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jasen

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Posts posted by jasen

  1. I'm finding this interesting conjecture, maybe I'm not the first to state this conjecture.
    I have tested this conjecture to 100.000 first positive integer.
    anybody can provide the prove or disprove this conjecture ?

    the conjectre is :
    Every positive integer can be written as additon of 1 triangle number, 1 square number, and 1 pentagonal number

    note :

    Triangle numbers are generated by the formula, tn = ½n(n+1).
    The first ten Triangle numbers are:
    0, 1, 3, 6, 10, 15, 21, 28, 36, 45, ...

    Square numbers are generated by the formula, Sn=n*n.
    The first tenSquare numbers are:
    0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100,...

    Pentagonal numbers are generated by the formula, Pn=n(3n-1)/2.
    The first ten pentagonal numbers are:
    0, 1, 5, 12, 22, 35, 51, 70, 92, 117,...

    100 first positive integer which follow the conjecture :

    Spoiler

    Numbers    =     Triangle + Square + Pentagon
    1    =    0    +    0    +    1
    2    =    0    +    1    +    1
    3    =    1    +    1    +    1
    4    =    0    +    4    +    0
    5    =    0    +    0    +    5
    6    =    0    +    1    +    5
    7    =    1    +    1    +    5
    8    =    3    +    0    +    5
    9    =    0    +    4    +    5
    10    =    0    +    9    +    1
    11    =    1    +    9    +    1
    12    =    0    +    0    +    12
    13    =    0    +    1    +    12
    14    =    0    +    9    +    5
    15    =    1    +    9    +    5
    16    =    0    +    4    +    12
    17    =    0    +    16    +    1
    18    =    1    +    16    +    1
    19    =    3    +    4    +    12
    20    =    3    +    16    +    1
    21    =    0    +    9    +    12
    22    =    0    +    0    +    22
    23    =    0    +    1    +    22
    24    =    1    +    1    +    22
    25    =    0    +    25    +    0
    26    =    0    +    4    +    22
    27    =    1    +    4    +    22
    28    =    0    +    16    +    12
    29    =    1    +    16    +    12
    30    =    0    +    25    +    5
    31    =    0    +    9    +    22
    32    =    1    +    9    +    22
    33    =    3    +    25    +    5
    34    =    3    +    9    +    22
    35    =    0    +    0    +    35
    36    =    0    +    1    +    35
    37    =    0    +    25    +    12
    38    =    0    +    16    +    22
    39    =    0    +    4    +    35
    40    =    1    +    4    +    35
    41    =    0    +    36    +    5
    42    =    1    +    36    +    5
    43    =    6    +    25    +    12
    44    =    0    +    9    +    35
    45    =    1    +    9    +    35
    46    =    10    +    1    +    35
    47    =    0    +    25    +    22
    48    =    0    +    36    +    12
    49    =    0    +    49    +    0
    50    =    0    +    49    +    1
    51    =    0    +    0    +    51
    52    =    0    +    1    +    51
    53    =    1    +    1    +    51
    54    =    0    +    49    +    5
    55    =    0    +    4    +    51
    56    =    1    +    4    +    51
    57    =    3    +    49    +    5
    58    =    0    +    36    +    22
    59    =    1    +    36    +    22
    60    =    0    +    9    +    51
    61    =    0    +    49    +    12
    62    =    1    +    49    +    12
    63    =    3    +    9    +    51
    64    =    0    +    64    +    0
    65    =    0    +    64    +    1
    66    =    1    +    64    +    1
    67    =    0    +    16    +    51
    68    =    1    +    16    +    51
    69    =    0    +    64    +    5
    70    =    0    +    0    +    70
    71    =    0    +    1    +    70
    72    =    1    +    1    +    70
    73    =    3    +    0    +    70
    74    =    0    +    4    +    70
    75    =    1    +    4    +    70
    76    =    0    +    25    +    51
    77    =    1    +    25    +    51
    78    =    28    +    49    +    1
    79    =    0    +    9    +    70
    80    =    1    +    9    +    70
    81    =    0    +    81    +    0
    82    =    0    +    81    +    1
    83    =    1    +    81    +    1
    84    =    0    +    49    +    35
    85    =    1    +    49    +    35
    86    =    0    +    16    +    70
    87    =    0    +    36    +    51
    88    =    1    +    36    +    51
    89    =    3    +    16    +    70
    90    =    3    +    36    +    51
    91    =    10    +    81    +    0
    92    =    0    +    0    +    92
    93    =    0    +    1    +    92
    94    =    1    +    1    +    92
    95    =    0    +    25    +    70
    96    =    0    +    4    +    92
    97    =    1    +    4    +    92
    98    =    3    +    25    +    70
    99    =    0    +    64    +    35
    100    =    0    +    49    +    51

     

  2. 2 hours ago, Thalia said:

    I started with making a 4x4 using the factor distribution numbers DejMar mentioned. After a combination of the above and a lot of guessing. . . 

