Given that \(a,{\text{ }}b,{\text{ }}c,{\text{ }}d \in \mathbb{Z}\), show that

\[(a - b)(a - c)(a - d)(b - c)(b - d)(c - d) \equiv 0(\bmod 3).\]

EITHER

we work modulo 3 throughout

the values of a, b, c, d can only be 0, 1, 2     R2

since there are 4 variables but only 3 possible values, at least 2 of the variables must be equal \((\bmod 3)\)     R2

therefore at least 1 of the differences must be \(0(\bmod 3)\)     R2

the product is therefore \(0(\bmod 3)\)     R1AG

OR

we attempt to find values for the differences which do not give \(0(\bmod 3)\) for the product

we work modulo 3 throughout

we note first that none of the differences can be zero     R1

a − b can therefore only be 1 or 2     R1

suppose it is 1, then b − c can only be 1

since if it is 2, \((a - b) + (b - c) \equiv 3 \equiv 0(\bmod 3)\)     R1

c − d cannot now be 1 because if it is

\((a - b) + (b - c) + (c - d) = a - d \equiv 3 \equiv 0(\bmod 3)\)     R1

c − d cannot now be 2 because if it is

\((b - c) + (c - d) = b - d \equiv 3 \equiv 0(\bmod 3)\)     R1

we cannot therefore find values of c and d to give the required result     R1

a similar argument holds if we suppose a − b is 2, in which case b − c must be 2 and we cannot find a value of c − d     R1

the product is therefore \(0(\bmod 3)\)     AG

[7 marks]

Most candidates who solved this question used the argument that there are four variables which can take only one of three different values modulo 3 so that at least two must be equivalent modulo 3 which leads to the required result. This apparently simple result, however, requires a fair amount of insight and few candidates managed it.