# Number System Tutorial Part III: Factors, Multiples, Unit Digit and Last Two Digits of Exponents

This is a continuation of article on number system. Here we will discuss few of the important concepts factors, multiples, applications of HCF and LCM in finding remainder, factorial etc. This will help you in solving aptitude problems related power, reminder etc very quickly. Also, we'll look at concepts and shortcuts in finding the unit digit (last digit) and last two digit of large exponent expression.

## Objectives

• Factors and Multiples
• Prime factorization method
• Number of factors of a given number
• Number of ways to express a number as the product of two factors
• Sum of all the factors of a given number
• Number of ways to express a number as the product of two co-prime factors
• Euler's \phi function
• Sum of all co-prime factors of given number
• LCM and HCF application on Remainders
• Factorial and its properties
• Largest power of a factor in n!
• Unit digit of an exponential expression
• Last two digits of an exponential expression

## Factors and Multiples

Factors and multiples are fully restricted in the set of natural numbers. That means any number other than a natural number can't be the factor or multiple of any number.

### Factors

If y is a factor of x, where x and y are natural numbers, then; x/y = k, k should be natural number.

### Multiple

If x is a multiple of y, where x and y are natural numbers, then; x/y = k, k should be natural number.

In general, if a and b are any two natural numbers and a/b is a natural number, then
-> a is a multiple of b
-> b is a factor of a

### Divisibility

If x is divisible by y, where x and y are any two integers (either positive or negative) and y != 0, then; x/y = k, where k is any integer.

16 is a multiple of 4 but not a multiple of -4.
16 is divisible by both 4 and -4.
Note: 0 (zero) is divisible by all integers except '0'.

## Important concepts on factors

### Concept I: Factors of natural number

Let us consider a number 12. Factors of 12 are 1, 2, 3, 4, 6 and 12. There are 6 factors for 12.
Similarly factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30. There are 8 factors for 30.
Key points

• All perfect squares have an odd number of factors
• Perfect square of any prime number has exactly 3 factors
• All non-perfect squares have even number of factors
• Perfect numbers: If the sum of all factors (except that number itself) of a number is equal to the number itself is called a perfect number

Eg: Factors of 6 except 6 -> 1, 2, 3
Sum of above factors = 1 + 2 + 3 = 6
28 and 496 are the other perfect numbers in the first 1000 natural numbers.

### Concept II: Prime factorization

All natural numbers except 1 can be expressed in its prime factorization form. Simply a natural number greater than 1 is the suitable combination of its prime factors.
Eg: 10 = 2^1 xx 5^1
16 = 2^4
30 = 2^1 xx 3^1 xx 5^1
100 = 2^2 xx 5^2

### Concept III: Number of factors of a natural number

If 'N' a natural number and N = a^p xx b^q xx c^r xx..., where a, b, c... are the distinct prime factors of N and p, q, r... are natural numbers which are representing the occurrence of corresponding prime factors in N.

Number of factors of N (including 1 and N) = (p +1) (q + 1) (r +1)...
Find the number of factors of 10, 30 100 and 1024
10 = 2^1 xx 5^1 -> "number of factors" = (1 +1) (1 + 1) = 2 xx 2 = 4
30 = 2^1 xx 3^1 xx 5^1 -> "number of factors" = (1 +1) (1 + 1) (1 + 1) = 2 xx 2 xx 2 = 8
100 = 2^2 xx 5^2 -> "number of factors" = (2 + 1) (2 + 1) = 3 xx 3 = 9
1024 = 2^10 -> "number of factors" = (10 + 1) = 11

### Concept IV: Number of ways to express a number as the product of two factors

If N is a non-perfect square (number of factors should be even)
Then the number of ways to express N as the product of two factors = "number of factors"/2
If N is a perfect square
Then the number of ways to express N as the product of any two factors = ("number of factors " + 1)/2
If N is a perfect square
Then the number of ways to express N as the product of two DISTINCT factors = ("number of factors " - 1)/2

Let's take a look at few examples.

Let us consider the number 30, it can be expressed as the product of two factors in the following ways: 1 xx 30, 2 xx 15, 3 xx 10, 5 xx 6
i.e. 30 can be expressed as the product of two factors in 4 ways. This is half the total number of factors of 30. i.e. the total number of factors of 30 is 8.

