Number System for CUET General Aptitude

Q: What are the formulas for the sum of the first n natural, odd and even numbers?

The sum of the first n natural numbers is (n(n + 1))/2, the sum of the first n odd numbers is n^2, and the sum of the first n even numbers is n(n + 1).

Q: How do you find the least number to add or subtract to make a number divisible?

Divide and find the remainder r. To subtract, take away r. To add, add the divisor minus r.

About 15 min to learn · 71 questions

CUET's number system questions cover types of numbers, place value, sums of series, divisibility rules, remainders, unit digits of large powers, factors and primes, and trailing zeros. Almost every question falls to one rule applied in under a minute: a divisibility test, a unit-digit cycle or a factor-counting formula. This lesson gives each rule with a worked example and the traps the wrong options are built from.

Syllabus area: General Mental Ability, Numerical Ability

On this page · 10 sections

01The core idea
02Types of numbers
03Sums of series
04Divisibility
05Remainders and unit digits
06Factors and primes
07Exam shortcuts
08Common mistakes
09Quick revision
10Check your understanding

The core idea

Number system questions look varied, but nearly all of them fall to one of a handful of rules: a divisibility test, a remainder rule, a unit-digit cycle, or the factor-counting formula. Learn the rules and the question becomes a one-line check.

Two habits help throughout. Break a number into its prime factors whenever you can, because divisibility, factors, squares and cubes all read straight off the factorisation. And when a question involves a huge number, work only with the part that matters: the last digit, the last two or three digits, the digit sum or the remainder.

Types of numbers

What are the different types of numbers?

The families nest inside each other: natural numbers inside whole numbers, whole numbers inside integers, and integers inside rational numbers. Irrational numbers sit beside the rationals, and together the two make up the real numbers.

Type	What it contains	Examples
Natural numbers	1, 2, 3, …	7, 150
Whole numbers	0 and the natural numbers	0, 7, 150
Integers	Whole numbers and their negatives	−12, 0, 45
Rational numbers	Anything that can be written $\tfrac{p}{q}$ , with $p$ and $q$ integers and $q \ne 0$	$\tfrac{3}{4}$ , −5, 0.25, 0.333…
Irrational numbers	Real numbers that cannot be written $\tfrac{p}{q}$	$\sqrt{2}$ , $\sqrt{15}$ , $\pi$

Rational or irrational?

A decimal that terminates, like 0.125, or recurs, like 0.4545…, is always rational: $0.4545\ldots = \tfrac{45}{99} = \tfrac{5}{11}$ .
The square root of a whole number is rational only when the number is a perfect square: $\sqrt{81} = 9$ is rational, and $\sqrt{20} = 2\sqrt{5}$ is not.
Simplify before you judge: $(3 + \sqrt{5})(3 - \sqrt{5}) = 9 - 5 = 4$ is rational.

What is the difference between prime, composite and co-prime numbers?

A prime has exactly two factors, 1 and itself. A composite number has more than two. The number 1 is neither. Two numbers are co-prime when their only common factor is 1, and they need not be prime themselves: 8 and 15 are co-prime.

2 is the only even prime.
There are 25 primes below 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
To test a number for primality, divide by the primes up to its square root. For 113, try 2, 3, 5 and 7; $11^2 = 121$ is already too big. None divides, so 113 is prime.

Watch out

Some numbers only look prime: $51 = 3 \times 17$ , $57 = 3 \times 19$ , $87 = 3 \times 29$ and $91 = 7 \times 13$ .

What is the difference between place value and face value?

The face value of a digit is the digit itself. Its place value is the digit times the value of its position. In 5,72,418 the 7 has face value 7 and place value 70,000, because it sits in the ten-thousands place. The Indian places run ones, tens, hundreds, thousands, ten-thousands, lakhs, ten-lakhs and crores.

Example 1 · Place and face value

In 9,14,736, find the place value of 4 minus the face value of 1.

The 4 is in the thousands place, so its place value is 4,000.
The face value of 1 is 1.
$4{,}000 - 1 = 3{,}999$ .

Answer

3,999

Sums of series

What are the formulas for the sum of the first n natural, odd and even numbers?

The sum of the first $n$ natural numbers is $\tfrac{n(n + 1)}{2}$ , the sum of the first $n$ odd numbers is $n^2$ , and the sum of the first $n$ even numbers is $n(n + 1)$ .

