A clock problem by Nihar and Julian (Part 2)
Take a 24 - 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?
Details And assumptions:
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As an explicit example , if the timing shown in regular clock is 0 6 : 1 9 , then it will be seen as 6 1 : 9 0 in the inverted upside down clock and is one of our required timings . So you have to find the number of such timings.
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If the timing is 0 3 : 1 6 , upside down it would be 9 1 : E 0 , and would not be accepted.
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Note that after every 2 3 : 5 9 , the clock shows 0 0 : 0 0 and not 2 4 : 0 0 .
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All digits look like this: (for clarifying the shape of digits)
This problem was created by me and Julian (the apple).
The answer is 476.
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2 solutions
There are 7 acceptable digits 0 , 1 , 2 , 5 , 6 , 8 , 9 . The numbers of hour digits and minute digits are as follows:
□ 0 0 0 0 0 0 0 1 1 1 1 1 1 1 2 2 2 □ 0 1 2 5 6 8 9 0 1 2 5 6 8 9 0 1 2 1 7 : × □ 0 1 2 5 4 □ 0 1 2 5 6 8 9 × 7 = 4 7 6
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 2 4 hours clock , we have the timing format as shown above.
So the first two digits can be 0 0 , 0 1 , 0 2 , 0 5 , 0 6 , 0 8 , 0 9 , 1 0 , 1 1 , 1 2 , 1 5 , 1 6 , 1 8 , 1 9 , 2 0 , 2 1 , 2 2 that is 1 7 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: 1 7 × 4 × 7 = 4 7 6 .