(i)     Given that \({(43)_n} \times {(56)_n} = {(3112)_n}\), show that \(3{n^3} - 19{n^2} - 38n - 16 = 0\).

(ii)     Hence determine the value of \(n\).

Determine the set of values of \(n\) satisfying \({(13)_n} \times {(21)_n} = {(273)_n}\).

Show that there are no possible values of \(n\) satisfying \({(32)_n} \times {(61)_n} = {(1839)_n}\).

(i)     \(n\) satisfies the equation \((4n + 3)(5n + 6) = 3{n^3} + {n^2} + n + 2\)     M1A1

\(3{n^3} - 19{n^2} - 38n - 16 = 0\)    (AG)

(ii)     \(n = 8\)     A1

Note:     If extra solutions \(( - 1,{\text{ }} - 2/3)\) are not rejected (them just not appearing is fine) do not award the final A1.

[3 marks]

\(n\) satisfies the equation \((n + 3)(2n + 1) = 2{n^2} + 7n + 3\)     A1

this is an identity satisfied by all \(n\)     (A1)

\(n > 7\) or \(n \geqslant 8\)     A1

[3 marks]

\(n\) satisfies the equation \((3n + 2)(6n + 1) = {n^3} + 8{n^2} + 3n + 9\)     A1

\({n^3} - 10{n^2} - 12n + 7 = 0\)    A1

roots are 11.03, 0.434 and –1.46     A1

since there are no integer roots therefore the product is not true in any number base     R1AG

Note:     Accept an argument by contradiction that considers the equation modulo \(n\), with \(n > 10\).

[4 marks]

Well answered.

The fact that this gave an identity was managed by most. Then some showed their misunderstanding by saying any real number. Few noticed that the digit 7 means that the base must be greater than 7.

The cubic equation was generally reached but many candidates then forgot what type of number \(n\) had to be. To justify that there are no positive integer roots you need to write down what the roots are. There were a couple of really neat solutions that obtained a contradiction by working modulo \(n\).