Parity and Divisibility Arguments

The parity of a sum depends only on the parities of what you added, never on the actual numbers. Two of a kind give an even total — odd plus odd is even, and even plus even is even. One of each gives an odd total. Subtraction follows exactly the same table, because seven take three is even in just the same way that seven plus three is. So whenever a question asks whether a total is odd or even, you can answer it without doing the arithmetic at all.

With a longer list, adding an even number never changes whether the running total is odd or even, so you can ignore every even number and just count the odd ones. In this list there are two odd numbers, seven and thirteen, and two is an even count, so the total is even. That gives the rule worth memorising: count the odd numbers, and if that count is even the total is even, while if the count is odd the total is odd. This is exactly what settles the card question on ACER's sample paper, and it takes about five seconds.

Multiplying behaves differently from adding, and the table is lopsided. A product is odd only when every single factor is odd; one even factor anywhere in the list drags the whole product even, because that factor contributes a two. So for a long product you do not need to check everything — you only need to find one even factor, and you can stop. If you cannot find one, the product is odd.

The last digit of a product depends only on the last digits of the numbers you multiplied. Twenty-three times seventeen must end in one, because three sevens are twenty-one. On a four-option question that immediately kills any option not ending in one. To choose between the survivors, estimate: twenty-three times seventeen is close to twenty times twenty, which is four hundred, so three hundred and ninety-one rather than three hundred and seventy-one. Last digit narrows, estimate decides, and together they beat doing the multiplication. The same trick works for sums.

A handful of divisibility tests run entirely in your head. A number is divisible by two if its last digit is even, and by five if the last digit is nought or five, and by ten if it is nought. For three, add the digits and see whether that total is a multiple of three — seven four one gives twelve, so yes. For nine, add the digits and look for a multiple of nine. For four, look at the last two digits only. And divisible by six means divisible by two and by three together, so check both rather than one. That is the lesson: parity of sums, counting the odd numbers, one even factor, the last digit, and digit sums. Quiz and exercises now.