Consider another number 36, it's factors are: 1, 2, 3, 4, 6, 9, 12, 18, 36 (Total 9 factors)
36 can be expressed as the product of two factors in the following ways: 1 xx 36, 2 xx 18, 3 xx 12, 4 xx 9, 6 xx 6. Hence the number of ways to express 36 as the product of two factors is 5. And the number of ways to express 36 as the product of two DISTINCT factors in 4 ways (excluding 6 xx 6)

### Concept V: Sum of all factors

If 'N' a natural number and N = a^p xx b^q xx c^r xx..., where a, b, c... are the distinct prime factors of N and p, q, r... are natural numbers which are representing the occurrence of corresponding prime factors in N.

Sum of all factors of N (including 1 and N)
Sum of all factors of N (including 1 and N) = (a^0 + a^1 + a^2 + "..." + a^p) (b^0 + b^1 + b^2 + "..." + b^q) (c^0 + c^1 + c^2 + "..." + c^r)"..."
OR
[(a^(p+1) - 1)/(a-1)][(b^(q+1) - 1)/(b-1)][(c^(r+1) - 1)/(c-1)]"..."
Find the sum of all factors of 60?
60 = 2^2 xx 3^1 xx 5^1
Sum of all factors of 60 = (2^0 + 2^1 + 2^2) (3^0 + 3^1) ( 5^0+ + 5^1)
= (1 + 2 + 4) ( 1 + 3) (1 + 5) = 168

### Concept VI: Number of ways to express a number as the product of two relatively prime factors

If 'N' a natural number and N = a^p xx b^q xx c^r xx..., where a, b, c... are the distinct prime factors of N and p, q, r.... are natural numbers which are representing the occurrence of corresponding prime factors in N.

Then the number of ways to express N as the product of two co-primes = 2^(n-1), where n is the number of distinct prime factors of N.
How many different ways 30 can be expressed as the product of two relatively prime factors ?
30 = 2^1 xx 3^1 xx 5^1: There are three distinct prime factors of 3, so n = 3
:.Therefore the number of ways to express N as the product of two co-primes = 2^(3-1) = 2^2 = 4
And the required expressions are 1 xx 30, 2 xx 15, 3 xx 10, 5 xx 6. Note: 1 is relatively prime to all natural numbers except 1.

### Euler's \phi function or relatively prime to N, which are less than N.

Euler's \phi (phy) function is defined as
\phi(N) = the number of all co-primes to N, which are less than N.

If 'N' a natural number and N = a^p xx b^q xx c^r xx..., where a, b, c... are the distinct prime factors of N and p, q, r.... are natural numbers which are representing the occurrence of corresponding prime factors in N.

\phi(N) = N[1 - 1/a][1 - 1/b][1 - 1/c]... 
If p is relatively prime to 24 and p < 24. How many such values are possible for p? OR Find the number of co-primes to 24, which are less than 24.?
24 = 2^3 xx 3^1
\phi(24) = 24[1 - 1/2][1 - 1/3] = 24 xx 1/2 xx 2/3 = 8
Hence there are 8 possible values for p(1, 5, 7, 11, 13, 17, 19, 23).

### Concept VIII: Sum of all relatively prime numbers to N, which are less than N.

This is the extension of the above result. Here we are finding the sum of all co-primes to N which are less than N.

"Sum" = N^2/2[1 - 1/a][1 - 1/b][1 - 1/c]... OR N/2 \phi(N)
Find the sum of all co-primes to 24, which are less than 24.?
"Sum" = 24^2/2[1 - 1/2][1 - 1/3] = 96

## Remainders - LCM & HCF application

The basic concept of LCM & HCF is clubbed with the concept of remainders is required for solving certain types of questions.

## LCM + Remainders

### Type I : Common remainder

Any number which when divided by x, y or z, leaving the same remainder r in each of the divisions, is of the form k( "LCM of x, y and z") + r, where k is any whole number. When k = 0, we will get the least such value but this is the remainder itself and is less than the divisor.

Find the least prime number which leaves a remainder of 4, when it is divided by 5 or 7.?
Least number which leaves the remainder 4 when divided by 5 or 7 is: k ("LCM of 5 and 7") + 4 = 35k + 4
if k = 0; 35k + 4 = 4 -> "not prime"
if k = 1; 35 k + 4 = 39 -> "not prime"
if k = 2; 35 k + 4 = 74 -> "not prime"
if k = 3; 35k + 4 = 109 -> "prime"
Therefore the required least prime = 109.
Find the largest three digit number that leaves a common remainder when it is divided by 9 or 11.?
N = k ("LCM of 9 and 11") + r = 99k + r
Largest three digit multiple of 99 is 990.:. N = 990 + r
Largest possible value for r = 8, because the least divisor is 9 and the remainder should be lesser than divisor.
:. N = 990 + 8 = 998

### Type II: Common Negative Remainders

When a number N divided by x, y, or z, it leaves the same negative remainder '-r', then the general form of such number N = k ("LCM of x, y and z") - r, where k is any natural number.

