NumberBaseCeption

Number Theory Level 4

$\large \overline{162}_{\overline{31}_{x}}=3 \left (\overline{65}_{\overline{31}_{x}} \right )$

Find the value of $x$ such that the above equation is true.

Details and Assumptions

The subscript represents number base.

The answer is 4.

2 solutions

William Isoroku
Jan 15, 2015

We need to find $the\:number\:base\:in\:which\:base\:31\:is\:in$ . To do this, use the method of how to express numbers in certain number base:

$31_{x}=3x+1$

Now substitute this into $162$ and $65$ (again, to express numbers in number bases):

$1(3x+1)^{2}+6(3x+1)+2(3x+1)^{0}=3(6(3x+1)+5(3x+1)^{0})$ $\longrightarrow$

$1(3x+1)^{2}+6(3x+1)+2=3(6(3x+1)+5)$ $\longrightarrow$

$9x^2-30x-24=0$

Solving this quadratic equation gives $x=4$ and $x=\frac{-2}{3}$ . Of course number bases can't be negative so $x=\boxed{4}$

Plug in $4$ and the answer equation is true.

You say "of course number bases can't be negative." Why not? Base -2, for example, is quite convenient and fun to work with.. you don't need negative signs to represent integers. ;)

Otto Bretscher - 6 years, 2 months ago

Gaurav Tiwari
Jan 17, 2015

Let 31(base)x = y y^2 +6y+2=3(6y+5) So, y=13 Therefore, 13 = 3x+1 And x is 4

NumberBaseCeption

The answer is 4.

2 solutions

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