Take Your Time!

My digital 24-hour clock displays the time of day as three numbers X : Y : Z X : Y : Z .

Let A A be the number of moments throughout a regular day (from midnight to midnight) when the digit sums of X , Y , Z X,Y,Z are all equal. Find 1 0 2 A \lceil 10^{-2} A \rceil .

Details & Assumptions :

  • Ignore 24 : 00 : 00 24:00:00 in favor of 00 : 00 : 00 00:00:00 .
Image Credit: http://www.prestonlibrary.net/blog/wp-content/uploads/2009/04/clock.gif


The answer is 6.

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1 solution

Alex Delhumeau
May 30, 2015

Hour 0 0 , only 1 1 moment ( 1 2 1^2 ): 00 : 00 : 00 00:00:00 ,

Hour 1 , 4 1, 4 moments ( 2 2 2^2 ): 01 : 01 : 01 , 01 : 01 : 10 , 01 : 10 : 01 , 01 : 10 : 10 01:01:01, 01:01:10, 01:10:01, 01:10:10 ,

Hour 2 , 9 2, 9 moments ( 3 2 3^2 ): 02 : 02 : 02 , 02 : 02 : 11 , 02 : 02 : 20 , etc 02:02:02, 02:02:11, 02:02:20, \text{ etc } \ldots ,

Hour 3 , 16 3, 16 moments ( 4 2 4^2 ): \ldots ,

There is clearly a pattern: the squares! However , there can not be anything above 36 36 moments for any given hour because we are dealing with time! Example:

Hour 17 , 36 17, 36 moments:

17 : ( 08 or 17 or 26 or 35 or 44 or 53 ) : ( 08 or 17 or 26 or 35 or 44 or 53 ) 17:(08 \text{ or } 17 \text{ or } 26 \text{ or } 35 \text{ or } 44 \text{ or } 53) : (08 \text{ or } 17 \text{ or } 26 \text{ or } 35 \text{ or } 44 \text{ or } 53) .

There is also the special case of 19 19 to consider— digsum ( 19 ) = 10 \text{digsum}(19)=10 , and the only possible combinations for this are 19 , 28 , 37 , 46 , 54 19, 28, 37, 46, 54 , so there are only 25 25 moments instead of 36 36 .

Hence, we have

1 + 4 + 9 + 16 + 25 + 36 + 36 + 36 + 36 + 36 + 4 + 9 + 16 + 25 + 36 + 1+4+9+16+25+36+36+36+36+36+4+9+16+25+36+ 36 + 36 + 36 + 36 + 25 + 9 + 16 + 25 + 36 = A = 580 36+36+36+36+25+9+16+25+36=A=580 , and 1 0 2 580 = 6 \lceil 10^{-2} \cdot 580 \rceil=\large{\boxed{6}} .

Explain moment clearly!!!

ashutosh yadav - 5 years, 6 months ago

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