Number System — AFCAT Numerical Ability Formulas and Traps

Types of Numbers: Sets, Rationals, Primes, Squares and Cubes

Rational or irrational, and terminating or repeating

Terminating test

\dfrac pq \text{ in lowest terms terminates} \;\text{means}\; q = 2^a \times 5^b

Primes, composites, even and odd numbers

Prime test

n > 1 \text{ is prime when no prime } p \le \sqrt n \text{ divides it}

Perfect squares and perfect cubes

Bracketing test

n^3 < N < (n+1)^3 \;\Rightarrow\; N \text{ is not a perfect cube}

Natural, whole, integer, rational, irrational and real numbers

Set	What it holds	In it	Not in it
Natural $\mathbb{N}$	The counting numbers 1, 2, 3, ...	$1,\ 15,\ 200$	$0,\ -4,\ 2.5$
Whole $\mathbb{W}$	The natural numbers and 0	$0,\ 9,\ 41$	$-1,\ \tfrac 34$
Integers $\mathbb{Z}$	The whole numbers and their negatives	$-12,\ 0,\ 8$	$\tfrac 12,\ 0.4$ A negative number is an integer but never a whole or natural number.
Rational $\mathbb{Q}$	$\tfrac pq$ with integers $p, q$ and $q \neq 0$	$\tfrac 34,\ -5,\ 0.25,\ 0.\overline{4}$	$\sqrt 3,\ \sqrt{11}$
Irrational	Decimals that never end and never repeat	$\sqrt 3,\ \sqrt{11},\ 2\sqrt 6$	$\sqrt{25},\ 0.\overline{4}$
Real $\mathbb{R}$	The rationals and irrationals together	Every point on the number line	The square root of a negative number

Common traps

Zero is whole, not natural

The natural numbers start at 1. Zero belongs to every other set in the chain, so 'zero is a natural number' is the false statement in an option list.

The chain runs one way

Every natural number is an integer, but not every integer is natural:

-3

is an integer and not natural. A statement that reverses the chain is false.

Simplify before you test

\tfrac{6}{15}

looks like a repeater because 15 has a factor 3, but it is

\tfrac 25 = 0.4

. Reduce to lowest terms first.

22/7 is not pi

\tfrac{22}{7}

is a fraction, so it is rational; it only approximates

\pi

. An option list asking for the irrational number wants

\pi

, not

\tfrac{22}{7}

.

A long decimal can still repeat

0.142857142857\ldots

looks endless, but the block 142857 repeats, so it is

\tfrac 17

and rational.

1 is not prime

A list that includes 1 is not a list of primes. A prime needs exactly two factors and 1 has only one.

Odd is not prime

51 = 3 \times 17

and

57 = 3 \times 19

are odd and composite. Test each candidate; do not stop at 'it is odd'.

A cube is not automatically a square

27 = 3^3

is a cube, but it lies between

5^2 = 25

and

6^2 = 36

, so it is not a square. Being both needs a sixth power.

Count the zeros

4,000 ends in three zeros, an odd number, so it cannot be a perfect square. 40,000 ends in four and is

200^2

.

Divisibility Rules from 2 to 12, and Counting Multiples

Learn this subtopic in the notes

Rules that add the digits: 3 and 9

Digit-sum rule

N \text{ divisible by } 9 \;\text{means}\; \text{its digit sum is divisible by } 9

Rules for 7 and 11

Rule for 11

(d_0 + d_2 + d_4 + \cdots) - (d_1 + d_3 + \cdots) = 0, \pm 11, \pm 22, \ldots

Divisors made of two rules: 6, 12, 15, 18, 36

Combining two rules

a \mid N,\ b \mid N,\ a \text{ and } b \text{ share no factor} \;\Rightarrow\; ab \mid N

A missing digit, or the least number to add or take away

Least number to add or subtract

N = dq + r:\quad \text{add } d - r,\ \text{or subtract } r

Counting and adding the multiples in a range

Multiples of d from d·m to d·n

\text{count} = n - m + 1,\qquad \text{sum} = \dfrac{\text{count} \times (dm + dn)}{2}

