Number System — SPI Mathematics Formulas and Traps

Number Types and Place Value

Place value and face value

Place value

\text{place value} = \text{digit} \times \text{value of its place}

Counting digits and numbers made from digits

Sum of all numbers from three different non-zero digits

\text{sum} = 2 \times (a + b + c) \times 111 \qquad \text{numbers from } m \text{ to } n = n - m + 1

Natural numbers, whole numbers, integers, even and odd

Family	What it holds	Examples
Natural numbers	The counting numbers, starting at 1	1, 2, 3, 4, ...
Whole numbers	The natural numbers and 0	0, 1, 2, 3, ... 0 is the smallest whole number. There is no smallest integer.
Integers	The whole numbers and their negatives	..., $-2$ , $-1$ , 0, 1, 2, ...
Even integers	Numbers of the form $2m$	$-4$ , 0, 6, 10
Odd integers	Numbers of the form $2m + 1$	$-3$ , 1, 7, 11
Powers of $-1$	1 for an even power, $-1$ for an odd power	$(-1)^4 = 1$ , $(-1)^7 = -1$

Each family contains the one above it: natural numbers are whole numbers, and whole numbers are integers.

Common traps

Calling 0 a natural number

The natural numbers start at 1. The set

\{0, 1, 2, 3, \ldots\}

is the whole numbers.

Losing track of alternating signs

In a long sum of powers of

-1

, pair the terms first. If a minus sign sits in front of every even power, then every term becomes

-1

, and the sum is minus the number of terms.

Confusing place value with face value

The face value of a digit never changes. Only the place value grows with the place, so read which one the question asks for.

Counting places from the left

Always count places from the right, starting with units. In a seven-digit number the second digit from the left is in the hundred thousands place.

The smallest five-digit number

It is 10000, not 11111 or 00000. A number cannot start with 0, so the first digit is 1 and the rest are zeros.

Counting a range one short

From 100 to 200 there are 101 numbers, not 100. With both ends included, the count is last minus first plus one.

Using the 111 rule when one digit is 0

A number cannot start with 0. From 0, 3 and 5 only four three-digit numbers exist (305, 350, 503, 530), with sum 1688. The rule

2 \times 8 \times 111 = 1776

wrongly adds 035 and 053.

Divisibility, Primes and Factorisation

Learn this subtopic in the notes

Prime factorisation

Prime factorisation and end zeros

n = p_1^{a_1} \times p_2^{a_2} \times \cdots \times p_k^{a_k} \qquad \text{zeros at the end of } 2^a \times 5^b \times \ldots = \text{the smaller of } a, b

p₁, p₂, ..., pₖthe different primes that divide n
a₁, a₂, ..., aₖhow many times each prime appears

Divisibility rules

Divisor	Rule	Example
2	Last digit is 0, 2, 4, 6 or 8	3576
3	Sum of the digits is divisible by 3	741: $7 + 4 + 1 = 12$
4	Last two digits form a number divisible by 4	5316: 16
5	Last digit is 0 or 5	8935
6	Divisible by both 2 and 3	912: even, $9 + 1 + 2 = 12$
8	Last three digits form a number divisible by 8	71240: 240
9	Sum of the digits is divisible by 9	6534: $6 + 5 + 3 + 4 = 18$
11	Alternating sum of the digits is 0 or a multiple of 11	3729: $3 - 7 + 2 - 9 = -11$ Add and subtract the digits in turn. The plain digit sum says nothing about 11.

Each example in the last column passes its rule.

Common traps

Taking away multiples of the product, not the LCM

Up to 50 there are 12 multiples of 4. Those that are also multiples of 6 are the multiples of 12 (12, 24, 36, 48), so 8 are divisible by 4 but not by 6. Taking away only the multiples of 24 gives 10, which is wrong.

Testing 6 with one rule

A number divisible by 3 is not always divisible by 6. It must also be even: 75 passes the rule for 3 but is odd.

Calling 1 a prime

1 has only one factor, so it is neither prime nor composite. The smallest prime is 2.

Numbers that look prime

91 (

7 \times 13

), 119 (

7 \times 17

) and 143 (

11 \times 13

) are odd and pass the rules for 3 and 5, yet they are composite. Test the primes up to the square root before deciding.

Stopping at a composite piece

2^2 \times 9 \times 5

is not finished, because 9 is

3^2

. Every factor in the final answer must be a prime.

Adding the powers of 2 and 5 for end zeros

2^3 \times 5^4

ends in 3 zeros, not 7. Each zero uses one 2 and one 5 together, so the smaller power decides.

Squaring the divisor

If a prime p divides

b^2

, then p divides b, but

p^2

need not divide b: 3 divides

3^2

, yet 9 does not divide 3.

HCF and LCM

Learn this subtopic in the notes

HCF and LCM from prime factors

HCF and LCM by powers

\text{HCF} = \prod p^{\text{lowest power}} \qquad \text{LCM} = \prod p^{\text{highest power}}

Euclid's division and the largest number that divides

Euclid's division

a = bq + r, \quad 0 \le r < b \qquad \text{HCF}(a, b) = \text{HCF}(b, r)

HCF times LCM equals the product

Product of two numbers

\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b

Common traps

Putting a prime in the HCF that one number lacks

The HCF uses only primes that divide every number. If 5 divides one number but not the other, 5 is not in the HCF.