Find the smallest number which when divided by 8 or 10, leaves the remainders 6 and 8 respectively.?
Divisors (d)Positive remainders (r)Negative remainders (r - d)
8 6 6 - 8 = -2
10 8 8 - 19 = -2
The required number N = "LCM"(8, 10) - 2 = 40 - 2 = 38
Find the largest three digit number which leaves remainders 20 and 25 when dived by 25 and 30 respectively.?
Negative remainder in each case; 20 - 25 = -5 and 25 - 30 = -5
Required number N = k("LCM of 25 and 30") - 5 = 150k - 5
When k = 6 -> N = 900 - 5 = 895
:.The largest such three digit number = 895.

### Type III: Distinct remainders for different divisors.

Find the smallest number which leaves remainders 2 and 7 when divided by 7 and 11 respectively.?
Let the number be 'N'; "Rem" (N/7) = 2
Therefore N = 7a + 2, where 'a' is the quotient and 'a' is a natural number.
"Rem" (N/11) = 7
Therefore N = 11b + 7, where 'b' is the quotient and 'b' is a natural number.
7a + 2 = 11b + 7
7a = 11b + 7 - 2
7a = 11b + 5
LHS of the equation is a multiple of 7 for any natural number value for 'a'. so the RHS also should be a multiple of 7 for some suitable natural number values for 'b'.
When b = 4 -> "RHS" = 11b + 5 = 11(4) + 5 = 49 , it is multiple of 7.
Hence 4 is the first such suitable value for 'b'.
So N = 11b + 7 = 11(4) + 7 = 51
The required least number = 51

## HCF + Remainders

### Type I: Distinct remainder for different dividends

The largest divisor 'd' with which the numbers X, Y and Z are divided leaves remainders a, b and c respectively is the HCF of (X -a), (Y - b) and (Z - c).

Find the largest divisor which gives the remainders 5, 9 and 6 when it divides 120, 170 and 190 respectively.?
Required divisor = "HCF" [ (120 - 5), ( 170 - 9), (190 - 6)]
= "HCF" [115, 161, 184] = 23.

### Type II: Common remainder

The largest divisor which is giving the same remainder when it divides X, Y and Z is; "HCF"[(X - Y),(Y - Z),(Z - X)] OR "HCF"[(X - Y),(Y - Z)]

Find the largest divisor which divides 239, 380 and 521, leaving the same remainder in each division.?
Required divisor = "HCF" [(380 - 239),(521 - 380)] = HCF [47,141] = 47

## Factorial (!)

Factorial is a well defined operator, which is defined as the "the product of a certain number of natural numbers starting from 1" The operator factorial is denoted by the symbol '!' (Exclamatory sign).
Examples: 4! = 1 xx 2 xx 3 xx 4 = 24
5! = 1 xx 2 xx 3 xx 4 xx 5 = 120
n! = 1 xx 2 xx 3 xx "...." xx (n - 2) xx (n - 1) xx n

Pre defined values
0! = 1
1! = 1
n! = n xx (n - 1)! = n xx (n - 1) xx (n - 2)! , and so on.

### Concept I: Largest power of a prime in N!

Let's look at these kind of problems with help of couple of examples.

Find the largest power of 2 in 5!?
5! = 1 xx 2 xx 3 xx 4 xx 5 = = 2^3 xx 3 xx 5
:.the largest power of 2 in 5! is 3.

If the question ask about a largest factorial value, then the above method is not easy to apply. So we have to think about an alternate method.

Long division method:

Find the largest power of 3 in 100!?
Do a long division of the number using the factor specified.

Largest power of 3 in 100! = sum of all quotient in all steps
-> 33 + 11 + 3 + 1 = 48.
If (76!)/(5^n) is an integer, find the maximum possible value for n.?
If (76!)/(5^n) is an integer, then n is the largest exponent of 5 in 76!

