A clock problem by Nihar and Julian (Part 1)

Logic Level 4

Take a 12 - hours digital clock. Now you must be bored seeing the same clock timings everyday. So now you keep that digital clock inverted upside-down. How many timings in the original clock are there such that they appear as "sensible numbers" when the clock is kept inverted upside-down?

Let $S$ be the number of such timings. Input your answer as $S+28$ .

Details And assumptions:

As an explicit example , if the timing shown in regular clock is $06:19$ , then it will be seen as $61:90$ in the inverted upside down clock and is one of our required timings . So you have to find the number of such timings.
If the timing is $03:16$ , upside down it would be $91:E0$ , and would not be accepted.
Note that after every $11:59$ , the clock shows $00:00$ and not $12:00$ .
All digits look like this: (for clarifying the shape of digits)

This problem was created by me and Julian (the apple).

The answer is 280.

2 solutions

Nihar Mahajan
Jun 20, 2015

First observe that $0,1,2,5,6,8,9$ remain sensible numbers when they appear in an inverted placed clock.

_ _ : _ _

Since we have a $12$ hours clock , we have the timing format as shown above.

So the first two digits can be $00 , 01 , 02 , 05 , 06 , 08 , 09 , 10 , 11$ that is $9$ choices. The third digit can be $0,1,2,5$ with $4$ choices . The fourth digit can be $0, 1 , 2 , 5 , 6 , 8, 9$ that is $7$ choices. So by product rule we have: $9 \times 4 \times 7 = 252 = S$ .

So , $S+28 = 252+28 = \boxed{280}$ .

I messed up on my first two tries because I forgot to add 28 and then added it wrong. Lucky for me, I got the third try right. I guess I should use paper for solving problems in the future and stop solving these in my head.

Shubham Bhargava - 5 years, 10 months ago

Chew-Seong Cheong
Jun 20, 2015

There are $7$ acceptable digits $0,1,2,5,6,8,9$ . The numbers of hour digits and minute digits are as follows:

$\begin{array} {ccccc} \Box & \Box & : & \Box & \Box \\ 0 & 0 & & 0 & 0 \\ 0 & 1 & & 1 & 1 \\ 0 & 2 & & 2 & 2 \\ 0 & 5 & & 5 & 5 \\ 0 & 6 & & & 6 \\ 0 & 8 & & & 8 \\ 0 & 9 & & & 9 \\ 1 & 0 & & & \\ 1 & 1 & & & \\ & 9 & \times & 4 & \times\space 7 & = 252 \end{array} \\ \Rightarrow S + 28 = \boxed{280}$

A clock problem by Nihar and Julian (Part 1)

This problem was created by me and Julian (the apple).

The answer is 280.

2 solutions

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