Number System
Aptitude Questions and Answers



1. Basic Formulae

1. $(a + b)^2 = a^2 + b^2 + 2ab$

2. $(a - b)^2 = a^2 + b^2 - 2ab$

3. $(a + b)^2 - (a - b)^2 = 4ab$

4. $(a + b)^2 + (a - b)^2 = 2 (a^2 + b^2)$

5. $(a^2 - b^2) = (a + b) (a - b)$

6. $(a + b + c)^2 = a^2 + b^2 + c^2 + 2 (ab + bc + ca)$

7. $(a^3 + b^3) = (a +b) (a^2 - ab + b^2)$

8. $(a^3 - b^3) = (a - b) (a^2 + ab + b^2)$

9. $(a^3 + b^3 + c^3 -3abc) = (a + b + c) (a^2 + b^2 + c^2 - ab - bc - ca)$

10. If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

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2. Types of Numbers

I. Natural Numbers

Counting numbers $1, 2, 3, 4, 5,...$ are called natural numbers

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II. Whole Numbers

All counting numbers together with zero form the set of whole numbers.

Thus,

(i) 0 is the only whole number which is not a natural number.

(ii) Every natural number is a whole number.

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III. Integers

All natural numbers, 0 and negatives of counting numbers i.e., ${..., - 3 , - 2 , - 1 , 0, 1, 2, 3,.....}$ together form the set of integers.

(i) Positive Integers: ${1, 2, 3, 4, .....}$ is the set of all positive integers.

(ii) Negative Integers: ${- 1, - 2, - 3,.....}$ is the set of all negative integers.

(iii) Non-Positive and Non-Negative Integers: 0 is neither positive nor negative.

So, ${0, 1, 2, 3,....}$ represents the set of non-negative integers,

while ${0, - 1 , - 2 , -3 , .....}$ represents the set of non-positive integers.

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IV. Even Numbers

A number divisible by 2 is called an even number, e.g.,$ 2, 4, 6, 8$, etc.

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V. Odd Numbers

A number not divisible by 2 is called an odd number. e.g.,$ 1, 3, 5, 7, 9, 11,$ etc.

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VI. Prime Numbers

A number greater than 1 is called a prime number, if it has exactly two factors, namely 1 and the number itself.

  • Prime numbers up to 100 are $: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.$
  • Prime numbers Greater than 100 : Let p be a given number greater than 100. To find out whether it is prime or not, we use the following method :

Find a whole number nearly greater than the square root of p. Let $k > *jp$. Test whether p is divisible by any prime number less than k. If yes, then p is not prime. Otherwise, p is prime. Example: We have to find whether 191 is a prime number or not. Now, $14 > V191$.

Prime numbers less than 14 are $2, 3, 5, 7, 11, 13.$

191 is not divisible by any of them. So, 191 is a prime number.

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VII. Composite Numbers

Numbers greater than 1 which are not prime, are known as composite numbers, e.g., $4, 6, 8, 9, 10, 12.$

Note:

(i) 1 is neither prime nor composite.

(ii) 2 is the only even number which is prime.

(iii) There are 25 prime numbers between 1 and 100.

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3. Remainder and Quotient

"The remainder is $r$ when $p$ is divided by k" means $p = kq + r$ the integer q is called the quotient.

For instance, "The remainder is 1 when 7 is divided by 3" means $7 = 3*2 + 1$. Dividing both sides of $p = kq + r$ by k gives the following alternative form $\dfrac{p}{k} = q + \dfrac{r}{k}$

Example:

The remainder is 57 when a number is divided by 10,000. What is the remainder when the same number is divided by 1,000?

(A) 5 (B) 7 (C) 43 (D) 57 (E) 570

Solution:

Since the remainder is 57 when the number is divided by 10,000, the number can be expressed as $10,000 n + 57$, where $n$ is an integer.