The largest possible value for n = 15 + 3 = 18

### Concept II: Largest power of a composite number in N!

Find the largest power of 6 in 15!.?
6 = 3 xx 2; Largest prime factor of 6 is 3.
For finding the largest power of 6 in 15!, it is enough to find the largest power of 3 in 15!

Largest power of n = 5 + 1 = 6
So the largest power of 6 in 15! is 6

### Concept 3: Number of zeros at the end of N!

In the product of first 'n' natural numbers, a 'zero' will get generated as per the combination of 2 and 5. Therefore finding the number of zeros at the end of N!, find the number of 5's in N!.

How many zeros at the end of 25! ?

i.e. There are 5 + 1 = 6 zeros at the end of 25!

## How to find unit digit of an exponential expression

The unit digit of the powers of natural numbers follows some interesting patterns. An understanding of the occurrence of unit digit will help you to score some points in your exams. This awareness is not only for using the typical questions from unit digit but also can use as the verification tool for basic arithmetic operations.

Unit digit of the different exponents of all numerals follows a certain cyclic order. It is possible to tabulate the cyclic pattern of unit digits in the following manner.

"Base"^nUnit DigitGeneral Form of Exponent
0^n 0 Any natural number
1^n 1 Any natural number
2^1 2 4k + 1
2^2 4 4k + 2
2^3 8 4k + 3
2^4 6 4k + 0
3^1 3 4k + 1
3^2 9 4k + 2
3^3 7 4k + 3
3^4 1 4k + 0
4^1 4 Odd
4^2 6 Even
5^n 5 Any natural number
6^n 6 Any natural number
7^1 7 4k + 1
7^2 9 4k + 2
7^3 3 4k + 3
7^4 1 4k + 0
8^1 8 4k + 1
8^2 4 4k + 2
8^3 2 4k + 3
8^4 6 4k + 0
9^1 9 Odd
9^2 1 Even

Let's look at how to use this table with the help of few examples.

Find the unit digit of 1020^34 ?
1020 ends in 0. From the table, there is only one unit digit for the pattern of 0. i.e. Any exponent of a number which ends in 0 has the unit digit '0'.
Therefore the unit digit of 1020^34 is 0.
Find the unit digit of 1542^345 ?
Exponent = 345
"Rem"[345/4] = 1, i.e 345 is of the form 4k + 1
From the pattern of 2 in the table, if the exponent in the form 4k + 1 then unit digit is 2.
Find the unit digit of 3689^3475 ?
From the table, 9^"odd" will end in 9 itself.
Therefore the unit digit of 3689^3475 is 9.

## How to find last two digits of a^b

If 'a' and 'b' are two natural numbers, then it is possible to find out the last two digits of the expression a^b. We can classify the situations as per the unit digit of the base value 'a'.

### Last two digits of exponents of a number which ends in '0'

Example: 20^n
If n = 1 -> last two digits are the last two digits of the base itself, i.e. 20
If n > 1 -> last two digits are always '00'.

### Last two digits of exponents of a number which ends in '1'

• Unit digit of a^b is always 1.
• Tens digit of a^b = "unit digit of"[ "tens digit of 'a'" xx "unit digit of 'b'"]
Find the last two digits of 21^34?
Unit digit = 1
Tens digit = 2 xx 4 = 8
Therefore the last two digits of 21^34 are 81.
Find the last two digits of 1341^73?
Unit digit = 1
Tens digit = 4 xx 3 = 12 -> 2,(unit digit of 12)
Therefore last two digits of 1341^73 = 21

### Last two digits of exponents of a number which ends in '5'

• If the exponent is 1, then the last two digits are the last two digits of the given number itself.
• If the exponent is greater than 1, then the last two digits are always '25'.
Find the last two digits of 1375^46
1375 ends in 5 and the exponent is greater than 1, therefore the last two digits of the given expression are 25.

### Last two digits of exponents of a number which ends in 3, 7 or 9

• 3^4, 7^4 and 9^2 are ending in 1.
• Convert the base in the form of ending with 1, through the conversion of the exponent.
• After changing the unit digit of the base to 1, then apply the method which we discussed for numbers ending in 1.

