(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 $n$.