Date | November 2017 | Marks available | 3 | Reference code | 17N.3dm.hl.TZ0.5 |
Level | HL only | Paper | Paper 3 Discrete mathematics | Time zone | TZ0 |
Command term | Convert | Question number | 5 | Adapted from | N/A |
Question
The decimal number 1071 is equal to \(a\)060 in base \(b\), where \(a > 0\).
Convert the decimal number 1071 to base 12.
Write the decimal number 1071 as a product of its prime factors.
Using your answers to part (a) and (b), prove that there is only one possible value for \(b\) and state this value.
Hence state the value of \(a\).
Markscheme
EITHER
using a list of relevant powers of 12: 1, 12, 144 (M1)
\(1071 = 7 \times {12^2} + 5 \times {12^1} + 3 \times {12^0}\) (A1)
OR
attempted repeated division by 12 (M1)
\(1071 \div 12 = 89{\text{rem}}3;{\text{ }}89 \div 12 = 7{\text{rem}}5\) (A1)
THEN
\(1071 = {753_{12}}\) A1
[3 marks]
\(1071 = 3 \times 3 \times 7 \times 17\) A1
[1 mark]
in base \(b\) \(a060\) ends in a zero and so \(b\) is a factor of 1071 R1
from part (a) \(b < 12\) as \(a060\) has four digits and so the possibilities are
\(b = 3,{\text{ }}b = 7\) or \(b = 9\) R1
stating valid reasons to exclude both \(b = 3\) eg, there is a digit of 6
and \(b = 9\) eg, \(1071 = {(1420)_9}\) R1
\(b = 7\) A1
Note: The A mark is independent of the R marks.
[4 marks]
\(1071 = {(3060)_7} \Rightarrow a = 3\) A1
[1 mark]