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Sequence of numbers divisible by 7.

  1. Jan 19, 2013 #1
    1. The problem statement, all variables and given/known data

    A sequence of 3-digit numbers divisible by 7 is given. Find:
    a) first three and last three numbers.
    b) how many numbers are there in that sequence.
    c) sum of all the numbers in the sequence.

    2. Relevant equations



    3. The attempt at a solution

    Cause they are 3 digit numbers that means: 100-999.
    In order for a number to be divisible by 7 you need to get the last digit, double it and subtract it from the number without the last digit. If that number is divisible by 7, you stop there, if not, repeat the steps. I tried this way, but it got me nowhere. So it came to my mind that 7,14,21,28 are all divisible by 7 cause the distance between those numbers is 7. So with a bit of trial and error I found that 105 is divisible by 7. So the first three numbers of the sequence are 105, 112, 119. The last two are 994, 987, 980.
    Now how do I find how many numbers between 100 and 999 are divisible by 7?
    And for the sum I know the formula. Sn = ((a1 + an) / 2) * n
     
  2. jcsd
  3. Jan 19, 2013 #2

    Mark44

    Staff: Mentor

    You can write the numbers as {15*7, 16*7, 17*7, ..., 142*7}. Writing them this way will make it easier to count them.

    If the sequence consisted of multiples of, say, 5, like this: {15, 20, 25, 30, ..., 95} = {3*5, 4*5, 5*5, 6*5, ..., 19*5}, there are (19 - 3) + 1 = 17 elements in the set. It's important to add 1 so as to count the first number.
     
  4. Jan 19, 2013 #3

    Mark44

    Staff: Mentor

    I shoud add that in my example, if you don't want to write the numbers as multiples of 5, then you need to divide by 5.

    (95 - 15)/5 + 1 = 80/5 + 1 = 16 + 1 = 17, same as before.
     
  5. Jan 21, 2013 #4
    You have found the first term is 105, the next is 105+7, the next 105+2*7, the next 105+3*7

    You have an arithemtic sequence. Where a0 is first term, an the nth term, an-1 last term, n the number of terms and d the common difference

    a0=105
    an-1 = 994
    d=7

    an = a0 +(n-1)d use this to find n
    sum of all terms is n(a0 +an-1)/2
     
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