Base of numerical system where 32+135=211 is valid?
Please help.no idea how to solve.
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.
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.