Rules that read the last digits: 2, 4, 5, 8, 10 and 25

Divisor	Test	Example that passes
2	Last digit is 0, 2, 4, 6 or 8	5,318
4	Last two digits form a multiple of 4	7,316 (16) An even last digit is not enough: 1,326 ends in 26, which is not a multiple of 4.
8	Last three digits form a multiple of 8	41,128 (128)
5	Last digit is 0 or 5	6,035
10	Last digit is 0	4,560
25	Last two digits are 00, 25, 50 or 75	3,175

Common traps

Two digits are not enough for 8

3,140 ends in 40, a multiple of 8, but the rule for 8 reads three digits: 140 is not a multiple of 8, so 3,140 is not divisible by 8.

Divisible by 3 is not divisible by 9

99,999,996 has digit sum 69, a multiple of 3 but not of 9. It is divisible by 3 only.

11 passes as well as 0

An alternating sum of 11 or

-11

means divisible, just as 0 does. Only a result that is not a multiple of 11 means no.

Subtract twice the last digit

For 7 the doubled last digit is taken away, not added: 504 gives

50 - 8 = 42

. Adding gives 58 and the wrong verdict.

The pieces must share nothing

12 is divisible by 4 and by 6 but not by 24. Test 24 as

3 \times 8

, never as

4 \times 6

.

Add the divisor minus the remainder

4,327 leaves remainder 1 on division by 7, so 6 must be added. Adding the remainder itself gives 4,328, which leaves 2.

Count both ends

From

4 \times 1

to

4 \times 10

there are

10 - 1 + 1 = 10

multiples of 4, not 9. Subtracting the two multipliers forgets one end.

Both 4 and 6 means 12

4 and 6 share the factor 2, so a number divisible by both is a multiple of 12, not of 24. Multiply only when the two share no factor.

Prime Factorisation, and Counting Divisors from It

Learn this subtopic in the notes

Writing a prime factorisation

Prime factorisation

N = p_1^{a_1} \times p_2^{a_2} \times \cdots \times p_k^{a_k} \quad (p_i \text{ prime})

Reading a factorisation: distinct primes, repeats and the number itself

For N = pᵃ × qᵇ × rᶜ

\text{distinct primes} = 3,\qquad \text{primes with repeats} = a + b + c

How many divisors a number has

Number of divisors

N = p^a q^b r^c \;\Rightarrow\; d(N) = (a + 1)(b + 1)(c + 1)

Odd divisors, even divisors and the sum of the divisors

Odd and even divisors of 2ᵃ × qᵇ × rᶜ

\text{odd} = (b + 1)(c + 1),\qquad \text{even} = a(b + 1)(c + 1)

Common traps

Keep going until every factor is prime

2^2 \times 3 \times 15

is not finished:

15 = 3 \times 5

, so the factorisation is

2^2 \times 3^2 \times 5

.

Add exponents when pieces share a prime

3600 = 36 \times 100

has

2^2

from each piece, so the power of 2 is

2^4

, not

2^2

.

Distinct is not counted with repeats

2^3 \times 5

has 2 distinct prime factors but 4 counted with repeats. Read which one the question asks for.

A bare exponent counts once

In

2^4 \times 3 \times 5^2

the 3 has exponent 1. Leaving it out makes the sum of exponents 6 instead of 7.

Add 1 to each exponent first

2^3 \times 3^2

has

(3 + 1)(2 + 1) = 12

divisors. Multiplying the exponents themselves gives 6, which is wrong.

Factorise fully before counting

Written as

4 \times 25

, 100 seems to give

(1 + 1)(1 + 1) = 4

divisors. Only prime powers work:

2^2 \times 5^2

gives 9.

Even divisors are not half the total

2^3 \times 3 = 24

has 8 divisors: 2 odd (1 and 3) and 6 even. Split the count with the formula, never in half.

Factors and Multiples: HCF, LCM and Co-prime Numbers

Learn this subtopic in the notes

Factors, multiples and the common ones

Factor pairs

n = a \times b \;\Rightarrow\; a \text{ and } b \text{ are both factors of } n

HCF, by primes or by Euclid, and co-prime numbers

HCF from prime factors

\text{HCF} = \text{product of the common primes, each to its lowest power}

LCM, and the least number leaving a given remainder

LCM from prime factors

\text{LCM} = \text{product of every prime present, each to its highest power}

HCF × LCM = product of the two numbers

Two numbers

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

Common traps

Factors versus multiples

The factors of 24 run from 1 to 24; its multiples start at 24 and go on (48, 72, ...). An option list of multiples offered as factors is the wrong answer.