Picking a common multiple that is not the least

72 is a common multiple of 12 and 18, but the LCM is 36. Options often include larger common multiples; the LCM is the smallest.

Taking the HCF before removing the remainder

When a remainder is left, subtract it from each number first. The answer divides the reduced numbers exactly, not the original ones.

An answer smaller than the remainder

A divisor must be larger than the remainder it leaves. If your answer is 4 and the remainder is 5, something went wrong.

Using the product rule for three numbers

For 2, 4 and 8 the HCF is 2 and the LCM is 8, so HCF times LCM is 16, but the product is 64. The rule is for two numbers only.

Fractions and Rational Numbers

Learn this subtopic in the notes

Rational numbers and comparing them

Rational number

\frac{p}{q}, \quad p, q \text{ integers}, \quad q \ne 0 \qquad \text{a number between } x \text{ and } y: \frac{x + y}{2}

Adding fractions and taking a fraction of a number

Fraction sums and parts

\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd} \qquad \frac{a}{b} \text{ of } N = \frac{a}{b} \times N

Recurring decimals as fractions

Pure recurring decimals

0.\overline{a} = \frac{a}{9} \qquad 0.\overline{ab} = \frac{ab}{99}

$ab$ the two-digit block that repeats, read as one number

Common traps

Reading negative fractions by size

-\frac{2}{3}

is less than

-\frac{1}{5}

, because it lies further left on the number line. Every number between two negatives is negative, so a positive option can never lie between them.

Thinking the numerator cannot be 0

\frac{0}{q}

is a rational number (it is 0) for any non-zero integer q. The rule that forbids 0 is for the denominator.

Taking a fraction of the whole instead of the rest

If half of the remaining people stay home, multiply

\frac{1}{2}

by the fraction that remains. Taking

\frac{1}{2}

of the whole group gives a wrong total.

Adding the denominators

\frac{1}{2} + \frac{1}{3}

is

\frac{5}{6}

, not

\frac{2}{5}

. Only the numerators add, and only after the denominators are made the same.

Putting the whole part into the block

3.\overline{18}

is

3 + \frac{18}{99} = \frac{35}{11}

, not

\frac{318}{99}

. Only the repeating digits go over the nines.

Mixing up where the bar starts

0.\overline{45}

is

0.4545\ldots

, but

0.4\overline{5}

is

0.4555\ldots

. They are different numbers, and both can appear among the options.

Irrational Numbers

Learn this subtopic in the notes

Sums and products with irrational numbers

Squares and products with a root

(a + \sqrt{b})^2 = a^2 + 2a\sqrt{b} + b \qquad (a + \sqrt{b})(a - \sqrt{b}) = a^2 - b

Which numbers are irrational

Number	Type	Reason
$\sqrt{2}$ , $\sqrt{3}$ , $\sqrt{5}$	Irrational	2, 3 and 5 are not perfect squares
$\sqrt{49}$	Rational	It equals 7
$\sqrt{1296}$	Rational	It equals 36
$\pi$	Irrational	Its decimal never ends and never repeats $\frac{22}{7}$ is close to $\pi$ but is a fraction, so it is rational.
$\frac{22}{7}$ , 3.14	Rational	Each is a fraction of two integers
$0.\overline{3}$	Rational	It repeats: $\frac{1}{3}$
0.101001000100001...	Irrational	It never ends and has no repeating block

A root sign alone decides nothing: check whether the number under it is a perfect square.

Common traps

Treating every root as irrational

\sqrt{2401} = 49

is a whole number. Before calling a root irrational, check whether the number under it is a perfect square.

Taking 22/7 for π

\frac{22}{7}

is a fraction, so it is rational. Only

\pi

itself is irrational.

Assuming two irrationals add to an irrational

(5 + \sqrt{2}) + (1 - \sqrt{2}) = 6

, which is rational. The sum of two irrationals can be rational or irrational.

Dropping the middle term when squaring

(1 + \sqrt{5})^2

is

6 + 2\sqrt{5}

, not

1 + 5 = 6

. The middle term

2a\sqrt{b}

keeps the result irrational.

Forgetting the zero case

Rational times irrational is irrational only when the rational is not 0. A question that says just 'a rational number' hides this exception.

Number System formulas

Number Types and Place Value

Place value and face value

Counting digits and numbers made from digits

Natural numbers, whole numbers, integers, even and odd

Common traps

Divisibility, Primes and Factorisation

Prime factorisation

Divisibility rules

Common traps

HCF and LCM

HCF and LCM from prime factors

Euclid's division and the largest number that divides

HCF times LCM equals the product

Common traps

Fractions and Rational Numbers

Rational numbers and comparing them

Adding fractions and taking a fraction of a number

Recurring decimals as fractions

Common traps

Irrational Numbers

Sums and products with irrational numbers

Which numbers are irrational

Common traps

More SPI Mathematics formula sheets