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Need help with AMATYC math problem

  1. Mar 23, 2009 #1
    Does anyone know how to solve this problem:

    "Let N be the smallest number divisible by 33 which is greater than 1,000,000 and whose digits are all 0’s and 1’s. What are N’s leading four digits?"

    I ran accross this problem in an old AMATYC contest question.

    Any help would be much appreciated.
  2. jcsd
  3. Mar 23, 2009 #2
    It is doubtful that the contest would have wanted my one-line Mathematica solution:

    In :=
    First /@ RealDigits /@
    Select[Range[1000000, 2000000], IntegerQ[#/33] &],
    Function[s, And @@ (If[# > 1, False, True] & /@ s)]]

    Out : =

    {1, 1, 0, 1, 1, 1, 1}

    Therefore the winner is 1101111 = 33367 * 33
  4. Mar 23, 2009 #3
    By a trivial extension of my method we also solve the problem for the case where the lower bound is a billion, so the answer is 1000011111, and when the lower bound is a trillion so that the answer is 1000000101111.
  5. Mar 23, 2009 #4
    Try using the divisibility rules for 11 and 3.
  6. Mar 24, 2009 #5
    Using Mathematica is definitely not an option, but very effective. Nice program. Thanks.

    I'll try working throught the divisibilty rules for 11 and 3 like aligatorman suggests and see what I can come up with.
  7. Mar 31, 2009 #6
    Using the rules:

    Division by three: The sum of the digits is divisible by 3.

    Division by eleven: The alternating sum of the digits is divisible by 11 (or is 0). i.e. for 1111, +1-1+1-1=0 so it is divisible by 11.

    First, I presumed that the answer is 7 digits long with the first digit being one.

    The digit sum could be 6 or 3.

    Presuming the digit sum is 6:

    If the numbers digits are 1,a,b,c,d,e,f then a+b+c+d+e+f = 5 (meaning that there is only one 0)

    This zero could go in any of the 6 places, but if you take into account for the division by 11 rule, the smallest number possible to make with 6 ones in it is:


    Presuming the digit sum is 3:

    We know that a+b+c+d+e+f=2 and that a-b+c-d+e-f = 0. If we want a smaller answer, then we also know that either a=0, or a=1 and b=0. I will check each possibility for solutions.

    if the answer is 1,10c,def:

    c+d+e+f=1 and -c+d-e+f = 0. There are NO solutions in this form! (because how can the sum of just one one be zero?)

    if the answer is 1,0bc,def:

    b+c+d+e+f=2 and b-c+d-e+f=-1. There are NO solutions in this form either! (because how can the sum of 2 ones be -1?)

    So the correct answer is the one mentioned before:


    Is this right?
  8. Apr 2, 2009 #7
    Sorry for late response.
    That was the answer: 1011, 111.
    Thanks. Your explanation was very helpful.
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