Number System — SSC CGL Quantitative Aptitude Formulas and Traps

Number Types, Primes and Squares

When a fraction has a terminating decimal

Terminating decimal test

\frac{p}{q} \ (\text{lowest terms}) \text{ terminates exactly when } q = 2^a \times 5^b

$q$ the denominator after cancelling
$a, b$ whole numbers, either may be 0

Perfect squares near a given number

Bracketing between squares

k^2 \le N < (k + 1)^2: \ \text{add } (k + 1)^2 - N \text{ or subtract } N - k^2 \qquad (a + b)^2 = a^2 + 2ab + b^2

$N$ the given number
$k$ the whole number whose square is just at or below N

The families of numbers and how they nest

Family	What it contains	Examples
Natural numbers	the counting numbers, starting at 1	1, 7, 250
Whole numbers	0 and every natural number	0, 1, 2, 15 0 is whole but not natural.
Integers	the whole numbers and their negatives	$-4$ , 0, 7
Rational numbers	every $\frac{p}{q}$ with $p, q$ integers and $q \ne 0$	$\frac{3}{4}$ , $-2$ , 1.5, $0.333\ldots$
Irrational numbers	real numbers that are not $\frac{p}{q}$	$\sqrt{2}$ , $\sqrt{7}$ , $\pi$
Real numbers	all rational and all irrational numbers	every point on the number line
Even numbers	integers that are multiples of 2	$-6$ , 0, 86, 88
Odd numbers	integers that leave remainder 1 on division by 2	$-7$ , 3, 15

Natural numbers sit inside the whole numbers, whole numbers inside the integers, integers inside the rationals; the rationals and the irrationals together make the real numbers.

Common traps

Calling 0 a natural number

Natural numbers start at 1. Zero is a whole number, an integer and a rational number, but not a natural number.

Saying rational times irrational is always irrational

It is irrational only when the rational number is not zero. Since

0 \times \sqrt{3} = 0

, the honest answer is 'sometimes rational, sometimes irrational'.

Taking a repeating decimal for an irrational number

0.333\ldots = \frac{1}{3}

is rational. A decimal is irrational only when it never stops and never settles into a repeating block.

Judging the denominator before cancelling

\frac{6}{15}

looks recurring because 15 has a factor 3, but it cancels to

\frac{2}{5} = 0.4

. Always reduce to lowest terms first.

Counting 1 as a prime

A prime has exactly two factors. The number 1 has only one, so it is neither prime nor composite.

Letting 51, 57 or 91 through as prime

They look prime but are not:

51 = 3 \times 17

,

57 = 3 \times 19

,

91 = 7 \times 13

. Run the digit-sum check for 3 and try 7 before you accept a two-digit number.

Going to the wrong square

The least number to add takes you UP to the next square; the least number to subtract takes you DOWN to the square below. Bracket the number first so you know both.

Trusting the last digit as proof

A number ending in 1, 4, 5, 6 or 9 may still not be a square (26 is not). The last digit can only strike options out; confirm a square by bracketing or squaring.

Division Algorithm and Multiples

Learn this subtopic in the notes

The division algorithm

Division algorithm

\text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder}, \qquad 0 \le \text{Remainder} < \text{Divisor}

The nearest multiple of a number

Distance to the nearest multiples

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

$N$ the given number
$d$ the divisor
$r$ the remainder when N is divided by d

Common traps

Accepting a remainder as large as the divisor

A remainder is always less than the divisor. If your working gives a remainder equal to or bigger than the divisor, the quotient is too small.

Scaling the quotient of a division that had a remainder

Multiplying the quotient by the ratio of the two factors is exact only when the original division left remainder 0. Otherwise rebuild the dividend and divide again.

Adding the remainder instead of the gap

To reach the next multiple you add the divisor minus the remainder. Adding the remainder itself overshoots or falls short: with

650 = 30 \times 21 + 20

, you add 10, not 20.

Landing on a multiple with the wrong number of digits

For the smallest 5-digit multiple, start from 10000, not from 9999. A multiple just below 10000 has only four digits, so check the digit count of your answer before you mark it.

