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4th puzzle post

  1. Feb 19, 2006 #1
    a number puzzle.....

    tell me a the lowest number for which
    the number just below it is divisible by 2 ,the one below that 3, below that 4 ,and below that 5 ,and below that 6 .
     
  2. jcsd
  3. Feb 19, 2006 #2
    Is it 59 ?
     
  4. Feb 20, 2006 #3
    damm to quickly solved yep ur right ill tell u the method i used then compare it with urs ..i wan t to know wat urs was aswell ..assume has 2 digits ..number ending in 5 has either o or 5 on its end ..but add 3 and it must be divisible by 2 so it must end in 5 , add 2 and it must be divisble by 3 ..so sum of didgits must sum to a multiple of 3 so its 1st digit is either 27or 57 or 87 since subtracting 1 must be divisible by 4 it must be 57 so add 2 and u get ....59
     
  5. Feb 20, 2006 #4
    I just worked it out in my head just multiplying 6 by progressive integers until I found one where adding 1 was a multiple of 5, adding 2 a multiple of 4 etc..
     
  6. Feb 20, 2006 #5
    I solved it by taking the LCM of 2, 3, 4, 5, and 6.Then that number -2 is divisible by 2,, that number -3 is divisible by 3, etc. So the answer is LCM - 1.
     
  7. Feb 20, 2006 #6

    AKG

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    I like daveb's solution. My solution:

    We're looking for a number such that:

    x-1 = 0 (mod 2)
    x-2 = 0 (mod 3)
    ...

    giving

    x = 1 (mod 2)
    x = 2 (mod 3)
    ...
    x = 5 (mod 6)

    We see that x = 5 (mod 6) makes x = 1 (mod 2) and x = 2 (mod 3) redundant, so we just need to solve:

    x = 3 (mod 4)
    x = 4 (mod 5)
    x = 5 (mod 6)

    The Chinese remainder theorem guarantees a unique solution (mod 30) to the last two congruences. It's easy to solve. The second gives x = 4 + 5y. Plugging into the third gives:

    4 + 5y = 5 (mod 6)
    5y = 1 (mod 6)
    y = 5 (mod 6)

    So x = 4 + 5(5) = 29 is the unique solution (mod 30). We see that 29 is not congruent to 3 (mod 4), so we add 30, and get 59, which is congruent to 3 (mod 4), so we're done. I'm not sure if daveb's solution proves that LCM - 1 is the least possible solution, but if it does, then it's a nice solution.

    EDIT: Actually, yes it does. If x is the solution, then the conditions of the problem require that x+1 is divisible by 2, 3, 4, 5, and 6. This means that LCM | x+1. The smallest possible choices for x+1 is thus, clearly, LCM, so the smallest choice for x is LCM - 1.
     
    Last edited: Feb 20, 2006
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