Concepts: Number Play (part 1)

1.1 Multiples, factors and divisibility

A multiple of a number is what you get by multiplying it by a whole number. The multiples of 6 are 6, 12, 18, 24 and so on. A factor of a number divides it exactly, with no remainder. Since 6 × 7 = 42, the numbers 6 and 7 are factors of 42, and 42 is a multiple of both.

Divisibility is the same idea stated as a test. One number is divisible by another when the division leaves no remainder. The chapter starts with a simple question: can every natural number be written as a sum of consecutive natural numbers? Some can be written in more than one way. For example, 15 = 7 + 8, and also 4 + 5 + 6, and also 1 + 2 + 3 + 4 + 5. Exploring such questions and then explaining the pattern is the theme of Number Play.

The tools are the ones used throughout the chapter: algebraic expressions for general numbers, parity, remainders, place value and divisibility shortcuts. None of the shortcuts in this chapter has to be memorised blindly; each one can be proved with the place-value expansion.

1.2 Sums of consecutive numbers

Four consecutive numbers can be written as n, n + 1, n + 2 and n + 3. Their sum is n + (n + 1) + (n + 2) + (n + 3) = 4n + 6. The expression 4n + 6 is always even, because 4n is even and 6 is even, and the sum of two even numbers is even. This is why four consecutive numbers always add to an even number, even though the individual numbers alternate between odd and even.

The same algebra answers the textbook questions. If the sum is 34, then 4n + 6 = 34, so 4n = 28 and n = 7. The four numbers are 7, 8, 9 and 10.

If the greatest of five consecutive numbers is p, the other four are p - 1, then p - 2, next p - 3, and finally p - 4. There is no need to introduce a new letter for the smallest number: naming the greatest one and counting backwards is enough.

Odd numbers always split into two consecutive numbers: 7 = 3 + 4, 11 = 5 + 6, 21 = 10 + 11. In general, 2m + 1 = m + (m + 1). Even numbers sometimes split into consecutive numbers and sometimes do not. The textbook examples show 10 = 1 + 2 + 3 + 4 and 12 = 3 + 4 + 5, while 2 and 4 have no such split into two or more positive consecutive numbers.

1.3 Parity and plus or minus signs

Parity is the property of being odd or even. The useful rules are: odd odd then even even even even next odd even and odd, and the same rules hold for subtraction. An expression with plus and minus signs changes its value by an even amount when one sign is switched, because changing +b to -b subtracts 2b, and 2b is even.

Take four consecutive numbers 3, 4, 5 and 6 and place plus or minus signs in all possible ways. One expression is 3 + 4 - 5 + 6 = 8, and another is 3 - 4 - 5 - 6 = -12. Every result is even. The algebra explains why: start from any expression a ± b ± c ± d and switch one sign. The value changes by 2 times that number, which is even, so the old and new values have the same parity. Since every expression can be reached from any other by a sequence of sign switches, all eight expressions have the same parity.

The same argument works for any number of terms. The sum or difference of several numbers always has the parity obtained by counting the odd terms: an even number of odd terms gives an even result, and an odd number of odd terms gives an odd result.

1.4 Even numbers and multiples of 4

Every even number has the form 2m. Split even numbers by their remainder on division by 4. A multiple of 4 is 4p and leaves remainder 0. An even value that 4 does not divide is 4q + 2 and leaves remainder 2.

Now add two even numbers. If both are multiples of 4, the sum is 4p + 4q = 4(p + q), a multiple of 4. If both leave remainder 2, the sum is (4p + 2) + (4q + 2) = 4(p + q + 1), again a multiple of 4. If one is a multiple of 4 and the other leaves remainder 2, the sum is 4(p + q) + 2, which leaves remainder 2 and is not a multiple of 4.

Worked example 1. Are 12 + 16 and 6 + 10 multiples of 4? Both are: 12 and 16 are multiples of 4, and 6 and 10 each leave remainder 2, so 6 + 10 = 16. The pair 12 + 6 = 18 leaves remainder 2, so it is not a multiple of 4.

1.5 Divisibility reasoning with algebra

When a divides both m and n then M = ap and N = aq for some whole numbers p and q. Then M + N = a(p + q) and M - N = a(p - q), so a divides both the sum and the difference. This one result powers many of the statements in the exercise.

A second useful result: if a divides b, then every multiple of b is divisible by a. For example, since 8 divides 56, every multiple of 56 is a multiple of 8. A third: if a number is divisible by both b and c, then it is a multiple of the LCM of b and c. The number 24 is divisible by 4 and 6, so it must be divisible by LCM(4, 6) = 12; checking 12 is enough to find the common multiple when the two factors share a factor.

Worked example 2. Two numbers leave remainders 3 and 5 when divided by 7. What remainder does their sum leave? Write them as 7a + 3 and 7b + 5. The sum is 7(a + b) + 8 = 7(a + b + 1) + 1, so the remainder is 1. The difference is 7(a - b) - 2 = 7(a - b - 1) + 5, so the remainder is 5. The algebra explains both answers without computing either number.

1.6 Always, sometimes or never

A statement about numbers can be always true, sometimes true or never true. To prove "always", use algebra that works for every integer. To prove "sometimes", give one example where it holds and one where it fails. To prove "never", show that the algebra contradicts the statement for every case.

Consider the statements from the chapter. "The sum of two multiples of 8 is a multiple of 8" is always true: 8a + 8b = 8(a + b). "When 8 divides a number then any two numbers that add up to it are each divisible by 8" is sometimes true, because 72 = 48 + 24 uses two multiples of 8, but 72 = 50 + 22 does not. "When 7 divides a number then all its multiples are divisible by 7" is always true: (7j) × m = 7(jm).

Choosing examples and non-examples is part of the reasoning. A single counterexample is enough to disprove "always", while one example is not enough to prove it. The exercises use this method repeatedly.