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**SOME BASIC FORMULAE****1.** (a + b)(a - b) = (a^{2} - b^{2})**2.** (a + b)^{2} = (a^{2} + b^{2} + 2ab)**3.** (a - b)^{2} = (a^{2} + b^{2} - 2ab)**4.** (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca)**5.** (a^{3} + b^{3}) = (a + b)(a^{2} - ab + b^{2})**6.** (a^{3} - b^{3}) = (a - b)(a^{2} + ab + b^{2})**7. **(a^{3} + b^{3} + c^{3} - 3abc) = (a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc - ac)**8.** When a + b + c = 0, then a^{3} + b^{3} + c^{3} = 3abc.

Question 1:What is the square of 109?

**Visual Representation of some Formulas**

**1. (A + B) ^{2} = **A

**2. (A + B + C) ^{2} = **A

**IMPORTANT NUMBER SYSTEM FORMULAS****1. **Highest power of n in m! is **[m/n] + [m/n ^{2}] + [m/n^{3}] +â€¦..** where, [x] is the greatest integer less than or equal to x.

The answer is 4.

**3.** If all possible permutations of n distinct digits are added together with the sum =** (n-1)! * (sum of n digits) * (11111â€¦ n times)****Example: ****What would be the sum of all the numbers which can be formed by using the digits 1, 3, 5, 7 all at a time and which have no digits repeated?****Solution:** Sum of the numbers formed by taking all the given n digits is (sum of all the n digits) * (n-1) !* (111â€¦..n times).

Here n = 4, and Sum of 4 digits = 16

The sum of all the numbers which can be formed by using the digits 1, 3, 5, 7 is

= (16) * (4 â€“ 1)! * ( 1111)

= 16 * 3! * 1111.

**4.** If the number can be represented as N = a^{p}âˆ— b^{q}âˆ— c^{r}. **Then,** the number of factors of N is **(p+1) * (q+1) * (r+1)**

**5. **Sum of the factors:

Question 2:How many factors does 60 has?

**Example 1: ****Find the number of factors of 98 and also find the sum and product of all factors.****Solution:** First write the number 98 into prime factorization

98 = 2 Ã— 49 = 2 Ã— 7 Ã— 7

98 = 2^{1} x 7^{2 }

Here A = 2 , B = 7 , p= 1 , q = 2

Number of factors for 98

= (p + 1)(q +1) = 2 Ã— 3 = 6

Sum of all factors of 98

= 3 Ã— 57 = 171

Product of all factors of number 98

= (98)^{6/2} = 941192

**Example**** ****2. Find the number of odd, even, perfect square, perfect cube factors of 4500.****Solution.**

4500 = 45 Ã— 100 = 9 Ã— 5 Ã— 10 Ã— 10

4500 = 3 Ã— 3 Ã— 5 Ã— 5 Ã— 2 Ã— 5 Ã— 2

4500 = 2^{2} Ã— 3^{2} Ã— 5^{3}

Here consider A = 2 , B = 3 , C = 5 , p= 2 , q = 2 and r = 3

â‡’ Here identifying that odd number are 3 and 5.

Numbers of odd factors of number 4500

= (q + 1) (r + 1) = 3 Ã— 4 = 12

âˆ´ Total number of factors

= (p + 1)(q +1)(r +1) = 3 Ã— 3 Ã— 4 =36

Numbers of even factors of number

= Total number of factors â€“ Numbers of odd factors

= 36 â€“ 12 = 24

Number of perfect square factors of number 4500 = 2 x 2 x 2 = 8

Number of perfect cube factors of number 4500 = 1 x 1 x 2 = 2

**6. **If the number of factors is odd then N is a **perfect square**.

**7. **If there are n factors, then the number of pairs of factors would be **n/2**. If N is a perfect square then the number of pairs (including the square root) is **(n+1)/2.**

**Example****: ****In how many ways you can express 36 as the product of two of its factors?****Solution: ****Step 1**: Prime factorization of 36 i.e. we write 36 = 2^{2} Ã— 3^{2}**Step 2**: Number of factors of 36 will be (2+1)(2+1) = 9 (i.e. factors are 1, 2,3, 4, 6, 9, 12, 18, 36)**Step 3**: Since we are asked total number of ways hence we include square root of 36 i.e. 6 as well. Thus number of ways you can express 36 as the product of two of its factors is (9+1)/2 = 5.