Divisibility Tests

Learn this subtopic in the notes

The alternating-sum test for 11

Test for 11

D = (\text{sum of digits in odd places}) - (\text{sum of digits in even places}), \quad 11 \text{ divides } N \text{ exactly when } D = 0, \pm 11, \pm 22, \ldots \qquad \overline{abcabc} = \overline{abc} \times 7 \times 11 \times 13

$D$ the alternating difference, places counted from the right
$\overline{abc}$ the three-digit number with digits a, b, c

Testing a composite divisor by its co-prime parts

Common co-prime splits

6 = 2 \times 3 \quad 12 = 3 \times 4 \quad 18 = 2 \times 9 \quad 24 = 3 \times 8 \quad 33 = 3 \times 11 \quad 36 = 4 \times 9 \quad 44 = 4 \times 11 \quad 45 = 5 \times 9 \quad 72 = 8 \times 9 \quad 99 = 9 \times 11

Remainders from the digit tests

Remainders from digits

N \div 9 \text{ leaves the remainder of (digit sum)} \div 9 \qquad N \div 11 \text{ leaves the remainder of } D \div 11

Divisibility tests for 2, 3, 4, 5, 7, 8, 9, 10, 11 and 25

Divisor	Test	Example
2	last digit is 0, 2, 4, 6 or 8	3574
3	digit sum is divisible by 3	5124: digit sum 12
4	last two digits form a multiple of 4	7316: 16
5	last digit is 0 or 5	6845
7	twice the last digit taken from the rest gives a multiple of 7	2149: $214 - 18 = 196 = 7 \times 28$
8	last three digits form a multiple of 8	41136: $136 = 8 \times 17$ Three digits, not two: 8 is where most slips happen.
9	digit sum is divisible by 9	7236: digit sum 18
10	last digit is 0	4590
11	the two sums of alternate digits differ by 0 or a multiple of 11	2816: $(6 + 8) - (1 + 2) = 11$
25	last two digits are 00, 25, 50 or 75	3475

Each test says whether the number divides exactly. The same digits also give the remainder, which a later concept on this page uses.

Common traps

Testing 8 on the last two digits

4 needs the last two digits; 8 needs the last three. 316 ends in 16, a multiple of 8, yet

316 = 8 \times 39 + 4

.

Treating divisible by 3 as divisible by 9

A digit sum of 12, 21 or 24 gives divisibility by 3 only. For 9 the digit sum itself must be 9, 18, 27 and so on.

Rejecting a difference of 11

The difference does not have to be 0. A difference of 11,

-11

or 22 also means the number is divisible by 11.

Skipping a zero digit

In 30712 the 0 sits in place 4. Leave it out and every digit after it changes side, and the test gives the wrong verdict.

Splitting a divisor into parts that share a factor

Testing 4 and 6 does not test 24, because 4 and 6 share the factor 2. Use co-prime parts:

24 = 3 \times 8

.

Running only one of the two tests

For 36 a number must pass both 4 and 9. 1242 has digit sum 9 but ends in 42, which is not a multiple of 4, so 36 does not divide it.

Reading the 11 difference from the left

For the remainder, count places from the right. From the left, a number with an even count of digits gives the difference with the wrong sign: 4072 read from the left gives 9, but its remainder is 2.

Giving a negative difference as the remainder

A remainder is never negative. A difference of

-9

on division by 11 means a remainder of

-9 + 11 = 2

.

Missing Digits

Learn this subtopic in the notes

A missing digit from a digit-sum test

Hidden digit from a digit sum

S + x = 9k \ (\text{or } 3k), \quad 0 \le x \le 9

$S$ the sum of the digits you can see
$x$ the hidden digit

A missing digit from a last-digits test

Last-digit tests for 8 and 4

\overline{abc} \text{ is a multiple of } 8: \ a \text{ even} \Rightarrow 8 \text{ divides } \overline{bc}; \ a \text{ odd} \Rightarrow 8 \text{ divides } \overline{bc} + 4 \qquad 4 \text{ divides } \overline{bc} \text{ exactly when } 4 \text{ divides } 2b + c

$\overline{abc}$ the last three digits, with a in the hundreds place
$\overline{bc}$ the last two digits read as a two-digit number

Common traps

Accepting a value that is not a digit

Options sometimes include 10 or 12. A missing digit lies between 0 and 9, so throw those out before testing anything.

Missing the second answer for 9

If the visible digits already add to a multiple of 9, the hidden digit can be 0 or 9. Read whether the stem wants the least or the greatest.

Letting a multiple of 8 and 5 end in 5

A number ending in 5 is odd, and no odd number is a multiple of 8. Divisible by both 5 and 8 means the last digit is 0.

Running the 8 test on two digits

The tail for 8 is three digits long. 136 and 236 share the last two digits 36, but only 136 is a multiple of 8; the hundreds digit decides.

Setting the difference to 0 and stopping

If

D = 0

needs a digit of 10 or more, try

D = 11

and

D = -11

. A digit that makes the difference 11 is just as valid.

