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About DeeGee

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  1. DeeGee added a post in a topic Find the number   

    Could not edit the above post but there is a correction to be made. You need to check the the -1 numbers for divisibility as well. Although it does not make a difference for 4 digit numbers, for higher digit numbers it will be needed.
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  2. DeeGee added a post in a topic Find the number   

  3. DeeGee added a post in a topic Help me again   

  4. DeeGee added a topic in New Logic/Math Puzzles   

    I saw this puzzle (similar to the one about catching rabbit that moves into an adjacent hole each night) on the internet somewhere and the solution to it is quite easy (even I was able to figure it out easily, so I guess it would be easy for you too). Here goes:

    A spy lives on the number line which extends from 0 to infinity. Each night the spy moves n digits ahead. For example, if one day he was at 10 and each night he moves 2 digits ahead, next day he would be at 12. Also, each day the agents can choose to check any one digit to catch him if he is there.
    The agents know that he is at 0 today but somehow they cant get there today to catch him. Is there a strategy to make sure he would be caught?

    This is the original puzzle which is easy enough. Thinking further, if I decided to add a twist to make it more interesting.

    Lets say the number line extends from -infinity to +infinity and the spy moves randomly either +n or -n on any given day. Now, is there a better strategy to catch him rather than just waiting for him at 0 to turn up eventually some day?

    Thinking even further (I really should stop thinking now!), how about the thief is on the line from -infinity to + infinity and each day he decides randomly which direction to go in and with roll of a dice to decide how many steps to go. That is if the dice shows 5 and he decides to go forward, he moves +5 from the previous number; next day the roll shows 4 and he decides to go backwards, he goes -4 from the previous number. Now, what is the strategy to catch him if it is known that he started at 0?

    Since the wording is not the original one, I may have missed something or may not be very clear. let me know if you need clarification. This is not a trick question so don't try to play around with the wording :-)
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  5. DeeGee added a post in a topic   

    In the case of silver foils or even simple paper its not friction but the static charge that causes difficulty in separation.
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  6. DeeGee added a post in a topic   

    There are 4 total possible solutions. Of these, two are words!

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  7. DeeGee added a post in a topic   

    I am not a programmer and know little about algorithms and coding. But here's an idea of how it could be done

    The only problem is that the computer might take time to process this program as it involves too many numbers
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  8. DeeGee added a post in a topic   

    You are right. I didnt add the second 23*2. It is indeed 95 mins and 19 trains then.
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  9. DeeGee added a post in a topic   

    I didnt see the spoiler button, so admin pls put this in a spoiler.

    The time to travel between stations is 1 min and the train stops for 1 min and loop time is also 1 min.

    For a given train, starting from station 1, and coming back to station 1, the total time to go "full circle" is
    1 (stop at station 1) + 23*2 (reach last station #24) + 1 (turn around) + 23*2 (go back to station 1) + 1 (trun around) = 49 mins

    Since a train arrives at every 5 mins, the total number of trains running is 49/5 = 10 trains
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  10. DeeGee added a post in a topic Quick Draw Aces   

    I agree with Bushindo's solution. Here's a loose point to keep going on this - How do you implement it?
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  12. DeeGee added a post in a topic   

    Really, are you sure? Because after 80, the increments were very small.
    Did you subtract the cases where previous flips are to be rejected? I didnt do it beyond 80 as I was doing it on excel and had to slightly modify formula for ech flip and eventually I got tired of doing it!
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