) + ( 10 date reached ) 42\textrm{ days }=\left(\underset{\textrm{going to Feb }1}{3}\right)+\left(\underset{\textrm{going to March }1}{29}\right)+\left(\underset{\textrm{date reached}}{10}\right)

This gives us 10 : 30 PM 10:30\textrm{ PM } of 1 0 th 10^{\textrm{th}} March 2015 ^\prime2015 . We still have to add the remaining 2 2 hrs. Since we have new day after 12 : 00 AM 12:00\textrm{ AM} and the final time we get for next telecast is 12 : 30 AM 12:30\textrm{ AM} , we add another day to our obtained date.

This gives us the date of 1 1 th 11^{\textrm{th}} March 2015 ^\prime2015 and time of 12 : 30 AM 12:30\textrm{ AM} as date and time of next telecast.

Prasun Biswas - 6 years, 1 month ago

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ah.. its a very nice and systematic solution.

Rishabh Tripathi - 6 years, 1 month ago

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It is almost equivalent to yours though. By the way, I noticed that I forgot to upvote your solution earlier. Well, better late than never. :)

Prasun Biswas - 6 years, 1 month ago

By the way, there's are minor typos in few lines of your solution (mostly LaTeX \LaTeX problems). I suggest you edit them properly to make your solution cleaner.

Prasun Biswas - 6 years, 1 month ago

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Removed some typos . Thanks for telling.

Rishabh Tripathi - 6 years, 1 month ago

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