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Thread: Problem of warriors

  1. #1
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    Unhappy Problem of warriors

    Hello. I'm trying to solve this excersise:
    There are 3 warriors who are shooting in order to be more trained. The first warrior shoots a rifle 1 time in 5 secunds, the second one - in 6 secunds, the third - in 7 sec. They all started at the same time. How many times did everyone shoot? We know that the second man was shooting four minutes longer than the first one and 3 min less than the the third one. All shooting time was less than an hour.


    I think that we can consider that all shooting time is equal to the time of the third shooter, because he was training 7 minutes more comparing to the first and 3 min more comparing with the second. But the main problem, I suppose, is that we don't know exact time of how much they trained for. Just less than an hour. So perhaps I should make an inequality, but I'm not sure. Any help would be apreciated. Thanks!
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  2. #2
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    Re: Problem of warriors

    Quote Originally Posted by vilius123 View Post
    Hello. I'm trying to solve this excersise:
    There are 3 warriors who are shooting in order to be more trained. The first warrior shoots a rifle 1 time in 5 secunds, the second one - in 6 secunds, the third - in 7 sec. They all started at the same time. How many times did everyone shoot?
    With what ending condition? Without some kind of ending condition, they could just keep firing forever!

    We know that the second man was shooting four minutes longer than the first one and 3 min less than the the third one. All shooting time was less than an hour.


    I think that we can consider that all shooting time is equal to the time of the third shooter, because he was training 7 minutes more comparing to the first and 3 min more comparing with the second. But the main problem, I suppose, is that we don't know exact time of how much they trained for. Just less than an hour. So perhaps I should make an inequality, but I'm not sure. Any help would be apreciated. Thanks!
    5 divides into 60 min= 1 hour 12 times so in one hour the first person will have fired 12 times. 7 divides into 60 8 times with remainder 4 minutes so the second person will have fired 8 times in one hour. 9 divides in 60 6 times with remainder 7 minutes so the third person will have fired 6 times in one hour. All three will have fired a total or 12+ 8+ 6= 26 times.
    Last edited by HallsofIvy; Dec 7th 2017 at 02:30 PM.
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  3. #3
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    Re: Problem of warriors

    Let's convert to seconds. Suppose person 1 gets off a shots, person 2 b shots and 3 c shots. Then:
    5a+240=6b
    6b+180=7c
    7c<3600
    We know 5a is divisible by 6 (since 240 is already). This means that a is divisible by 6. We also know that
    5a+420 = 7c

    So a is divisible by 7 (since 420 already is).

    Thus, a is a multiple of 42.

    Next,
    7c-180=6b and 7c-420 = 5a implies c is a multiple of 30.

    Then, 6b-240=5a and 6b+180=7c implies b is divisible by 5 and b is congruent to 5 mod 7. So, b is congruent to 5 mod 35.

    Now, we can rewrite:
    a=42x
    b=35y+5
    c=30z

    Let's solve with respect to z:
    5a+420=7c
    210x+420=210z
    x+2=z
    6b+180=7c
    210y+210=210z
    y+1=z

    Therefore, z is a minimum of 3 and maximum of 17. This gives person 1 getting off 42(z-2) shots, person b getting off 35(z-1)+5 shots and person 3 getting off 30z shots. There are 15 solutions to this problem.
    Last edited by SlipEternal; Dec 7th 2017 at 04:31 PM.
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