Rewriting 10,000 as $1,000*10$ yields $10,000 n + 57 = 1,000(10 n ) + 57$

Now, since $n$ is an integer, $10 n$ is an integer. Letting $10 n = q$ , we get

$10,000 n + 57 = 1,000*q + 57$

Hence, the remainder is still 57 (by the $p = kq + r$ form) when the number is divided by 1,000. The answer is (D).

Method II (Alternative form):

Since the remainder is 57 when the number is divided by 10,000, the number can be expressed as $10,000 n + 57$. Dividing this number by 1,000 yields

$\dfrac{10,000 n + 57}{1000}$ $= \dfrac{10,000 n}{1000} + \dfrac{57}{1000}$ $ = 10 n + \dfrac{57}{1000}$

Hence, the remainder is 57 (by the alternative form $\dfrac{p}{k} = q + \dfrac{r}{k}$ ), and the answer is (D).

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4. Even, Odd Numbers

A number n is even if the remainder is zero when $n$ is divided by $2: n = 2z + 0$, or $n = 2z$.

A number $n$ is odd if the remainder is one when $n$ is divided by $2: n = 2z + 1$.

The following properties for odd and even numbers are very useful - you should memorize them:

$$\begin{align*} \text{even * even} &= \text{even}\\ \text{odd * odd} &= \text{odd}\\ \text{even * odd} &= \text{even}\\ \text{even + even} &= \text{even}\\ \text{odd + odd} &= \text{even}\\ \text{even + odd} &= \text{odd} \end{align*} $$


Example:

If $n$ is a positive integer and $(n + 1)(n + 3)$ is odd, then $(n + 2)(n + 4)$ must be a multiple of which one of the following?

(A) 3 (B) 5 (C) 6 (D) 8 (E) 16

Solution:

$(n + 1)(n + 3)$ is odd only when both $(n + 1)$ and $(n + 3)$ are odd. This is possible only when $n$ is even.

Hence, $n = 2m$, where $m$ is a positive integer. Then,

$\begin{align*} (n + 2)(n + 4) &= (2m + 2)(2m + 4)\\ &= 2(m + 1)2(m + 2)\\ &= 4(m + 1)(m+ 2) \end{align*}$

$=\text{4 * (product of two consecutive positive integers, one which must be even)}$ $=\text{4 * (an even number), and this equals a number that is at least a multiple of 8}$

Hence, the answer is (D).

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5. Tests of Divisibility

5.1. Divisibility By 2

A number is divisible by 2, if its unit's digit is any of $0, 2, 4, 6, 8$.

Example:

84932 is divisible by 2, while 65935 is not.

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5.2. Divisibility By 3

A number is divisible by 3, if the sum of its digits is divisible by 3.

Example:

592482 is divisible by 3, since sum of its digits $= (5 + 9 + 2 + 4 + 8 + 2) = 30$, which is divisible by 3.

But, 864329 is not divisible by 3, since sum of its digits $=(8 + 6 + 4 + 3 + 2 + 9) = 32$, which is not divisible by 3.

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5.3. Divisibility By 4

A number is divisible by 4, if the number formed by the last two digits is divisible by 4.

Example:

892648 is divisible by 4, since the number formed by the last two digits is 48, which is divisible by 4. But, 749282 is not divisible by 4, since the number formed by the last two digits is 82, which is not divisible by 4.

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5.4. Divisibility By 5

A number is divisible by 5, if its unit's digit is either 0 or 5. Thus, 20820 and 50345 are divisible by 5, while 30934 and 40946 are not.

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5.5. Divisibility By 6

A number is divisible by 6, if it is divisible by both 2 and 3.

Example:

The number 35256 is clearly divisible by 2.Sum of its digits $ = (3 + 5 + 2 + 5 + 6) = 21$, which is divisible by 3. Thus, 35256 is divisible by 2 as well as 3. Hence, 35256 is divisible by 6.

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5.6. Divisibility By 8

A number is divisible by 8, if the number formed by the last Three digits of the given number is divisible by 8.

Example:

953360 is divisible by 8, since the number formed by last three digits is 360, which is divisible by 8. But, 529418 is not divisible by 8, since the number formed by last three digits is 418, which is not divisible by 8.