Do not drop 1 or the number itself

A factor list of 28 without 1 or 28 is incomplete, and so is a list of common factors without 1.

Co-prime is not prime

8 and 15 are co-prime though neither is prime. Two odd numbers need not be co-prime: 15 and 21 share the factor 3.

Lowest powers for HCF

For

2^3 \times 5

and

2 \times 5^2

the HCF is

2 \times 5 = 10

. Taking the highest powers gives the LCM instead.

The LCM is not always the product

8 and 14 share the factor 2, so their LCM is 56, not

8 \times 14 = 112

. The product is the LCM only for co-prime numbers.

Add the remainder, do not subtract it

A number leaving remainder 3 on 4, 9 and 14 is 3 more than a common multiple: 255, not 249.

Two numbers only

For 2, 4 and 8 the HCF is 2 and the LCM is 8, so HCF × LCM is 16, but the product is 64. The rule holds for a pair, not for three numbers.

Remainders and Quotients: Division, Remainder Arithmetic and Powers

Learn this subtopic in the notes

Dividend = divisor × quotient + remainder

Division algorithm

N = d \times q + r,\qquad 0 \le r < d

Remainders of sums, multiples and products

Working with remainders

a \equiv r_a,\ b \equiv r_b \pmod d \;\Rightarrow\; a + b \equiv r_a + r_b,\quad ab \equiv r_a r_b \pmod d

Remainders of large powers

Splitting the exponent

a^k \equiv 1 \pmod d \;\Rightarrow\; a^{kq + s} \equiv a^s \pmod d

Common traps

The remainder is smaller than the divisor

Dividing 500 by 16, the answer 30 remainder 20 is wrong because 20 is more than 16: one more 16 fits, giving 31 remainder 4.

Multiply back to check

d \times q + r

must give the dividend exactly.

16 \times 31 + 4 = 500

confirms the division above.

Reduce the answer

If

N \equiv 7 \pmod 9

, then

2N

gives

14

, which is not a remainder on 9. Reduce it:

14 - 9 = 5

.

The digit-sum shortcut is for 3 and 9 only

123 has digit sum 6, and it does leave 6 on division by 9. On division by 7 it leaves 4, not 6.

Minus 1 means one less than the divisor

12^{25} \equiv -1 \pmod{13}

, and the remainder to choose is 12. A negative number is never the answer.

Use the full cycle length

The powers of 3 on 7 repeat every 6, not every 3. Splitting 41 by 3 gives

3^2

and the wrong remainder 2; by 6 it gives

3^5

and 5.

Number System formulas

Types of Numbers: Sets, Rationals, Primes, Squares and Cubes

Rational or irrational, and terminating or repeating

Primes, composites, even and odd numbers

Perfect squares and perfect cubes

Natural, whole, integer, rational, irrational and real numbers

Common traps

Divisibility Rules from 2 to 12, and Counting Multiples

Rules that add the digits: 3 and 9

Rules for 7 and 11

Divisors made of two rules: 6, 12, 15, 18, 36

A missing digit, or the least number to add or take away

Counting and adding the multiples in a range

Rules that read the last digits: 2, 4, 5, 8, 10 and 25

Common traps

Prime Factorisation, and Counting Divisors from It

Writing a prime factorisation

Reading a factorisation: distinct primes, repeats and the number itself

How many divisors a number has

Odd divisors, even divisors and the sum of the divisors

Common traps

Factors and Multiples: HCF, LCM and Co-prime Numbers

Factors, multiples and the common ones

HCF, by primes or by Euclid, and co-prime numbers

LCM, and the least number leaving a given remainder

HCF × LCM = product of the two numbers

Common traps

Remainders and Quotients: Division, Remainder Arithmetic and Powers

Dividend = divisor × quotient + remainder

Remainders of sums, multiples and products

Remainders of large powers

Common traps

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