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Proposed Problem

Discussion in 'Mathematics' started by AVDBA, Sep 20, 2019.

  1. AVDBA

    AVDBA New commenter

    Show the number:

    1 1 ... 1 1 2 2 ... 2 2 5

    where 1 occurs n - 1 times and 2 occurs n times, is a perfect square.
     
  2. AVDBA

    AVDBA New commenter

    I would like to propose the following hints:

    * To write n = 1 1 ... 1 1 2 2 ... 2 2 5

    * To multiply n by 10

    * To work the expression keeping in mind the representation of numbers using base 10
     
  3. AVDBA

    AVDBA New commenter

    This problem seems to be interesting for working with decimal representation of numbers. Base 10 representation has common usage within Engineering courses, as well as within different scientific fields, so the reason for giving background on the subject for the students.

    The following approach of defining a name for the considering the given number and working with the named number throughout is suggested to reach the solution.

    Defining:

    n = 11111 ... 11222222 ... 225

    and multiplying n by 10, we have:

    10n = 11111 ... 11222222 ... 2250

    with the number 1 (digit) appearing (n - 1) times and the number 2 (digit) appearing n times.

    Now, subtracting n from 10n, i.e.:

    10n - n = 9n = 11111 ... 11222222 ... 2250 - 11111 ... 11222222 ... 225

    leads to (using the well known subtraction algorithm):

    111111 ... 12222222 ... 2250
    - 11111 ... 11222222 ... 2225
    ----------------------------
    100000 ... 01000000 ... 0025

    It is important to infer that multiplying n by 10 "pushes" the original number one position to the left - being the numbers, now, paired for writing the usual algorithm for subtracting two numbers, as above.

    One infers that:

    * The first sequence of zeros of the result of (10n - n) (the explicitly written zeros here from the third line of the subtraction algorithm above: 100000 ... 01 ... 25) is paired with the sequence of digits '1' of n (in the above subtraction algorithm, comparing the second and third lines of this algorithm), except for the last digit '1' of n, so there occur (n - 1) - 1 = n - 2 zeros in this first sequence

    * The second sequence of zeros of the result of (10n - n) (the explicitly written zeros here from the third line of the subtraction algorithm above: 1 ... 1000000 ... 0025) is paired with the sequence of digits '2' of n (in the above subtraction algorithm, comparing the second and third lines of this algorithm), except for the last digit '2' of n, so there occur (n) - 1 = n - 1 zeros in this second sequence

    By these observations, one reaches:

    9n = 100000 ... 01000000 ... 0025 =>
    9n = 1 x 10^[2 + (n - 1) + 1 + (n - 2))] + 1 x 10^[2 + (n - 1)] + 25 =>
    9n = 10^(2n) + 10^(n + 1) + 25 =>
    9n = 5^2 + 10 x 10^(n) + 10^(2n) =>
    9n = 5^2 + 2 x 5 x 10^(n) + (10^n)^2 =>
    9n = (5 + 10^n)^2

    Any number of the form 2 + 10^n, with n natural, is a multiple of 3, since adding up the its digits gives 3. Henceforth, 3 + 2 + 10^n = 5 + 10^n is also a multiple of 3. Hence, the number:

    m = (5 + 10^n)/3

    is some natural number not equal to 0, and:

    m^2 = [(5 + 10^n)/3]^2

    is a perfect square. Since this is n, we have showed that n = 11111 ... 11222222 ... 225 is a perfect square.
     

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