Understanding Number Bases and the Problem

The question asks us to find the value of the base 'b' in the given equation involving numbers represented in different bases. The equation is: $(146)_b + (313)_{b-2} = (246)_8$.

To solve this, we need to convert each number from its respective base into the standard base 10 (decimal) system. Once all terms are in base 10, we can solve the resulting algebraic equation for 'b'.

An important constraint for any number base system is that the base must be greater than the largest digit used in that number. In base $b$, the digits are 1, 4, and 6, so $b$ must be greater than 6 ($b > 6$). In base $b-2$, the digits are 3, 1, and 3, so $b-2$ must be greater than 3 ($b-2 > 3$, which means $b > 5$). Also, a base must be at least 2, so $b-2 \ge 2$, implying $b \ge 4$. In base 8, the digits are 2, 4, and 6, and 8 is greater than 6, which is valid. Combining these constraints, the base $b$ must be an integer such that $b > 6$.

Converting Numbers to Base 10

To convert a number from base $r$ to base 10, we use the formula: $(d_n d_{n-1} \ldots d_1 d_0)_r = d_n \times r^n + d_{n-1} \times r^{n-1} + \ldots + d_1 \times r^1 + d_0 \times r^0$.

Converting $(146)_b$ to Base 10:

$(146)_b = 1 \times b^2 + 4 \times b^1 + 6 \times b^0$

$(146)_b = b^2 + 4b + 6$ in base 10.

Converting $(313)_{b-2}$ to Base 10:

$(313)_{b-2} = 3 \times (b-2)^2 + 1 \times (b-2)^1 + 3 \times (b-2)^0$

$(313)_{b-2} = 3 \times (b^2 - 4b + 4) + 1 \times (b-2) + 3 \times 1$

$(313)_{b-2} = 3b^2 - 12b + 12 + b - 2 + 3$

$(313)_{b-2} = 3b^2 - 11b + 13$ in base 10.

Converting $(246)_8$ to Base 10:

$(246)_8 = 2 \times 8^2 + 4 \times 8^1 + 6 \times 8^0$

$(246)_8 = 2 \times 64 + 4 \times 8 + 6 \times 1$

$(246)_8 = 128 + 32 + 6$

$(246)_8 = 166$ in base 10.

Setting up and Solving the Equation in Base 10

Now substitute the base 10 equivalents back into the original equation:

$(b^2 + 4b + 6) + (3b^2 - 11b + 13) = 166$

Combine like terms:

$b^2 + 3b^2 + 4b - 11b + 6 + 13 = 166$

$4b^2 - 7b + 19 = 166$

Subtract 166 from both sides to set the quadratic equation to zero:

$4b^2 - 7b + 19 - 166 = 0$

$4b^2 - 7b - 147 = 0$

We have a quadratic equation in the form $ab^2 + cb + d = 0$, where $a=4$, $c=-7$, and $d=-147$. We can solve for $b$ using the quadratic formula:

$b = \frac{-c \pm \sqrt{c^2 - 4ad}}{2a}$

$b = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(4)(-147)}}{2(4)}$

$b = \frac{7 \pm \sqrt{49 + 16 \times 147}}{8}$

$b = \frac{7 \pm \sqrt{49 + 2352}}{8}$

$b = \frac{7 \pm \sqrt{2401}}{8}$

Let's find the square root of 2401. We know that $40^2 = 1600$ and $50^2 = 2500$. The number 2401 ends in 1, so its square root must end in 1 or 9. Let's try 49:

$49 \times 49 = 2401$

So, $\sqrt{2401} = 49$.

Substitute this back into the formula for $b$:

$b = \frac{7 \pm 49}{8}$

This gives us two possible values for $b$:

• $b_1 = \frac{7 + 49}{8} = \frac{56}{8} = 7$
• $b_2 = \frac{7 - 49}{8} = \frac{-42}{8} = -\frac{21}{4}$

Checking the Validity of Solutions for Base b

We established earlier that the base $b$ must be an integer greater than 6 ($b > 6$). Let's check our potential solutions:

• For $b_1 = 7$:
  • Is $b_1$ an integer? Yes, 7 is an integer.
  • Is $b_1 > 6$? Yes, $7 > 6$.
  • The base $b$ is 7. The number $(146)_7$ is valid as all digits (1, 4, 6) are less than 7.
  • The base $b-2$ is $7-2=5$. The number $(313)_5$ is valid as all digits (3, 1, 3) are less than 5.
  • The base 8 for $(246)_8$ is valid as all digits (2, 4, 6) are less than 8.
  • Since $b=7$ satisfies all conditions, it is a valid solution.
• For $b_2 = -\frac{21}{4}$:
  • Is $b_2$ an integer? No, $-\frac{21}{4}$ is a fraction.
  • Also, bases of number systems must be positive. This value is negative.
  • Therefore, $b_2 = -\frac{21}{4}$ is not a valid solution for a number base.

The only valid value for $b$ is 7.

Conclusion

By converting the numbers from their respective bases to base 10 and solving the resulting quadratic equation, we found the possible values for $b$ to be 7 and $-\frac{21}{4}$. Applying the constraints required for number bases (must be an integer greater than the largest digit used in the numbers), we determine that only $b=7$ is a valid solution.

Revision Table: Number Base Equation Solution

Additional Information: Properties of Number Bases

• A number base (or radix) $r$ is the number of unique digits, including zero, used to represent numbers in a positional numeral system.
• The digits used in a base $r$ system range from 0 to $r-1$.
• For a number represented as $(d_n d_{n-1} \ldots d_0)_r$, each digit $d_i$ must satisfy $0 \le d_i < r$.
• The value of the number in base 10 is calculated as $\sum_{i=0}^{n} d_i \times r^i$.
• Common bases include base 2 (binary), base 8 (octal), base 10 (decimal), and base 16 (hexadecimal).
• When solving equations involving unknown bases, it is crucial to convert all numbers to a common base (usually base 10) and then solve the resulting algebraic equation. Remember to verify that the calculated base value satisfies the constraints derived from the digits used in the original numbers.