# Number Systems #2 Representing a number as prime factors helps in analysing problems.

N = pa × qb × rc

where p, q, r are prime numbers and a, b, c are the number of times each prime number occurs.

1. Number of Factors

Number of Factors = (a + 1)(b + 1)(c + 1)

Example: Find the number of factors of 24 × 32

Number of factors = (4 + 1)(2 + 1) = 5(3) = 15

2. Number of Ways of Expressing a Given Number as a Product of Two Factors

When a number is having even number of factors then, (a+1)(b+1)(c+1)/2

But if a number have odd number of factors then, [(a+1)(b+1)(c+1)-1]/2

3. Sum of the factors of a number

The sum of all the factors of the number are given by the formula:

Sum of Factors = (a(p+1) - 1)(b(q+1) - 1)(c(r+1) - 1)/(a-1)(b-1)(c-1)

### Concept of Cyclicity: Power Cycle

At times there are questions that require you to find the units digit in case of the numbers occurring in powers. Concept of cyclicity is used to find unit's digit in case the numbers are occurring in powers. The last digit of a number of the form ab falls in a particular sequence that depends on the unit digit of the number (a) and the power the number is raised to (b).

Consider the power cycle of 2:

• 2= 2
• 22 = 4
• 23 = 8
• 24 = 16
• 25 = 32
• 26 = 64

You can see that the unit digit gets repeated after every fourth power of 2. Hence, you can say that 2 has a power cycle of 2, 4, 8, 6 with cyclicity 4. This is applicable for all numbers ending in 2.

 Unit Digit Power Cycle Cyclicity 0 0 1 1 1 1 2 2, 4, 8, 6 4 3 3, 9, 7, 1 4 4 4, 6 2 5 5 1 6 6 1 7 7, 9, 3, 1 4 8 8, 4, 2, 6 4 9 9, 1 2

Example: Find the last digit of (173)99

The exponent is 99. On dividing 99 by 4 we get 24 as the quotient and 3 as the remainder.

Now these 24 pairs of 4 each do not affect the number at the units place.

So, (173)99 ≈ (173)3

Now, the number at the units place is 33 = 27. i.e. 7.

### Applications of Factorial

Factorial is defined for any positive integer. It is defined as the product of all the integers from 1 to n.

Meaning of Factorial

• N! = 1 × 2 × 3 × ... × (N-1) × N
• Examples: 3! = 6; 5! = 120

Maximum power of p (prime number) in n! (n factorial)

To find the highest power of a prime number (p) in a factorial (n!), keep dividing n by p and add all the quotients.

Alternatively, use the formula:

{n/p} + {n/p2} + {n/p3} + ...