Consider the equation (43)x = (y3)8 where x and y are unknown. The number of possible solution is
5
Solving the Base Conversion Equation (43)x = (y3)8
This problem asks us to find the number of possible solutions for the equation (43)x = (y3)8, where x and y represent unknown digits or bases.
Step 1: Convert the Equation to Base 10
To solve this equation, we first convert both sides from their respective bases into base 10 (decimal).
- The term
(43)xin base 10 is: $$ (43)_x = 4 \times x^1 + 3 \times x^0 = 4x + 3 $$ - The term
(y3)8in base 10 is: $$ (y3)_8 = y \times 8^1 + 3 \times 8^0 = 8y + 3 $$
Equating the base 10 expressions gives us:
$$ 4x + 3 = 8y + 3 $$Step 2: Simplify the Equation
Subtracting 3 from both sides of the equation simplifies it:
$$ 4x = 8y $$Dividing both sides by 4, we get the relationship between x and y:
Step 3: Analyze Constraints on Bases and Digits
For the equation to be valid, we must consider the constraints imposed by number bases:
- Base x Constraint: In the number
(43)x, the digits used are 4 and 3. In any number system, the base must be strictly greater than any digit present in the number. Therefore, the basexmust satisfy: $$ x > 4 $$ - Base 8 Constraint: The number
(y3)8is in base 8. This means the base is valid (8 is greater than 3). The digitymust be less than the base. Thus,ymust satisfy: $$ 0 \le y < 8 $$ Sinceyrepresents a digit, it must be an integer. So, the possible values foryare 0, 1, 2, 3, 4, 5, 6, and 7.
Step 4: Find Possible Solutions
We use the relationship x = 2y and the constraints x > 4 and 0 <= y <= 7 to find the valid pairs of (x, y).
Let's test each possible integer value for y from 0 to 7:
| Possible Value of y | Calculate x using $x = 2y$ | Resulting Base x | Check Validity of Base x (Is $x > 4$?) | Is this a Valid Solution? |
|---|---|---|---|---|
| 0 | $x = 2 \times 0 = 0$ | 0 | No (Base must be greater than 0) | No |
| 1 | $x = 2 \times 1 = 2$ | 2 | No ($2$ is not greater than $4$) | No |
| 2 | $x = 2 \times 2 = 4$ | 4 | No ($4$ is not strictly greater than $4$) | No |
| 3 | $x = 2 \times 3 = 6$ | 6 | Yes ($6 > 4$) | Yes |
| 4 | $x = 2 \times 4 = 8$ | 8 | Yes ($8 > 4$) | Yes |
| 5 | $x = 2 \times 5 = 10$ | 10 | Yes ($10 > 4$) | Yes |
| 6 | $x = 2 \times 6 = 12$ | 12 | Yes ($12 > 4$) | Yes |
| 7 | $x = 2 \times 7 = 14$ | 14 | Yes ($14 > 4$) | Yes |
Step 5: Count the Number of Solutions
By analyzing the table, we found the following valid pairs (x, y) that satisfy the original equation and the constraints:
- (6, 3)
- (8, 4)
- (10, 5)
- (12, 6)
- (14, 7)
Counting these pairs, we find there are 5 possible solutions.
Important Questions from Binary Operations
Multiplication of 111 2by 101 2is
Let * be binary operation defined on R by \(\rm p * q=\frac{p+q}{2}, \forall \) p, q ∈ R. The operation is:
Convert 29 into binary.
A. 10101
B. 11110
C. 11101
D. 11001
The product of the two binary numbers 011 and 110 is:
If G is the set of integer numbers, and a.b = a – b, ∀ a, b ∈ G, then G is
