Question

Two numbers in the base system $B$ are $2061_B$ and $601_B$. The sum of these two numbers in decimal system is 432. Find the value of $1010_B$ in decimal system.

• A. 110
• B. 120
• C. 130
• D. 140
• E. 150

Hint:

When dealing with unknown bases, always expand the digits into powers of $B$, form an equation with the given decimal value, and solve for $B$. Often, patterns like binomial expansions simplify the equation.

Answer 1

The Correct Option is C

Step 1: Express both numbers in decimal.
\[ 2061_B = 2B^3 + 0B^2 + 6B + 1, \] \[ 601_B = 6B^2 + 0B + 1. \] Step 2: Add them and equate to 432.
\[ 2061_B + 601_B = (2B^3 + 6B + 1) + (6B^2 + 1), \] \[ = 2B^3 + 6B^2 + 6B + 2. \] Given this equals $432$: \[ 2B^3 + 6B^2 + 6B + 2 = 432. \] Step 3: Simplify equation.
\[ 2B^3 + 6B^2 + 6B = 430, \] \[ B^3 + 3B^2 + 3B = 215. \] Notice left side is $(B+1)^3 - 1$. \[ (B+1)^3 - 1 = 215 \;\;\Rightarrow\;\; (B+1)^3 = 216. \] \[ B+1 = 6 \;\;\Rightarrow\;\; B = 5. \] Step 4: Find decimal value of $1010_B$.
\[ 1010_B = 1..... B^3 + 0..... B^2 + 1..... B + 0, \] \[ = B^3 + B. \] With $B=5$, \[ 1010_5 = 5^3 + 5 = 125 + 5 = 130. \] \[ \boxed{130} \]