Date | May 2015 | Marks available | 3 | Reference code | 15M.1.hl.TZ0.9 |
Level | HL only | Paper | 1 | Time zone | TZ0 |
Command term | Show that | Question number | 9 | Adapted from | N/A |
Question
An integer \(N\) given in base \(r\), can be expressed in base \(s\) in the form
\(N = {a_0} + {a_1}s + {a_2}{s^2} + {a_3}{s^3} + \ldots \) where \({a_0},{\text{ }}{a_1},{\text{ }}{a_2},{\text{ }} \ldots \in \{ 0,{\text{ }}1,{\text{ }}2,{\text{ }} \ldots ,{\text{ }}s - 1\} \).
In base \(5\) an integer is written \(1031\). Express this integer in base \(10\).
Given that \(N = 365,{\text{ }}r = 10\) and \(s = 7\) find the values of \({a_0},{\text{ }}{a_1},{\text{ }}{a_2},{\text{ }} \ldots \).
(i) Given that \(N = 899,{\text{ }}r = 10\) and \(s = 12\) find the values of \({a_0},{\text{ }}{a_1},{\text{ }}{a_2},{\text{ }} \ldots \)
(ii) Hence write down the integer in base \(12\), which is equivalent to \(899\) in base \(10\).
Show that \(121\) is always a square number in any base greater than \(2\).
Markscheme
\(1031 = 1 \times {5^3} + 0 \times {5^2} + 3 \times 5 + 1\) M1
\( = 125 + 0 + 15 + 1\)
\( = 141\) A1
\(365 = 1 \times {7^3} + 0 \times {7^2} + 3 \times 7 + 1\) M1
\( \Rightarrow {a_0} = 1,{\text{ }}{a_1} = 3,{\text{ }}{a_2} = 0,{\text{ }}{a_3} = 1\) A1
(i) \(899 = 6 \times {12^2} + 2 \times 12 + 11\) M1
\( \Rightarrow {a_0} = 11,{\text{ }}{a_1} = 2,{\text{ }}{a_2} = 6\) A1
(ii) \({(899)_{10}} = {(62B)_{12}}\)
(where \(B\) represents the digit in base \(12\) given by \({a_0} = 11\)) A1
Note: Accept any letter in place of \(B\) provided it is defined
\(121\) in base \(r\) is \(1 + 2r + {r^2}\) M1A1
\( = {(r + 1)^2}\) A1
which is a square for all \(r\) AG
Examiners report
This question was very well answered in general, although some candidates failed to see that \(121 = {11^2}\) in all number bases greater than \(2\).
This question was very well answered in general, although some candidates failed to see that \(121 = {11^2}\) in all number bases greater than 2.
This question was very well answered in general, although some candidates failed to see that \(121 = {11^2}\) in all number bases greater than 2.
This question was very well answered in general, although some candidates failed to see that \(121 = {11^2}\) in all number bases greater than 2.