19 votes 19 votes If $73_x$ (in base-x number system) is equal to $54_y$ (in base $y$-number system), the possible values of $x$ and $y$ are $8, 16$ $10, 12$ $9, 13$ $8, 11$ Digital Logic gatecse-2004 digital-logic number-representation easy + – Kathleen asked Sep 18, 2014 edited Jun 19, 2018 by Pooja Khatri Kathleen 6.3k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 30 votes 30 votes Answer is D. $x\times 7 + 3 = 5 \times y + 4 \implies 7x = 5y + 1$. Only option satisfying this is D. Aditi Dan answered Dec 21, 2014 edited Jun 26, 2018 by Milicevic3306 Aditi Dan comment Share Follow See all 2 Comments See all 2 2 Comments reply rohith1001 commented Oct 12, 2019 reply Follow Share I guess there are infinitely many solutions to this. (Not sure though. I don't know how to prove that there are infinitely many solutions. May be use mathematical induction?) Some of the solutions are: x y 8 11 18 25 43 60 68 95 I just plotted the graph for 5y = 7x - 1 here - link to get these solutions 0 votes 0 votes goxul commented Nov 21, 2019 reply Follow Share @rohith1001 Yes, there can be infinitely many solutions. The only conditons are that x > 7 and y > 5. 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes Convert both numbers to base 10: \[73_x = 7x^1 + 3x^0 = 7x + 3\] \[54_y = 5y^1 + 4y^0 = 5y + 4\] Equate the base 10 values: \[7x + 3 = 5y + 4\] Solve for \(x\) in terms of \(y\): \[7x = 5y + 1\] \[x = \frac{5y + 1}{7}\] Check the answer choices: Option (A): \(x = 8, y = 16 \) \(\rightarrow x = \frac{5 \times 16 + 1}{7} = \frac{81}{7}\) (not an integer) Option (B): \(x = 10, y = 12 \) \(\rightarrow x = \frac{5 \times 12 + 1}{7} = \frac{61}{7}\) (not an integer) Option (C): \(x = 9, y = 13 \) \(\rightarrow x = \frac{5 \times 13 + 1}{7} = \frac{66}{7}\) (not an integer) Option (D): \(x = 8, y = 11 \) \(\rightarrow x = \frac{5 \times 11 + 1}{7} = \frac{56}{7} = 8\) (integer) Therefore, the only possible values of \(x\) and \(y\) are 8 and 11, respectively. rajveer43 answered Jan 12 rajveer43 comment Share Follow See all 0 reply Please log in or register to add a comment.