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Homework Help: [Sets] Cardinality problem

  1. Jan 17, 2007 #1
    [Resolved][Sets] Cardinality problem

    1. The problem statement, all variables and given/known data

    let A be a Set of all natural numbers from 1 to 6000 that are divsible by 3 or 7 but not 105.

    1.What is the cardinality of A?

    2.How many numbers in A give 2 as the remained of division by 3.

    2. Relevant equations
    3. The attempt at a solution

    1. My thinking was like this

    How many multiples of 3 are from 1 to 6000. Well 6000/3 = 2000. How many multiples of 7? Well 6000/7 = 857. How many multiples of 7 are also multiples of 3? I would say 2000 / 7 = 285 that is in that 2000 multiples of 3 , 285 multiples are multiples of 7 aswell. That gives

    +2000 multiples of 3
    +857 multiples of 7 -285 multiples of 7 that are multiples of 3 aswell

    that gives a total of 2572 numbers

    now to put !divison by 105 into picture. Soo they must not be multiples of 105. We just substract multiples of 105, right? that is 2572 / 105 = 24, that is

    the cardinality of A is 2572 - 24 = 2548.

    Is any of the above correct or even close to a solution?

    Now for the second part my thinking is like this, what do we have in those 2548 numbers, well we have something like this multiples of 3 and 7 but not 105. Now multiples of 3 will ofcorse not give the proper remainder, now for 7,

    7 // remainders 3*2 + 1 rem 1
    14 // rem 2
    21 // 0
    28 // 1
    35 // 2
    42 // 0
    49 // 1
    56 // 2
    63 // 0

    Now I noticed a pattern that multiples of 7 that give remainder 2 have a pattern, meanin starting from 2nd multiple which gives 2 as a remainder then every 2*k + 1 multiple gives 2 as the remainder. Meaning if I take 3 successive multiples of 7 one of them is guarantted to give 2 as the remainder. Meaning if there are 572 multiples of 7 (not multiples of 3) then there are 572 / 3 = 190 that will give 2 as the remainer.

    Any good? At least close?
     
    Last edited: Jan 17, 2007
  2. jcsd
  3. Jan 17, 2007 #2
    I think I got it. Thanks anyway.
     
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