Factorisation, HCF and LCM

Learn this subtopic in the notes

Prime factorisation and counting divisors

Divisors from the prime factorisation

N = p^a q^b r^c: \quad \text{number of divisors} = (a + 1)(b + 1)(c + 1) \qquad \text{sum of divisors} = (1 + p + \cdots + p^a)(1 + q + \cdots + q^b)(1 + r + \cdots + r^c) \qquad 1 + 2 + \cdots + n = \frac{n(n + 1)}{2}

$p, q, r$ the different prime factors of N
$a, b, c$ how many times each prime appears

Taking a common power out of a sum of powers

Sums and differences of powers

a^m + a^{m+1} + \cdots + a^{m+k} = a^m(1 + a + \cdots + a^k) \qquad a^n - b^n \text{ is divisible by } a - b \qquad a^n - b^n \text{ is divisible by } a + b \ (n \text{ even}) \qquad a^n + b^n \text{ is divisible by } a + b \ (n \text{ odd})

HCF, LCM and the least square multiple

HCF, LCM, tiles and the mean proportional

\text{HCF} \times \text{LCM} = a \times b \qquad \sqrt{ab} = \sqrt{\text{HCF} \times \text{LCM}} \qquad \text{tiles} = \frac{L}{h} \times \frac{B}{h}, \ h = \text{HCF}(L, B)

$a, b$ the two numbers
$L, B$ the length and breadth of the floor, in one unit
$h$ the side of the largest square tile

Common traps

Multiplying the powers instead of the powers plus one

2^3 \times 3^2

has

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

divisors, not

3 \times 2 = 6

. The plus one is the choice of taking none of that prime.

Forgetting the factors of the power you took out

3^8 + 3^9 = 3^8 \times 4

is divisible by 4, but also by 3, by 81 and by 12. When more than one option divides the sum, check the factors of

3^8

as well as the bracket.

Using a + b on an odd power of a difference

a^n - b^n

is divisible by

a + b

only for even

n

.

8^3 - 1 = 511 = 7 \times 73

is not divisible by

8 + 1 = 9

.

Using HCF times LCM for three numbers

HCF

\times

LCM equals the product only for two numbers. For 2, 4 and 8 the HCF is 2 and the LCM 8, but the product is 64.

Squaring the LCM to get a square multiple

The least square multiple of

12 = 2^2 \times 3

is

12 \times 3 = 36

, not

12^2 = 144

. Multiply only by the primes whose power is odd.

Remainders by Congruence and Cyclicity

Learn this subtopic in the notes

Replacing a number by its remainder

Swapping numbers for remainders

N = dk + r \Rightarrow N \equiv r \pmod d \qquad a \equiv r, \ b \equiv s \pmod d \Rightarrow a + b \equiv r + s, \ ab \equiv rs \pmod d

$d$ the divisor
$r, s$ the remainders of a and b on division by d

Remainders of large powers

Powers when the base leaves 1 or one less than the divisor

a \equiv 1 \pmod d \Rightarrow a^n \equiv 1 \qquad a \equiv -1 \pmod d \Rightarrow a^n \equiv 1 \ (n \text{ even}), \ a^n \equiv d - 1 \ (n \text{ odd})

$a$ the base
$d$ the divisor
$n$ the exponent

Common traps

Moving to a divisor that does not divide the first one

A remainder on division by 45 gives the remainder on division by 15, 9, 5 or 3, because they divide 45. It says nothing exact about division by 12.

Leaving the answer bigger than the divisor

After swapping, reduce again. Remainder 5 on division by 7, tripled, is 15, and the remainder is

15 - 14 = 1

, not 15.

Writing minus one as the remainder

A remainder is never negative. When the working gives

-1

on division by 13, the remainder is

13 - 1 = 12

.

Ignoring whether the exponent is odd or even

A base one less than the divisor leaves 1 for an even exponent but leaves one less than the divisor for an odd one. Check the exponent's parity before you write the answer.

Number System formulas

Number Types, Primes and Squares

When a fraction has a terminating decimal

Perfect squares near a given number

The families of numbers and how they nest

Common traps

Division Algorithm and Multiples

The division algorithm

The nearest multiple of a number

Common traps

Divisibility Tests

The alternating-sum test for 11

Testing a composite divisor by its co-prime parts

Remainders from the digit tests

Divisibility tests for 2, 3, 4, 5, 7, 8, 9, 10, 11 and 25

Common traps

Missing Digits

A missing digit from a digit-sum test

A missing digit from a last-digits test

Common traps

Factorisation, HCF and LCM

Prime factorisation and counting divisors

Taking a common power out of a sum of powers

HCF, LCM and the least square multiple

Common traps

Remainders by Congruence and Cyclicity

Replacing a number by its remainder

Remainders of large powers

Common traps

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