Show that when p = 2 , N is even if and only if its least significant digit, \({a_0}\) , is 0.

Show that when p = 3 , N is even if and only if the sum of its digits is even.

\(N = {a_n} \times {2^n} + {a_{n - 1}} \times {2^{n - 1}} + ... + {a_1} \times 2 + {a_0}\)     M1

If \({a_0} = 0\) , then N is even because all the terms are even.     R1

Now consider

\({a_0} = N - \sum\limits_{r = 1}^n {{a_r} \times {2^r}} \)     M1

If N is even, then \({a_0}\) is the difference of two even numbers and is therefore even.     R1

It must be zero since that is the only even digit in binary arithmetic.     R1

[5 marks]

\(N = {a_n} \times {3^n} + {a_{n - 1}} \times {3^{n - 1}} + ... + {a_1} \times 3 + {a_0}\)

\( = {a_n} \times ({3^n} - 1) + {a_{n - 1}} \times ({3^{n - 1}} - 1) + ... + {a_1} \times (3 - 1) + {a_n} + {a_{n - 1}} + ... + {a_1} + {a_0}\)     M1A1

Since \({3^n}\) is odd for all \(n \in {\mathbb{Z}^ + }\) , it follows that \({3^n} - 1\) is even.     R1

Therefore if the sum of the digits is even, N is the sum of even numbers and is even.     R1

Now consider

\({a_n} + {a_{n - 1}} + ... + {a_1} + {a_0} = N - \sum\limits_{r = 1}^n {{a_r}({3^r} - 1)} \)     M1

If N is even, then the sum of the digits is the difference of even numbers and is therefore even.     R1

[6 marks]

The response to this question was disappointing. Many candidates were successful in showing the ‘if’ parts of (a) and (b) but failed even to realise that they had to continue to prove the ‘only if’ parts.

The response to this question was disappointing. Many candidates were successful in showing the ‘if’ parts of (a) and (b) but failed even to realise that they had to continue to prove the ‘only if’ parts.