**8.** If the number can be expressed as **N = 2 ^{p} âˆ— a^{q} âˆ— b^{r} . . .** where the power of 2 is p and a, b are prime numbers

- Then the number of even factors of N = p (1+q) (1+r) . . .
- The number of odd factors of N = (1+q) (1+r)â€¦

**9.** When we are asked to calculate how many positive integral solutions are possible for the equation **X ^{2}â€“ Y^{2}= N**, there can be 4 cases.

**CASE 1**: **N is an odd number and not a perfect square**

In this case, total number of positive integral solutions will be

= (Total number of factors of N) / 2

**Example. ****How many positive integral solutions are possible for the equation X ^{2 }â€“ Y^{2 }= 135?**

So, total number of positive integral solutions = 8/2 = 4.

In this case, total number of positive integral solutions will be = [(Total number of factors of N) â€“ 1] / 2

**Example.**** ****How many positive integral solutions are possible for the equation X ^{2}â€“ Y^{2 }= 121?**

So, total number of positive integral solutions = (3-1)/2 = 1

In this case, total number of positive integral solutions will be = [Total number of factors of (N/4)] / 2

**Example.**** ****How many positive integral solutions are possible for the equation X ^{2}â€“ Y^{2 }= 160?**

So, total number of positive integral solutions = 8/2 = 4.

**NOTE: **If a given number is of the form 4k+2, it cannot be expressed as difference of two squares.

**CASE 4****:** **N is an even number and a perfect square**

In this case, total number of positive integral solutions will be = {[Total number of factors of (N/4)] â€“ 1 } / 2

**Example.**** ****How many positive integral solutions are possible for the equation X ^{2}â€“ Y^{2}= 256?**

So (7-1)/2 = 3 positive integral solutions.

**10.** Number of digits in** a ^{b} = [b log_{m}(a)] + 1** ; where m is the base of the number and [.] denotes the greatest integer function

**11. **Even number which is not a multiple of 4, can never be expressed as a difference of 2 perfect squares.

**12. ****Basic summation properties**

(a) Sum of first n odd numbers is **n**^{2}

(b) Sum of first n even numbers is **n(n+1)**

- (1
^{Â²}+ 2^{Â²}+ 3^{Â²}+ â€¦.. + n^{Â²}) = n ( n + 1 ) (2n + 1) / 6 - (1
^{Â³}+ 2^{Â³}+ 3^{Â³}+ â€¦.. + n^{Â³}) = (n(n + 1)/ 2)Â²

**13. **The product of the factors of N is given by **N ^{a/2},** where

**14.** The last two digits of a^{2}, (50 â€“ a)^{2}, (50 + a)^{2}, (100 â€“ a)^{2} . . .. . . . . are same. Every number can be written as (50n Â± x), where x is a number from 0 to 25. Here by 50n, i just mean any multiple of 50 i.e. 0, 50, 100, 150, â€¦â€¦

0 to 25 = (0 to 25) itself.

25 to 50 = 50 â€“ (25 to 0)

50 to 75 = 50 + (0 to 25)

75 to 100 = 100 â€“ (25 to 0)

100 to 125 = 100 + (0 to 25)

125 to 150 = 150 â€“ (25 to 0) and so on.

But how does this help us in finding the last two digits of any square?

(50n Â± x)^{2} = 2500n^{2} Â± 100nx + x^{2}

The last two digits of each of 2500n^{2} and 100nx will be 00. Thus the last two digits of the RHS, and hence of the LHS, will be the last two digits of x^{2}.

**Example 1.**** ****What are the last two digits of 268 ^{2}?**

Thus, the last two digits of 268

**Example 2.**** ****What are the last two digits of 278 ^{2}?**

278 = 50 Ã— 6 â€“ 22, to be x in the range 0 to 25.

Thus, the last two digits of 278

- When n is odd, the last 2 digits are 24.
- When n is even, the last 2 digits are 76.

