number system basic formula

Basic Formulae

1. (a+b)2=a2+b2+2ab$(a+b{)}^2=a^2+b^2+2ab$

2. (a−b)2=a2+b2−2ab$(a-b{)}^2=a^2+b^2-2ab$

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

4. (a+b)2+(a−b)2=2(a2+b2)$(a+b{)}^2+(a-b{)}^2=2(a^2+b^2)$

5. (a2−b2)=(a+b)(a−b)$(a^2-b^2)=(a+b)(a-b)$

6. (a+b+c)2=a2+b2+c2+2(ab+bc+ca)$(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$

7. (a3+b3)=(a+b)(a2−ab+b2)$(a^3+b^3)=(a+b)(a^2-ab+b^2)$

8. (a3−b3)=(a−b)(a2+ab+b2)$(a^3-b^3)=(a-b)(a^2+ab+b^2)$

9. (a3+b3+c3−3abc)=(a+b+c)(a2+b2+c2−ab−bc−ca)$(a^3+b^3+c^3-3abc)=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

10. If a+b+c=0$a+b+c=0$, then a3+b3+c3=3abc$a^3+b^3+c^3=3abc$.

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2. Types of Numbers

I. Natural Numbers

Counting numbers 1,2,3,4,5,...$1,2,3,4,5,...$ 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,.....$...,-3,-2,-1,0,1,2,3,.....$ together form the set of integers.

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

(ii) Negative Integers: −1,−2,−3,.....$-1,-2,-3,.....$ 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,....$0,1,2,3,....$ represents the set of non-negative integers,

while 0,−1,−2,−3,.....$0,-1,-2,-3,.....$ 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$2,4,6,8$, 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,$1,3,5,7,9,11,$ 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.$:2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97.$
• Prime numbers Greater than 100: Let p$p$ 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$p$. Let k>∗jp$k>\ast jp$. Test whether p$p$ is divisible by any prime number less than k$k$. If yes, then p$p$ is not prime. Otherwise, p$p$ is prime. Example: We have to find whether 191 is a prime number or not. Now, 14>V191$14>V191$.

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

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.$4,6,8,9,10,12.$

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

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

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 10,000n+57$10,000n+57$, where n$n$ is an integer.

Rewriting 10,000 as 1,000∗10$1,000\ast 10$ yields 10,000n+57=1,000(10n)+57$10,000n+57=1,000(10n)+57$

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

10,000n+57=1,000∗q+57$10,000n+57=1,000\ast q+57$

Hence, the remainder is still 57 (by the p=kq+r$p=kq+r$ 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 10,000n+57$10,000n+57$. Dividing this number by 1,000 yields

10,000n+571000${\displaystyle \frac{10,000n+57}{1000}}$ =10,000n1000+571000$={\displaystyle \frac{10,000n}{1000}}+{\displaystyle \frac{57}{1000}}$ =10n+571000$=10n+{\displaystyle \frac{57}{1000}}$

Hence, the remainder is 57 (by the alternative form pk=q+rk${\displaystyle \frac{p}{k}}=q+{\displaystyle \frac{r}{k}}$ ), 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$n$ is divided by 2:n=2z+0$2:n=2z+0$, or n=2z$n=2z$.

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

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

even * evenodd * oddeven * oddeven + evenodd + oddeven + odd=even=odd=even=even=even=odd$$\begin{array}{rl}\text{even * even}& =\text{even}\\ \text{odd * odd}& =\text{odd}\\ \text{even * odd}& =\text{even}\\ \text{even + even}& =\text{even}\\ \text{odd + odd}& =\text{even}\\ \text{even + odd}& =\text{odd}\end{array}$$

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Example:

If n$n$ is a positive integer and (n+1)(n+3)$(n+1)(n+3)$ is odd, then (n+2)(n+4)$(n+2)(n+4)$ 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)$(n+1)(n+3)$ is odd only when both (n+1)$(n+1)$ and (n+3)$(n+3)$ are odd. This is possible only when n$n$ is even.

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

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

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

Hence, the answer is (D