Show that \({2^n} \equiv {( - 1)^n}(\bmod 3)\), where \(n \in \mathbb{N}\).

Hence show that an integer is divisible by 3 if and only if the difference between the sum of its binary (base 2) digits in even-numbered positions and the sum of its binary digits in odd-numbered positions is divisible by 3.

Express the hexadecimal (base 16) number \({\text{ABB}}{{\text{A}}_{{\text{16}}}}\) in binary.

METHOD 1

\({2^n} = {(3 - 1)^n}\)    M1

\( = {3^n} + n{3^{n - 1}}( - 1) + \frac{{n(n - 1)}}{2}{3^{n - 2}}{( - 1)^2} + {\text{ }} \ldots {\text{ }} + {( - 1)^n}\)    A1

since all terms apart from the last one are divisible by 3     R1

\({2^n} \equiv {( - 1)^n}(\bmod 3)\)    AG

METHOD 2

attempt to reduce the powers of \(2(\bmod 3)\)     M1

\({2^0} = 1(\bmod 3);{\text{ }}{2^1} =  - 1(\bmod 3);{\text{ }}{2^2} = 1(\bmod 3);{\text{ }}{2^3} =  - 1(\bmod 3){\text{ }} \ldots \)    A1

since \(1(\bmod 3) \times 2 =  - 1(\bmod 3)\) and \( - 1(\bmod 3) \times 2 = 1(\bmod 3)\) the result can be generalized     R1

\(2n \equiv {( - 1)^n}(\bmod 3)\)    AG

[3 marks]

the binary number \(N = {({a_n}{a_{n - 1}} \ldots {a_2}{a_1}{a_0})_2}\) has numerical value

\({a_0} \times 1 + {a_1} \times 2 + {a_2} \times {2^2} +  \ldots  + {a_n} \times {2^n}\)    A1

\(N = \left( {{a_0} - {a_1} + {a_2} -  \ldots {{( - 1)}^n}{a_n}} \right)(\bmod 3)\)    M1A1

hence divisibility of \(N\) by 3 coincides with statement of question     AG

[3 marks]

\({\text{ABB}}{{\text{A}}_{16}} = 10 \times {16^3} + 11 \times {16^2} + 11 \times 16 + 10 \times 1\)    (A1)

\(N = {(1010)_2} \times {2^{12}} + {(1011)_2} \times {2^8} + {(1011)_2} \times {2^4} + {(1010)_2} \times {2^0}\)     (M1)(A1)

Note:     Award M1 for expressing A and B in binary.

\(N = {(1010101110111010)_2}\)    A1

[4 marks]

Many candidates were able to make a beginning to this question and attempted a solution to part (a). Some were let down by being unable to fully explain their reasoning.

In part (b) a number of fully correct answers were seen but some candidates appeared to be completely unaware of what constituted a sensible approach.

Again in part (c) a number of correct answers were seen but a significant number also appeared to have little idea on how to start.