Date | May 2013 | Marks available | 6 | Reference code | 13M.3dm.hl.TZ0.3 |
Level | HL only | Paper | Paper 3 Discrete mathematics | Time zone | TZ0 |
Command term | Rewrite | Question number | 3 | Adapted from | N/A |
Question
When numbers are written in base n, \({33^2} = 1331\).
By writing down an appropriate polynomial equation, determine the value of n.
Rewrite the above equation with numbers in base 7.
Markscheme
the equation can be written as
\({(3n + 3)^2} = {n^3} + 3{n^2} + 3n + 1\) M1A1
any valid method of solution giving n = 8 (M1)A1
Note: Attempt to change at least one side into an equation in n gains the M1.
[4 marks]
METHOD 1
as decimal numbers,
\({(33)_8} = 27,{\text{ }}{(1331)_8} = 729\) A1A1
converting to base 7 numbers,
\(27 = {(36)_7}\) A1
7)729 M1
7)104(1
7) 14(6
7) 2(0
7) 0(2
therefore \(729 = {(2061)_7}\) A1
the required equation is
\({36^2} = 2061\) A1
METHOD 2
as a decimal number, \({(33)_8} = 27\) A1
converting to base 7,
\(27 = {(36)_7}\) A1
multiplying base 7 numbers
36
× 36
1440 M1
321 A1
2061 A1
the required equation is
\({36^2} = 2061\) A1
Note: Allow M1 for showing the method of converting a number to base 7 regardless of what number they convert.
[6 marks]
Examiners report
Part (a) was a good indicator of overall ability. Many candidates did not write both sides of the equation in terms of n and thus had an impossible equation, which should have made them realise that they had a mistake. The answers that were given in (a) and (b) could have been checked, so that the candidate knew they had done it correctly.
Part (b) was not well answered and of those candidates that did, some only gave one side of the equation in base 7. The answers that were given in (a) and (b) could have been checked, so that the candidate knew they had done it correctly.