Sum of the first $n$	Formula	With $n = 12$
Natural numbers, $1 + 2 + \cdots + n$	$\tfrac{n(n + 1)}{2}$	$\tfrac{12 \times 13}{2} = 78$
Odd numbers, $1 + 3 + 5 + \cdots$	$n^2$	144
Even numbers, $2 + 4 + 6 + \cdots$	$n(n + 1)$	156

Any evenly spaced run

\text{Sum} = \frac{\text{number of terms} \times (\text{first} + \text{last})}{2} \qquad \text{Number of terms} = \frac{\text{last} - \text{first}}{\text{gap}} + 1

Example 2 · Odd numbers in a range

Find the sum of all odd numbers from 21 to 59.

Number of terms: $(59 - 21) \div 2 + 1 = 20$ .
Sum: $20 \times (21 + 59) \div 2 = 20 \times 40 = 800$ .

Answer

800

Example 3 · A run that does not start at 1

Find the sum of the natural numbers from 31 to 60.

Sum from 1 to 60: $60 \times 61 \div 2 = 1{,}830$ .
Sum from 1 to 30: $30 \times 31 \div 2 = 465$ .
Difference: $1{,}830 - 465 = 1{,}365$ .

Answer

1,365

Watch out

Counting terms: divide the span by the gap, then add 1. From 21 to 59 in steps of 2 is 20 terms, not 19.

Divisibility

What are the divisibility rules from 2 to 11?

Each rule checks divisibility from the last few digits or the digit sum, without doing the division.

Divisor	Rule
2	Last digit is 0, 2, 4, 6 or 8
3	Digit sum is divisible by 3
4	The last two digits form a number divisible by 4
5	Last digit is 0 or 5
6	Divisible by both 2 and 3
8	The last three digits form a number divisible by 8
9	Digit sum is divisible by 9
10	Last digit is 0
11	Sum of the digits in odd places minus sum in even places is 0 or a multiple of 11
25	Last two digits are 00, 25, 50 or 75

The test for 11

For 8,316, the odd places from the left give $8 + 1 = 9$ and the even places $3 + 6 = 9$ . The difference is 0, so 8,316 is divisible by 11; it is $11 \times 756$ . A difference of 11 or 22 passes too.

Watch out

Divisibility by 4 is not divisibility by 8. 6,124 passes the test for 4, since 24 is a multiple of 4, but fails the test for 8, since $124 \div 8 = 15.5$ .

How do you test divisibility by 12, 15, 18, 36 or 72?

Split the divisor into co-prime factors and apply both tests: 12 is $3 \times 4$ , 15 is $3 \times 5$ , 18 is $2 \times 9$ , 24 is $3 \times 8$ , 36 is $4 \times 9$ , 45 is $5 \times 9$ , and 72 is $8 \times 9$ .

Example 4 · Divisible by 36?

Is 3,924 divisible by 36?

$36 = 4 \times 9$ , and 4 and 9 are co-prime.
The last two digits, 24, are divisible by 4.
The digit sum, $3 + 9 + 2 + 4 = 18$ , is divisible by 9.

Answer

Yes

Indeed $3{,}924 = 36 \times 109$ .

Watch out

The factors must be co-prime. 18 is divisible by 2 and by 6, but not by 12, because 2 and 6 share a factor.

How do you find a missing digit so that a number is divisible?

Write the rule as an equation in the missing digit, remembering that it must be a single digit from 0 to 9. For a composite divisor, satisfy both co-prime tests and keep only a digit that passes both.

Example 5 · Missing digit for 9

Find the digit ☐ if 5☐714 is divisible by 9.

Digit sum: $5 + 7 + 1 + 4 + \square = 17 + \square$ .
The next multiple of 9 is 18, so ☐ = 1. The one after, 27, would need ☐ = 10, which is not a digit.

Answer

Example 6 · Missing digit for 11

Find the digit ☐ if 81☐3 is divisible by 11.

Odd places: $8 + \square$ . Even places: $1 + 3 = 4$ .
The difference $(8 + \square) - 4 = 4 + \square$ must be 0 or 11, so ☐ = 7.

Answer

Check: $8{,}173 = 11 \times 743$ .

Exam shortcut

When several digits work, read whether the question wants the smallest, the largest or the sum of all of them.

How do you find the least number to add or subtract to make a number divisible?

Divide and find the remainder $r$ . To subtract, take away $r$ . To add, add the divisor minus $r$ .

Example 7 · Least number to add

What is the least number to add to 1,000 to make it divisible by 23?

$23 \times 43 = 989$ , so the remainder is $1{,}000 - 989 = 11$ .
Add $23 - 11 = 12$ .