**16.**** Divisibility****â€¢ **Divisibility by 2 â†’ Last digit divisible by 2**â€¢ **Divisibility by 4 â†’ Last two digits divisible by 4**â€¢ **Divisibility by 8 â†’ Last three digits divisible by 8**â€¢ **Divisibility by 16 â†’ Last four digit divisible by 16**â€¢ **Divisibility by 3 â†’ Sum of digits divisible by 3**â€¢ **Divisibility by 9 â†’ Sum of digits divisible by 9**â€¢ **Divisibility by 27 â†’ Sum of blocks of 3 (taken right to left) divisible by 27**â€¢ **Divisibility by 7 â†’ Remove the last digit, double it and subtract it from the truncated original number. Check if number is divisible by 7**â€¢ **Divisibility by 11 â†’ (sum of odd digits) - (sum of even digits) should be 0 or divisible by 11**17.**** Cyclicity**

To find the last digit of a_{n} find the cyclicity of a.**For Example**: if a=2, we see that**â€¢ **2^{1 }= 2**â€¢ **2^{2 }= 4**â€¢ **2^{3 }= 8**â€¢ **2^{4 }= 16**â€¢ **2^{5 }= 32

Hence, the last digit of 2 repeats after every 4th power. Hence cyclicity of 2 = 4. Hence if we have to find the last digit of a^{n}. The steps are:

1. Find the cyclicity of a, say it is x

2. Find the remainder when n is divided by x, say remainder r

3. Find a^{r} if r > 0 and a^{x }when r = 0

**18. ****HCF and LCM****â€¢ **HCF * LCM of two numbers = Product of two numbers**â€¢ **The greatest number dividing a, b and c leaving remainders of x_{1}, x_{2} and x_{3} is the HCF of (a - x_{1}), (b - x_{2}) and (c - x_{3}).**â€¢ **The greatest number dividing a, b and c (a<b<c) leaving the same remainder each time is the HCF of (c - b), (c - a), (b - a).**â€¢ **If a number, N, is divisible by X and Y and HCF (X, Y) = 1. Then, N is divisible by X*Y

**Example 1****: Find the HCF of 96, 36 and 18.****Solution:**

96 =2 Ã— 3 Ã— 2 Ã— 2 Ã— 2 Ã— 2

36 = 2 Ã— 3 Ã— 2 Ã— 3

18 = 2 Ã— 3 Ã— 3

Therefore, the HCF of 96, 36 and 18 is the product of the highest number of common factors in the given numbers i.e., 2 Ã— 3 = 6. In other words, 6 is the largest possible integer, which can divide 96, 36 and 18 without leaving any remainder.

**Example 2.**** Find the HCF of 42 and 70.****Solution. **

42 = 3 Ã— 2 Ã— 7

70 = 5 Ã— 2 Ã— 7

Hence, HCF is x 7 = 14.

**Example 3.**** Find the HCF of numbers 144, 630 and 756.****Solution.**

144 = 2^{4} Ã— 3^{2}

630 =2 Ã— 3^{2 }Ã— 5 Ã— 7

756 = 2^{2} x 3^{3} x 7

Hence, HCF of 144, 630, 756 = 2 Ã— 3^{2} = 18.

Important:HCF of two prime numbers is always 1

HCF of co-prime numbers is always 1

**Example 4.**** Find the LCM of 96, 36 and 18.****Solution.**

96 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 =2^{5 }Ã— 3^{1}

36 = 2 x 2 x 3 x 3 = 2^{2} x 3^{2}

18 = 2x 3 x 3= 2^{1} x 3^{2}

Therefore, LCM of 96, 36 and 18 is the product of the highest powers of all the prime factors, i.e. 2^{5} x 3^{2}= 32 x 9 = 288.

That is, 288 is the smallest integer which is divisible by 96, 36 and 18 without leaving any remainder.

**Example 5.**** Find the LCM of 42 and 70.****Solution.**

42 = 3 Ã— 2 Ã— 7

70 = 5 Ã— 2 Ã— 7

Hence, LCM is 2 Ã— 3 Ã— 5 Ã— 7 = 210.

Apart from the method of prime factorization, there is another method of finding the LCM of given numbers and the method is known as the long division method. This method is quite helpful in getting LCM quickly if there are three or more than three numbers.

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