Tuesday’s Logic Puzzles

A brief break from reading this afternoon to tackle two logic/math problems below. See if you can deduce the answer on your own. Leave a comment if you know the answer!

1) Consider an analog clock with both an hour hand and a minute hand. What is the first time after 6PM that the hour hand and the minute hand are exactly coincident (i.e., on top of one another)? NOTEYour answer should be in this format: HH:MM:SS.DDD, where HH = hour, MM = minutes, SS = seconds, and DDD is the 1/1000th of a second decimal equivalent. (HINT: the first thing that comes to mind, 6:30PM, is an incorrect response).

2) Consider a room  with a very large table on which stand 100 lamps, each with an on/off switch. The lamps are arranged in a straight line, and each one is numbered 1, 2, 3, …, 99, 100. At the beginning of the experiment, all the lamps are turned off.

This room has an entry door and a separate door for an exit. One hundred people are recruited to participate in this experiment. Each of the 100 participants is also numbered 1 to 100, inclusive. When participant number 1 enters the room, he turns on EVERY lamp, and exits. When participant 2 enters the room, he flips the switch for every second lamp (thus, turning off lamps 2, 4, 6, 8, 10, and so on because participant 1 has turned all the lamps on his turn). Participant 2 exits and then participant 3 enters. Participant 3 flips the switch on every third lamp (thus changing the on/off state of the lamps which are numbered 3, 6, 9, 12, and so on). This process continues until all 100 participants have taken their turn and passed through the room.

Assume each participant can properly count and doesn’t make any mistakes in changing the on/off state of the lamp(s) he’s assigned to change the state of. Here is the question: after the 100th participant completes his journey through the room, how many lamps are illuminated? And which of those lamps (i.e., reference by number) are they?

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UPDATE (1:30PM): Both questions have been answered in the comments. To make up for these relatively simple questions, I’ll post a much more challenging logic question in the evening. It will have to do with a deck of cards.

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9 thoughts on “Tuesday’s Logic Puzzles

          • I started out what you would probably call “brute force” and changed to the easier way. I knew 6:30 wasn’t correct, but I knew it’d be between 6:30 and 6:35. So I calculated the percentage around the clock that the minute hands and hour hands would be at 6:32 and 6:33 and realized it wasn’t going to be a clean number. So I created a formula where the minute hand percentage around the clock would equal the hour hand percentage around the clock and solved for minutes:
            (M/60 +6)/12 = m/60
            5(M/60+6) = M
            30 + M/12 = M
            30 = 11M/12
            360 = 11M
            32.7272
            6:32:43.636363…

            • Excellent! Though I would actually classify your approach as the brute force approach!

              Here is the quicker, more elegant solution. At noon/midnight, the hands are exactly coincident. Going around the clock, the hands would be coincident again at exactly 10 other times (shortly after 1:05, 2:10, 3:15, 4:20, 5:25, 6:30, etc). So the equal parts occur at 1/11th of the 12 hours.

              For the question I posed, the answer is six times the 1/11th increment of the 12 hours. So 6*12/11=6.545454… Then, 0.545454*60 represents minutes after 6PM, so that’s 32.72727272. So we have 6:32:SS.DDD to deduce. Then just subtract 32 from there and again multiply by 60 to get 43.636363. So 6:32:43.636363 as you attained as well!

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