If any composite number N which can be expressed as N=$a^x\cdot b^y\cdot c^z...$a^x·b^y·c^z... , where a,b,c are the prime numbers and x,y,z are the powers of a,b,c respectively and so on. then the number of factors of N =(x+1)×(y+1)×(z+1)...

Properties of Averages

1. Average of N things/quantities is equal to the sum of all the things/quantities divided by number of things/quantities.

2. Average of N number of things/quantities always lies between the lowest and the highest quantities.

3. If each quantity is increased by a certain value K, then the new average is increased by K.

4. If each quantity is decreased by a certain value K, then the new average is decreased by K.

5. If each quantity is multiplied by a certain value K, then the new average becomes K times the the original average.

6. If each quantity is divided by a certain value K, then the new average becomes $\frac{1}{K}$1K  times the the original average.

1. When the remainders are same for all the divisors
   
   In this case the required number will be the LCM × N + remainder, where N is any natural number
   
   N=1, will give smallest such number
   
   N=2, will give second smallest such number and so on

2. When the remainders are different for different divisors but the respective difference between the divisors and the remainders remain constant.
   
   In this case the required number will be LCM × N +difference of any (divisor - remainder)
   
   refer to example 2.

3. When neither the divisors are same nor the respective difference between the divisors and the remaniders remain constant.
   
   In this case solve the question by forming equations and solving them.

Rules for Finding Last Two Digits of square of a number

In order to find the last two digits of square of any two digit number number, we can write it as difference or sum from 50 or 100, whichever is closer. In general the last two digits of a number (50±k)² or (100-k)² will be determined by square of k.

Rules for Finding Last Two Digits of any power of a number

1. For a natural number N^k, where N is a natural number ending with 0 and K is a natural number greater than 2,
   
   Last two digits will always be 00.

2. For a natural number N^k, where N is a natural number ending with 5 and K is a natural number greater than 2,
   
   Last two digits will always be 25.

3. For a natural number (2×m)^40k+1, where m is a odd number not ending with 5 and K is a natural number,
   
   Last two digits will always be (2×m +50).

4. For all the remaining cases, the last two digits of N^40k+x,
   
   Last two digits will always be equal to the last 2 digits of N^x

Unit digit follows a periodic pattern that is after a particular period it repeats in a cyclic form, this is called cyclicity.

• Unit digits of numbers ending with 0,1,5,6 is always the same irrespective of their powers raised on them.

• Unit digit of numbers ending with 4,9 follows the pattern with cyclicity of 2

• Unit digit of numbers ending with 2,3,7,8 follows the pattern with cyclicity of 4

Rule 1 : x is divided by y When we have to divide a number by another number, then we can write x in terms of y, i.e. (y×a)+b. This means that the remainder is b.

Rule 2 : x ± y is divided by z When we have to divide the sum or difference of two number by another number, then we write both the numbers as the product of the number with which we have to divide.

Rule 3 : x × y is divided by z When we have to divide the product of two numbers by another number, then the remainder will be equal to the remainder of each of the number divided separately.

Rule 4 : x^y is divided by z When we have to divide a number to some power with another number, then keep dividing the number with the divisor till the remainder is not less than the divisor.

Rule of Negative Remainder When a number x divided by y, then we can say that the remainder is -a if y×k=x+a, where k can be any natural number and a is smaller than y. This in turn means that the remainder is a times smaller than y.