Answer

Check: $1{,}012 = 23 \times 44$ .

Key idea

The same idea finds the largest or smallest number of a given length with a divisor. The largest four-digit multiple of 16: 9,999 leaves remainder 15 on division by 16, so it is $9{,}999 - 15 = 9{,}984$ .

How do you count the multiples of a number in a range?

Find the first and the last multiple in the range, divide each by the divisor, subtract, and add 1. To count numbers divisible by neither of two divisors, count those divisible by either, adding the two counts and subtracting the overlap, and take that from the total.

Example 8 · Counting multiples

How many multiples of 9 lie between 200 and 400? And how many numbers from 1 to 150 are divisible by neither 3 nor 5?

The first multiple of 9 is $207 = 9 \times 23$ and the last is $396 = 9 \times 44$ , so there are $44 - 23 + 1 = 22$ .
From 1 to 150: 50 are divisible by 3, 30 by 5, and 10 by both, that is, by 15.
Divisible by 3 or 5: $50 + 30 - 10 = 70$ . Neither: $150 - 70 = 80$ .

Answer

22 and 80

Remainders and unit digits

How do you find the remainder of a sum, product or power?

The remainder of a sum is the sum of the remainders, and the remainder of a product is the product of the remainders, each reduced by the divisor at the end. Powers follow from products.

$23 \times 38$ divided by 5: the remainders are 3 and 3, their product is 9, so the remainder is 4.
If $N$ leaves 4 on division by 9, then $N^2$ leaves 16, which reduces to 7.
If $N = 60k + 43$ , then $N$ divided by 12 leaves the remainder of 43, which is 7, because 12 divides 60.
25 leaves $-1$ on division by 26, so an even power of 25 leaves 1 and an odd power leaves $-1$ , which is 25.

The digit-sum shortcut

A number leaves the same remainder on division by 9 as its digit sum does, and the same holds for 3. For 7,26,548 the digit sum is 32, which leaves 5, so 7,26,548 leaves 5.

Watch out

A remainder must be less than the divisor: a "remainder" of 14 on division by 8 still reduces to 6. And the shortcut for a new divisor works only when it divides the old one.

How do you find the unit digit of a large power?

Unit digits of powers repeat in short cycles. Divide the power by the cycle length, and 4 works for every digit, then use the remainder to pick the position in the cycle. A remainder of 0 means the last position.

Last digit of the base	Cycle of unit digits	Length
0, 1, 5, 6	Always itself	1
4	4, 6	2
9	9, 1	2
2	2, 4, 8, 6	4
3	3, 9, 7, 1	4
7	7, 9, 3, 1	4
8	8, 4, 2, 6	4

Example 9 · One power

Find the unit digit of $8^{51}$ .

$51 \div 4$ leaves 3.
The 3rd term of 8, 4, 2, 6 is 2.

Answer

Example 10 · A product of powers

Find the unit digit of $2^{36} \times 3^{22}$ .

$36 \div 4$ leaves 0, so $2^{36}$ ends in the last term of 2, 4, 8, 6, which is 6.
$22 \div 4$ leaves 2, so $3^{22}$ ends in the 2nd term of 3, 9, 7, 1, which is 9.
$6 \times 9 = 54$ , which ends in 4.

Answer

Exam shortcut

For a difference, borrow 10 when the first unit digit is smaller: a number ending in 1 minus a smaller number ending in 7 ends in $11 - 7 = 4$ . And every factorial from $5!$ on ends in 0, so only $1! + 2! + 3! + 4! = 33$ decides the last digit of a factorial sum.

Factors and primes

How do you find the number of factors and the sum of factors?

Write the prime factorisation $N = a^p \times b^q \times c^r$ . The number of factors is $(p + 1)(q + 1)(r + 1)$ , and the sum of factors is the product of $1 + a + \cdots + a^p$ , $1 + b + \cdots + b^q$ and $1 + c + \cdots + c^r$ .

Factors of N = a^p × b^q × c^r

\text{Number} = (p + 1)(q + 1)(r + 1) \qquad \text{Sum} = (1 + a + \cdots + a^p)(1 + b + \cdots + b^q)(1 + c + \cdots + c^r)

Example 11 · Factors of 200

Find the number and the sum of the factors of 200.

$200 = 2^3 \times 5^2$ .
Number of factors: $(3 + 1)(2 + 1) = 12$ .
Sum of factors: $(1 + 2 + 4 + 8)(1 + 5 + 25) = 15 \times 31 = 465$ .

