Find the value of $x$ in $3115_x + 4514_x = 10632_x$.

Question

Find the value of $x$ in $3115_x + 4514_x = 10632_x$.

I don’t understand how to even begin to go about solving this, because the bases are unknown. I don’t know if I should try tackling the bases first or the actual numbers themselves, and I’m just a bit lost. Any help, please?

Answer 1

We know that $x \ge 7$ since $6$ occurs by itself.

Write $3115_x + 4514_x = 10632_x $ as $(3+4)x^3+(1+5)x^2+(1+1)x+(5+4) =x^4+6x^2+3x+2 $ or $7x^3+6x^2+2x+9 =x^4+6x^2+3x+2 $.

We must have $9 = 2+x$, so $x = 7$.

As a check, $7x^3 = x^4$ for $x = 7$.

Answer 2

Hint #1: The maximum number in this case is $6$, and you have only the numbers $0-6.$

Hint #2: Make an addition table to find out why $5 + 4 = 12$ and $3 + 4 = 10$ in this base.

Answer 3

The ones digit of $3115_x$ is $5$, and the ones digit of $4514_x$ is $4$.

The ones digit of their sum should be $9$, unless $x \le9$, in which case it should be $9-x$.

Can you take it from here?

Answer 4

THis is just $$(5+x+x^2+3x^3)+(4+x+5x^2+4x^3)=(2+3x+6x^2+x^4) $$ Thus because of $1+5=6$ (coefficients of $x^2$) and $3+4=0$ (coefficients of $x^3$) the base is $7$.