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a digital timer counts down from 3 min to zero, one second at a time.
for how many seconds does at least one of these digits show a 5?
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(a) First (leftmost) digit is never 5.
(b) Consider the set of numbers XY5 from 1 to 180 inclusive with the last
(rightmost) digit of 5 and the middle digit Y =/=5.
These numbers are
005, 015, 025, 035, 045, , 065, 075, 0.85, 0.95,
105, 115, 125, 135, 145, , 165, 175.
The crossed numbers are those with 5 in the middle position.
The amount of such "survived" numbers is 9 + 7 = 16.
The counter counts/adds 1 as each such a number appears.
(c) To it, we should add appearances X5Z.
There are exactly 10 + 10 = 20 such appearances
050, 051, 052, 053, 054, 055, 056, 057, 058, 059,
150, 151, 152, 153, 154, 155, 156, 157, 158, 159.
The counter adds 1 as each such a number appears.
(d) Thus the total (the last) number on the counter at the end is the sum 16 + 20 = 36. ANSWER
Solved.
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What I did to solve the problem is THIS:
I separated the set of numbers, containing at least one digit "5",
in two disjoint subsets (b) and (c) in a way that the counter adds 1
at every appearance of a number from each of the two subsets.