Let's look at an example to see how to find last two digits of an exponent of a number which ends in 7

Find the last two digits of 17^244?
First convert the base in such a way that it ends with 1.
17^244 = (17^4)^61
= (17^2 xx 17^2)^61
= (289 xx 289)^61
= ("___21")^61; Now base ends in 1, we can apply the procedure to find last two digits of exponents of a number which ends in 1
Unit digit = 1
Tens digit = 2 xx 1 = 2
:. "Last two digits of " 17^244 = 21

For finding the last two digits of the exponents of a number which ends in 3, 7 or 9, the following patterns will help you to make a quick approach.

#### Pattern I: Last two digits of 4th power of the numbers which end in 3.

03^4 -> 81 \leftarrow 53^4
13^4 -> 61 \leftarrow 63^4
23^4 -> 41 \leftarrow 73^4
33^4 -> 21 \leftarrow 83^4
43^4 -> 01 \leftarrow 93^4
Let's look at an example to see how to apply these patterns in practise.

Find the last two digits of 133^287 ?
"Last two digits of " 133^287 -> "Last two digits of " (33^4)^71 xx 33^3
From the above pattern it is easy to catch the last two digits of 33^4, and those are 21.
-> "Last two digits of " 21^71 xx 33^3
-> "Last two digits of " 21 xx 37 (because the last two digits of 21^71 are 21 and that for 33^3 = 33 xx 33 xx 33 -> 37)
:. "Last two digits of "133^287 " are " 77.

#### Pattern II: Last two digits of 4th power of the numbers which end in 7.

07^4 -> 01 \leftarrow 57^4
17^4 -> 21 \leftarrow 67^4
27^4 -> 41 \leftarrow 77^4
37^4 -> 61 \leftarrow 87^4
47^4 -> 81 \leftarrow 97^4

Find the last two digits of 157^45 ?
"Last two digits of " 157^45 -> "Last two digits of " (57^4)^11 xx 57
"Last two digits of " (01)^11 xx 57 (From the above pattern we can see that last two digits of 57^4 is 01)
= 57
Find the last two digits of 2067^158 ?
"Last two digits of " 2067^158 -> "Last two digits of " (67^4)^39 xx 67^2
-> "Last two digits of " 21^39 xx 89
-> "Last two digits of " 81 xx 89 = 09

#### Pattern III: Last two digits of square of the bases which ends in 9.

09^2 -> 81 \leftarrow 59^2
19^2 -> 61 \leftarrow 69^2
29^2 -> 41 \leftarrow 79^2
39^2 -> 21 \leftarrow 89^2
49^2 -> 01 \leftarrow 99^2

Find the last two digits of 39^121?
39^121 -> (39^2)^60 xx 39
->" Last two digits of " (21)^60 xx 39 (From the above pattern we can see that last two digits of 39^2 are 21)
->" Last two digits of " 01 xx 39
Therefore the last two digits of 39^121 are 39.

### Last two digits of exponents of a number which ends in 2, 4, 6 and 8

• Convert the given bases in the form of 2^10
• (2^10)^"even" will end in 76.
• (2^10)^"odd" will end in 24.
Find the last two digits of 2^70?
2^70 = (2^10)^7 = (2^10)^"odd"
Therefore the last two digits are 24.
Find the last two digits of 8^25?
8^25 = (2^3)^25 = 2^75
= (2^10)^7 xx 2^5
-> " Last two digits of " 24 xx 32 = 68
Find the last two digits of 108^24?
108^24 = (4 xx 27)^24
= (2^2)^24 xx 27^24
= 2^48 xx 27^24
=(2^10)^4 xx 2^8 xx 27^24
Last two digits of (2^10)^4 = 76
Last two digits of 2^8 = 56
Last two digits of 27^24 = (27^4)^6 -> " Last two digits of " (41)^6 -> 41
Therefore the last two digits of 108^24 = Last two digits of [76 xx 56 xx 41] = 96

### Tips for quick calculation

• 76^n is always end in 76 for any natural number values for n.
• 24^"even" will end in 76.
• 24^"odd" will end in 24.
• 76 xx 2^n will end in the last two digits of 2^n, where n >= 2.
Example :
76 xx 32 ends in 32.
76 xx 1024 ends in 24.
76 xx 16 ends in 16.
• 76 xx " any multiple of 4" will end in the last two digits of the multiple of 4.
Example:
76 xx 04 ends in 04.
76 xx 88 ends in 88.
• 76 xx " any even number but not a multiple of 4" will end in 50 + "last two digits of the number"
Example:
76 xx 2 ends in 50 + 2 = 52
76 xx 6 ends in 50 + 6 = 56
76 xx 22 ends in 50 + 22 = 72`

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