My digital 24-hour clock displays the time of day as three numbers .
Let be the number of moments throughout a regular day (from midnight to midnight) when the digit sums of are all equal. Find .
Details & Assumptions :
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Hour 0 , only 1 moment ( 1 2 ): 0 0 : 0 0 : 0 0 ,
Hour 1 , 4 moments ( 2 2 ): 0 1 : 0 1 : 0 1 , 0 1 : 0 1 : 1 0 , 0 1 : 1 0 : 0 1 , 0 1 : 1 0 : 1 0 ,
Hour 2 , 9 moments ( 3 2 ): 0 2 : 0 2 : 0 2 , 0 2 : 0 2 : 1 1 , 0 2 : 0 2 : 2 0 , etc … ,
Hour 3 , 1 6 moments ( 4 2 ): … ,
There is clearly a pattern: the squares! However , there can not be anything above 3 6 moments for any given hour because we are dealing with time! Example:
Hour 1 7 , 3 6 moments:
1 7 : ( 0 8 or 1 7 or 2 6 or 3 5 or 4 4 or 5 3 ) : ( 0 8 or 1 7 or 2 6 or 3 5 or 4 4 or 5 3 ) .
There is also the special case of 1 9 to consider— digsum ( 1 9 ) = 1 0 , and the only possible combinations for this are 1 9 , 2 8 , 3 7 , 4 6 , 5 4 , so there are only 2 5 moments instead of 3 6 .
Hence, we have
1 + 4 + 9 + 1 6 + 2 5 + 3 6 + 3 6 + 3 6 + 3 6 + 3 6 + 4 + 9 + 1 6 + 2 5 + 3 6 + 3 6 + 3 6 + 3 6 + 3 6 + 2 5 + 9 + 1 6 + 2 5 + 3 6 = A = 5 8 0 , and ⌈ 1 0 − 2 ⋅ 5 8 0 ⌉ = 6 .