     

      Hide contents

    10  3   20  45

    18  15  25  4

    75  12  6   5

    2   50  9   30

     

    If we change Thalia answer with ABC solution, than replacing ABC with permutation [2,3,5], we can get 6 diffrent answers, one of it is as picture bellow

    Spoiler

     

    ABC distribution 2.png

     

     

  3. To attack the problem, first  write the numbers into its prime factors
    2 = 2, 3 = 3, 4 = 2.2, 5 = 5
    6 = 2.3, 9 = 3.3, 10 = 2.5, 12 = 2.2.3
    15 = 3.5, 18 = 2.3.3, 20 = 2.2.5, 25 = 5.5
    30 = 2.3.5, 45 = 3.3.5, 50 = 2.5.5 75 = 3.5.5

    So all numbers are consist of 3 prime seed 2,3 and 5
    Now replace each seed with 3 symbols, A,B and C
    So what we get now  is A, B,C,AA,BB,CC,AB,AC,BC,ABB,ACC,BAA,BCC,CAA,CBB, and ABC
    There are 12A,12B, and 12C Total symbols, so each line must have 3A,3B and 3C

    What we have to do now is
    Put  A,B,C,AA,BB,CC,AB,AC,BC,ABB,ACC,BAA,BCC,CAA,CBB, and ABC to a 4x4 square table,
    so every horizontal, vertical and diagonal line have 3A,3B, and 3C.

  4. At a math class, a teacher gives his students 4 numbers, lets say the numbers are A, B, C and D
    Then the teacher ask his students to make an equallity equation from the numbers.

    Claudy says  2*a*b = c + 2*b
    Leon says    a*b*c = 12 * c
    Jemmy says   b*b + c*c = (a+b)^2
    Thomas says  30*d = 6*b*c
    Richard says a+b+c+d = a*a

    Then the teacher says "Ok students, you are smart, but one of you is wrong"

    If the numbers are positive integer not more than 100, Who is wrong, and what are the numbers ?
     

  5. 16 hours ago, DejMar said:
      Hide contents

    1. ABCDEFGHI     ABCDEFGHI
      124835796 and 421385796


    2. ABCDEFGHI     ABCDEFGHI
      137869524 and 317869524

     

    answer for 1st question is correct, but for 2nd question is wrong.

    13 * 786 = 10218 ,    31 * 786 = 24366 

  6. 2 Silmilar Questions

    1. Replace each letter (A to I) in equation bellow with number (1 to 9) to make them correct
    AB * CDE = CB * AED = FGHI 

    2. Replace each letter (A to I) in equation bellow with number (1 to 9) to make them correct
    AB * CDE = CE * ADB = FGHI 

  7. There is a number of kids in a stadium, the kids form 7 groups which each group contains equal number of kids.
    than 2 kids left the stadium, the remaining kids form 6 groups which each group contains equal number of kids.
    than 3 more kids left the stadium, the remaining kids form 5 groups which each group contains equal number of kids.
    than 5 more kids left the stadium, the remaining kids form 4 groups which each group contains equal number of kids.
    than 7 more kids left the stadium, the remaining kids form 3 groups which each group contains equal number of kids.

    If the initial number of kids is not more than 750 kids, how many kids in the stadium at the beginning ?
     

  8. 2 hours ago, bonanova said:

    Meanderings.
     

      Reveal hidden contents

    Suppose the volume and area are equal: V = abc = A = 2(ab+bc+ca)

    Then we have the Diophantine equation abc = 2(ab+bc+ca).

    abc - 2(ac+bc) = 2ab
    c(ab -2a-2b) = 2ab
    c = 2ab / (ab-2a-2b) = 2 + 4(a+b)/[ab-2(a+b)

    Search for integer values of (a,b) such that ab-2(a+b) divides 4(a+b) then calculate c.

    If we constrain (without loss of generality) a<=b<=c, then solutions can be found,
    by exhaustive search, for a=3, 4, 5, 6, as follows

    a   b   c  V=A
    = === === ====
    3   7  42  882
    3   8  24  576
    3   9  18  486
    3  10  15  450
    3  12  12  432
    4   5  20  400
    4   6  12  288
    4   8   8  256
    5   5  10  250
    6   6   6  216

    Interestingly, one of the solutions is a cube, for which V = A.
    Unfortunately, 4(6+6+6) is 72, not 216, a factor of 3 too small. So, (a,b,c) = (6,6,6) is not a solution.

    That is, 6:6:6 is not equal to 72:216:216.

    Another solution with b=c is 4 8 8.

    But again, 4(4+8+8) is 80, not the required 128.

    4:8:8 is not equal to 80:256:256.

    We've ruled out the cube (a=b=c) and we've ruled out V=A (b=c).

    The solution has a=b or all distinct a b c values.
    a=c is not possible unless a=b=c (cube)


          

     

    Bonanova, you are very close, your explaination is very cool, just quite not carefull.

    check again your exhaustive search list , you miss something.

  9. 16 hours ago, bonanova said:

    I'm confused about the ratios.
     

      Reveal hidden contents

    a:b:c I understand, because they all have dimensions of length. Since they have the same dimensions, their ratios are pure numbers, without units.

    But the sum of sides (which I take to mean a+b+c, but could mean 4a+4b+4c) has dimension of length, sum of surface area has dimension of area and volume has dimension of, um, volume. Normally ratios like that are meaningless without specifying the units used. The ratio of area to perimeter for a square for example is always a2/4a = a/4. But that is not a pure number, it's a measure. And it has, for a given square, say one that is 12 inches on a side, differing values: the ratio is 3 (inches), if the side is measured in inches, and 1/4 (foot), if the side is measured in feet.

    I guess we can ignore all this if we say there is a box, and its linear dimensions (in some units, never mind what they are) are a, b, and c. It's just that ratios of lengths, areas and volumes will not be pure numbers. But in fairness you did not say they were. ;)

     

    Sum of sides means sum of all 12 sides, sum of surface area means sum of all 6 areas.
    Ignore the dimensions. Let say length in m, area in m2 and volume in m3

    All length are  integer.

  10. I have a box
    Rasio of length, width and height of the box is a:b:c
    Rasio of sum of sides, sum of surface area and volume is also a:b:c
    find the length, width and height of the box
    a, b and c are integers & can be the same number.

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