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5.7. Divisibility By 9

A number is divisible by 9, if the sum of its digits is divisible by 9.

Example:

60732 is divisible by 9, since sum of digits $= (6 + 0 + 7 + 3 + 2) = 18$, which is divisible by 9.

But, 68956 is not divisible by 9, since sum of digits $= (6 + 8 + 9 + 5 + 6) = 34$, which is not divisible by 9.

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5.8. Divisibility By 10

A number is divisible by 10, if it ends with 0.

Example:

96410, 10480 are divisible by 10, while 96375 is not.

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5.9. Divisibility By 11

A number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places, is either 0 or a number divisible by 11.

Example:

The number 4832718 is divisible by 11, since $\text{:(sum of digits at odd places) - (sum of digits at even places) = }$

$= (8 + 7 + 3 + 4) - (1 + 2 + 8) = 11$, which is divisible by 11.

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5.10. Divisibility By 12

A number is divisible by 12, if it is divisible by both 4 and3.

Example:

Consider the number 34632.

(i) The number formed by last two digits is 32, which is divisible by 4,

(ii) Sum of digits $= (3 + 4 + 6 + 3 + 2) = 18$, which is divisible by 3. Thus, 34632 is divisible by 4 as well as 3. Hence, 34632 is divisible by 12.

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5.11. Divisibility By 14

A number is divisible by 14, if it is divisible by 2 as well as 7.

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5.12. Divisibility By 15

A number is divisible by 15, if it is divisible by both 3 and 5.

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5.13. Divisibility By 16

A number is divisible by 16, if the number formed by the last4 digits is divisible by 16.

Example:

7957536 is divisible by 16, since the number formed by the last four digits is 7536, which is divisible by 16.

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5.14. Divisibility By 24

A given number is divisible by 24, if it is divisible by both 3 and 8.

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5.15. Divisibility By 40

A given number is divisible by 40, if it is divisible by both 5 and 8.

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5.16. Divisibility By 80

A given number is divisible by 80, if it is divisible by both 5 and 16.

Note:

If a number is divisible by $p$ as well as $q$, where $p$ and $q$ are co-primes, then the given number is divisible by $pq$. If $p$ and $q$ are not co-primes, then the given number need not be divisible by $pq$, even when it is divisible by both $p$ and $q$.

Example:

36 is divisible by both 4 and 6, but it is not divisible by $(4*6) = 24$, since 4 and 6 are not co - primes.

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6. Progression

A succession of numbers formed and arranged in a definite order according to certain definite rule, is called a progression.

6.1. Arithmetic Progression (A.P.)

If each term of a progression differs from its preceding term by a constant, then such a progression is called an arithmetical progression. This constant difference is called the common difference of the A.P.

An A.P. with first term a and common difference d is given by $a, (a + d), (a + 2d),(a + 3d),.....$

The nth term of this A.P. is given by $T_n =a (n - 1) d$.

The sum of n terms of this A.P. $$\begin{align*} S_n &= \left(\dfrac{n}{2}\right) [2a + (n - 1)d]\\ &= \left(\dfrac{n}{2}\right)*(\text{first term + last term}). \end{align*}$$

Some Important Results:

(i) $(1 + 2 + 3 +.... + n) =\dfrac{n(n+1)}{2}$

(ii) $(l^2 + 2^2 + 3^2 + ... + n^2) = \dfrac{n (n+1)(2n+1)}{6}$

(iii) $(1^3 + 2^3 + 3^3 + ... + n^3) =n^2(n+1)^2$

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6.2. Geometrical Progression (G.P.)

A progression of numbers in which every term bears a constant ratio with its preceding term, is called a geometrical progression. The constant ratio is called the common ratio of the G.P.

A G.P. with first term a and common ratio r is $: a, ar, ar^2, .....$

In this G.P.$ n^{th} \text{ term, } T_n = ar^{n-1}$

sum of n terms, $S_n= \dfrac{a(1-r^n)}{(1-r)}$ when $r < 1$

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