Numerical base system?

Base of numerical system where 32+135=211 is valid?

Please help.no idea how to solve.

Comments

  • Let b be the base of the system (note that b ≥ 2):

    32 + 135

    = (3*b^1 + 2*b^0) + (1*b^2 + 3*b^1 + 5*b^0)

    = (3b + 2) + (b^2 + 3b + 5)

    = b^2 + 6b + 7

    211

    = 2*b^2 + 1*b^1 + 1*b^0

    = 2b^2 + b + 1

    b^2 + 6b + 7 = 2b^2 + b + 1

    b^2 - 5b - 6 = 0

    (b - 6)(b + 1) = 0

    b = 6 or b = -1

    Since b ≥ 2, b = 6

  • Let b be your base. Then 32+135=211 means

    (3b + 2) + (b^2 + 3b + 5) = 2b^2 + b + 1

    which simplifies to

    0 = b^2 - 5b - 6

    0 = (b+1)(b-6)

    So, b is either -1 or 6. However, a negative number base makes no sense, so your base must be 6.

  • If the base is x, the equation means:

    (3*x+2)+(1*x^2+3*x+5)=(2*x^2+1*x+1) =>

    x^2+6x+7=2x^2+x+1 =>

    0=x^2-5x-6 => from the quadratic formula:

    x1,2=(5+-sqrt(25+4*6))/2=(5+-7)/2 =>

    x1=6, x2=-1

    The base (-1) isn't good, because we use digits greater than 1(eg.: 5)

    So the base is 6.

    If the equation is in base 10, then in base 10 the equation is:

    20+59=79.

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