Answer

12 factors, summing to 465

The same factorisation answers the common variations.

Question about 200	Method	Answer
Odd factors	Ignore the power of 2: $2 + 1$	3: 1, 5 and 25
Even factors	Total minus odd: $12 - 3$	9
Factors that are multiples of 10	Count the factors of $200 \div 10 = 20$	6
Perfect-square factors	Even powers only: 2 choices for the 2, 2 for the 5	4: 1, 4, 25 and 100

Watch out

Do not drop the $+1$ . A prime to the power 1 still doubles the number of factors.

How do you check for perfect squares and perfect cubes?

In a perfect square every prime in the factorisation has an even power. In a perfect cube every power is a multiple of 3. To make a number a square or a cube by multiplying or dividing, fix only the primes whose powers break the rule.

$98 = 2 \times 7^2$ . To make a square, multiply by 2 to get $196 = 14^2$ , or divide by 2 to get $49 = 7^2$ .
$500 = 2^2 \times 5^3$ . To make a cube, multiply by 2 to get $1{,}000 = 10^3$ , or divide by 4 to get $125 = 5^3$ .

Exam shortcut

A perfect square never ends in 2, 3, 7 or 8, and never ends in an odd number of zeros. So 4,000 and 2,370 cannot be squares.

How many zeros are at the end of n!?

Count the factors of 5 in $n!$ , because each trailing zero needs a pair $2 \times 5$ and there are always more 2s. Add the whole-number parts of $\tfrac{n}{5}$ , $\tfrac{n}{25}$ , $\tfrac{n}{125}$ and so on.

Example 12 · Trailing zeros

How many zeros are at the end of $80!$ ?

$80 \div 5 = 16$ .
$80 \div 25 = 3.2$ , so 3.
$80 \div 125$ is less than 1, so stop. Total: $16 + 3 = 19$ .

Answer

Watch out

Forgetting the 25s and 125s. $50!$ ends in $10 + 2 = 12$ zeros, not 10.

Exam shortcuts

Factorise first. Factors, squares, cubes and divisibility all read off the prime factorisation.
Composite divisor. Split it into co-prime factors and test both.
Remainders. Reduce each part first, then add or multiply.
Unit digits. Divide the power by 4. A remainder of 0 means the last term of the cycle.
Divisible by 9 or 3. Use the digit sum, for the test and for the remainder.

Common mistakes

Calling a recurring decimal irrational. $0.777\ldots = \tfrac{7}{9}$ , and $\tfrac{22}{7}$ is rational too. It only approximates $\pi$ .
Calling 1 prime, or thinking co-prime numbers must be prime.
"Between" including the ends. The integers between −4 and 3 are −3 to 2: six of them, including 0.
Forgetting the +1 when counting terms.
Taking the test for 4 as the test for 8.
Overlapping factors for a composite divisor, such as 2 and 6.
Mixing up the remainder and the amount to add. Subtract the remainder. Add the divisor minus the remainder.
An unreduced remainder. It must be less than the divisor.
A cycle remainder of 0 read as the first term. It is the last.
Missing 25 and 125 in trailing zeros.

Quick revision

Types of numbers

Terminating or recurring decimal: rational
$\sqrt{n}$ is rational only for a perfect square
1 is neither prime nor composite. 2 is the only even prime
25 primes below 100

Series

$1 + \cdots + n = \tfrac{n(n + 1)}{2}$ . Odd: $n^2$ . Even: $n(n + 1)$
Terms $= \tfrac{\text{last} - \text{first}}{\text{gap}} + 1$

Divisibility

4: last two digits. 8: last three. 3 and 9: digit sum
11: odd places minus even places is 0 or a multiple of 11
Composite divisor: co-prime factors
Add divisor − remainder, or subtract the remainder

Remainders and unit digits

Remainders add and multiply
Cycles: 2, 3, 7, 8 of length 4. 4 and 9 of length 2. 0, 1, 5, 6 fixed
Remainder 0: the last term of the cycle

Factors

$a^p b^q$ : $(p + 1)(q + 1)$ factors
Odd factors: drop the power of 2
Square: all powers even. Cube: all powers multiples of 3
Zeros in $n!$ : $\tfrac{n}{5} + \tfrac{n}{25} + \cdots$

Highest common factors and lowest common multiples are worked in HCF and LCM, and fast arithmetic with these numbers in simplification and approximation.

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Check your understanding

71 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

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Types of numbers

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Types of numbers

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Divisibility rules

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Remainders and unit digits

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Factors and prime numbers

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