GATE CSE 2015 | Set 3 | Question: 35

Question

Consider the equation $(43)_x = (y3)_8$ where $x$ and $y$ are unknown. The number of possible solutions is _____

Comments

thanks

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https://www.youtube.com/watch?v=pGeDFJSSVYA

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So the number of possible solutions is 5

Answer 1

$$(43)_x=(y3)_8$$

Since a number in base$-k$ can only have digits from $0$ to $(k-1)$, we can conclude that: $x \geq 5$ and $y \leq 7$

Now, the original equation, when converted to decimal base gives:

$$\begin{align}4x^1 + 3x^0 &= y(8^1)+3(8^0)\\[0.8em]4x + 3 &= 8y + 3\\[0.8em]x&=2y\end{align}$$

So, we have the following constraints:

$$x\geq 5\\y \leq 7\\x=2y\\x,y \text{ are integers}$$

The set of values of $(x,y)$ that satisfy these constraints are:

$$\begin{align}(x&,y)\\\hline\\(6&,3)\\(8&,4)\\(10&,5)\\(12&,6)\\(14&,7)\end{align}$$

I am counting 5 pairs of values.

Comments

Don’t you think radix 1 system is also possible?

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@akagami

what is the maximum value possible in that system?

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by observation we can clearly find that x>= 5 and y<=7

on coverting both to decimal, x = 2y

the pairs become (6,3) (8,4) (10,5) (12,6) (14,7)

Answer 2

(43)_x =(y3)₈

From this equation it is clear x should be greater than 4 and y should be less than 8.

x>=5, and y<=7

Now convert this equation in decimal

_{=> 4*x+3=y*8+3}

_4x=8y

_{ie x=2y} 

With given constraint (x>=5 && y<=7)

possible values of x and y are (6,3), (8,4),(10,5),(12,6), (14,7) 

So possible number of solution is 5.

Answer 3

4x+3 = 8y+3

x = 2y  ....Equation 1

but x has to be greater than 4; coz the maximal number it represents is 4

y has to be less than 8; coz the maximal base possible is 8

hence, we get pairs of (x,y) from equation 1 as 

(2y, y)

(6, 3)
(8, 4)
(10, 5)
(12, 6)
(14, 7)
 

Answer 4

Analyze the exponents:

\[8 \text{ is a factor of } 3, \text{ so we can rewrite the equation as } (43)^x = [(y^3)^2]^4.\] This simplifies to \(43^x = y^6.\)

Consider possible constraints: Since \(43\) is prime, any factor of \(y^6\) must also include \(43.\) However, \(y\) itself cannot be \(43\) because then \(y^3\) wouldn't reach the \(8\)th power. Therefore, \(y\) must be divisible by \(43\), but not equal to \(43.\)

Determine valid combinations of \(x\) and \(y\):

Try different values for \(y\) that are multiples of \(43:\)

1. \(y = 86 - x = 2\) (invalid: \(86^6\) doesn't have a factor of \(43)\)
2. \(y = 129 - x = 3\) (valid: \(129^6\) has a factor of \(43)\)
3. \(y = 172 - x = 4\) (valid: \(172^6\) has a factor of \(43)\)
4. \(y = 215 - x = 5\) (valid: \(215^6\) has a factor of \(43)\)
5. \(y = 258 - x = 6\) (valid: \(258^6\) has a factor of \(43)\) 

Count the solutions:

We found \(5\) valid combinations (for \(y = 129, 172, 215, 258, \text{ and } 341)\). Therefore, there are \(5\) possible solutions for the equation \(43^x = (y^3)^8\).

Answer 5

Digit constraints: In any base-k number system, digits can only range from 0 to k-1. In base 8, this means digits can be from 0 to 7. Therefore, x must be greater than 5 and y must be less than 7.

Expanding the equation: We can convert the equation to decimal form:

$(43)_x = 4 * x^1 + 3 * x^0 = 4x + 3$

$(y3)_8 = y * 8^1 + 3 * 8^0 = 8y + 3$

Simplifying and solving: Combining the expressions and solving for x, we get:

$4x + 3 = 8y + 3$

$8y = 4x$

$x = 2y$

Finding possible solutions: Now, we substitute values for y within the constraints mentioned in step 1:

$y = 3  => x = 6$

$y = 4 => x = 8$

$y = 5 => x = 10$

$y = 6 => x = 12$

$y = 7 => x = 14$

Therefore, there are 5 possible solutions for the equation: (x, y) = (6, 3), (8, 4), (10, 5), (12, 6). (14, 7).