| Date | May 2016 | Marks available | 2 | Reference code | 16M.1.hl.TZ0.3 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find and Hence | Question number | 3 | Adapted from | N/A |
Question
Consider the Diophantine equation \(7x - 5y = 1,{\text{ }}x,{\text{ }}y \in \mathbb{Z}\).
Find the general solution to this equation.
Hence find the solution with minimum positive value of \(xy\).
Find the solution satisfying \(xy = 2014\).
Markscheme
one solution is \(x = - 2,{\text{ }}y = - 3{\text{ }}\left( {{\text{or }}(3,{\text{ }}4)} \right)\) (A1)
the general solution is
\(x = - 2 + 5N,{\text{ }}y = - 3 + 7N{\text{ }}({\text{or }}x = 3 + 5M,{\text{ }}y = 4 + 7M)\) M1A1
[3 marks]
a listing of small values of the product (M1)
\( \Rightarrow x = - 2,{\text{ }}y = - 3\) (the least positive value of \(xy\) is 6) A1
[2 marks]
use of “table” or otherwise to solve
\(35{N^2} - 29N + 6 = 2014{\text{ }}({\text{or }}35{M^2} + 41M + 12 = 2014)\) (M1)
obtain \(N = 8{\text{ }}({\text{or }}M = 7)\) (A1)
\(x = 38,{\text{ }}y = 53\) A1
[3 marks]
Examiners report
This was also a very successful question with many wholly correct answers seen. A small number of candidates made arithmetic errors in the calculations. Some candidates used unnecessarily long and complex methods for parts (b) and (c) which would have potentially left them short of time elsewhere.
This was also a very successful question with many wholly correct answers seen. A small number of candidates made arithmetic errors in the calculations. Some candidates used unnecessarily long and complex methods for parts (b) and (c) which would have potentially left them short of time elsewhere.
This was also a very successful question with many wholly correct answers seen. A small number of candidates made arithmetic errors in the calculations. Some candidates used unnecessarily long and complex methods for parts (b) and (c) which would have potentially left them short of time elsewhere.
