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IMPORTANT FORMULAE AND INSTRUCTIONS
NUMBERS
1. Every prime number greater than 3 can be written in the form of 6k+1 or 6k-1 where k is an integer.
2. Two numbers (which may be prime or composite) are said to be relative primes or co-primes if they do not have any common factor other than one. (Ex: 15 and 16)
Unity is a relative prime to all numbers.
3. A number is said to be a perfect number if sum of all its factors excluding itself (but including 1) is equal to the number itself. (Ex: 28,496,8128)
4. The brackets in an expression have to be opened in the order of vinculum, simplebrackets, curly brackets and square brackets, i.e.,[ { ( - ) } ] to be opened from inside outwards.
5. RULES FOR DIVISIBILITY:
1. Divisibility by 3:
A number is divisible by 3 if the sum of its digits is a multiple of 3.
6. To express a mixed recurring decimal into a vulgar fraction as below:
In the numerator write the entire given number formed by the recurring and non-recurring parts and subtract from it the part of the decimal that is not recurring. In the denominator, write as many nines as the period (i.e., as many nines as the number of digits recurring) and then place next to it as many zeroes as there are digits without recurring in the given decimal.
i.e. 0.ˉ̄
7. The given number N (which can be written as equal to …. Where a, b, c are prime factors of N and p, q, r …. are positive integers) can be expressed as the
product of two factors in different ways. The number of ways in which this can be done
is given by the expression
½ {(p+1) (q+1) (r+1) ….}
To find out the number of ways in which a perfect square can be expressed as a product of 2 factors, we have the following 2 rules
1. As a product of two DIFFERENT factors:
½ {(p+1)(q+1)(r+1) …. -1} ways
2. As a product of two factors :
½ {(p+1)(q+1)(r+1) … +1} ways
7. Sum of all the factors of a number:
If a number N = …. where a, b, c … are prime numbers and p, q, r … are positive integers, then, the sum of all the factors of N (including 1 and the number itself) is:
) ….
8. Numbers of ways of writing a number as product of two co-primes :
If N = …., then, the number of ways of writing N as a product of 2 co-primes is 2^(n-1), where n is the number of distinct prime factors of the given number N.
9. Number of co-primes to N, that are less than N:
If N is a number that can be written as …., then, the number of co-primes of N, which are less than N, represented by Φ (N) is,
N (1-(1/a)) (1-(1/b)) (1-(1/c)) ….
10. Sum of co-primes to N that are less than N:
The sum of the co-primes of N that are less than N is N/2. Φ (N).
11. INVOLUTION:



= + 3ab (a+b)
= -3ab (a-b)
(a+b) (a-b)
(a+b) (a-b)+
(a+b) (
(a-b) (
-3abc
3abc, if a+b+c
12. EVOLUTION:
Any even-root of a positive quantity may be negative or positive
No negative quantity can have an even-root.
Every odd-root will have the same sign as the quantity itself.

1. A number when divided by d leaving a remainder of r is of the form dq+r where q is some integer from 0, 1, 2…
2. The smallest multiple of d which is greater than or equal to N is obtained by adding (d-r) to N, i.e., N+ (d-r) will be the smallest multiple of d which is greater than N.
3. When a division is split into a sum of two divisions (with the same divisor as the original divisor), the original remainder will be equal to the sum of the remainders of the two individual divisions. Similarly, when a division is split into difference of two divisions, the original remainder will be equal to the difference of the remainders of the two divisions.
Ex:
15/6=8/6+7/6
Remainder 3= 2+1
15/6=29/6-14/6
Remainder 3=5-2
4. If the remainder in a division is negative, then add the divisor to the negative remainder to get the positive remainder.
Ex:
15/6=25/6-10/6
Remainder 1-4=-3
13. LCM AND HCF:
1. Product of n numbers = lcm*
2. HCF of fractions =
3. LCM of fractions=
4. HCF by long division:
Take two numbers. Divide the greater by the smaller; then divide the divisor by the remainder; divide the divisor of this division by the next remainder and so on until the remainder is zero. The last divisor is the HCF of the two numbers taken. By the same method find the HCF of this HCF and the third number. This will be the HCF of the three numbers.
5. Any number which when divided by p, q, r leaving the same remainder in each case will be of the form k (LCM of p, q, r)+s where k = 0, 1, 2, ….
6. Any number which when divided by p, q, r leaving respective remainders of s, t, u where (p-s) = (q-t)=(r-u)=v (say), will be of the form k (LCM p, q and r) – v.
7. The largest number with which the numbers p, q or r are divided giving remainders of s, t and u respectively will be the HCF of the three numbers (p – s), (q – t) and (v – u).
8. The largest number with which the numbers p, q or r are divided
9. HCF of (p ~ r) and (q ~ r)
14. THE LAST DIGIT OF ANY POWER:
1. The last digit of
The last digit of
The last digit of
The last digit of
2. The last digit of
The last digit of
The last digit of
The last digit of
3. The last digit of
The last digit of
The last digit of
The last digit of
4. The last digit of
The last digit of
The last digit of
The last digit of
Where ab… is a number ending with 2, 3, 7 or 8.
Suppose we want to find out the last digit of 2357 , we should look at a multiple of 4 which is less than or equal to the power 57. Since 56 is a multiple of 4 , the last digit of 2357 will be the same as the last digit of .

15. FINDING THE REMAINDER IN DIVISIONS INVOLVING POWERS OF NUMBERS:
PATTERN METHOD:
1. Remainder when is divided by 7 is 2
Remainder when is divided by 7 is 4
Remainder when is divided by 7 is 1

Suppose we want to find out the remainder of when divided by 7since 63 is divisible by 3 the remainder is 1.

REMAINDER THEOREM METHOD:
1. In the division /8, the dividend is and the divisor is 8. Since the numerator is in terms of powers of 3, express the denominator also in terms of powers of 3. In this case, 8 can be written as 9 – 1 which is – 1. Now the given problem reduces to finding out the remainder when is divided by – 1. Here, if we consider as x, it is equivalent to finding out the remainder when is divided by (x – 1) which as per remainder theorem, is f (1). So, the remainder will be , i.e., 1.
16. SUCCESSIVE DIVISION:
1. The smallest number which when successively divided by d1 ,d2 and d3 leaves respective remainders of r1, r2 and r3 is [{(r3*d2)+r2}*d1]+r1

D1 d2 d3 divisors
R1 r2 r3 remainders

17. LARGEST POWER OF A NUMBER IN N :
1. To find out the largest power of a number in n we have to successively divide n by that number. Sum of the quotients is the required number.
2. This method is applicable only if the number whose largest power is to be found out is a prime number. If the number is not a prime number, then we have to write the number as the product of prime numbers, find the largest power of each of the factors separately. Then the smallest among the largest powers of all these factors of the given number will give the largest power required.
18. RULES PERTAININGTO :

1. The following rules should be remembered for numbers in the form of
a. It is always divisible by a - b.
b. When n is even it is also divisible by a + b.
c. When n is odd it is divisible by a + b, if a + b is a factor of 2* .
2. The following rules should be remembered for the numbers in the form of .
a. When n is odd it is divisible by a + b.
b. When n is odd, it is divisible by a – b, when a – b is a factor of 2*
c. When n is even, it is divisible by a + b, if a + b is a factor of 2*