For the concept of identifying the unit digit, we have to first familiarize with the concept of cyclicity. Cyclicity of any number is about the last digit and how they appear in a certain defined manner. Let’s take an example to clear this thing:

The cyclicity chart of 2 is:

2^{1 }=2

2^{2} =4

2^{3 }=8

2^{4}=16

2^{5}=32

Have a close look at the above. You would see that as 2 is multiplied every-time with its own self, the last digit changes. On the 4^{th} multiplication, 2^{5} has the same unit digit as 2^{1}. This shows us the cyclicity of 2 is 4, that is after every fourth multiplication, the unit digit will be two.

Cyclicity table:

The cyclicity table for numbers is given as below:

How did we figure out the above? Multiply and see for yourself. It’s good practice.

Now let us use the concept of cyclicity to calculate the Unit digit of a number.

**What is the unit digit of the expression 4 ^{993}?**Now we have two methods to solve this but we choose the best way to solve it i.e. through cyclicity

We know the cyclicity of 4 is 2

Have a look:

4

^{1}=4

4

^{2 }=16

4

^{3 }=64

4

^{4 }=256

From above it is clear that the cyclicity of 4 is 2. Now with the cyclicity number i.e. with 2 divide the given power i.e. 993 by 2 what will be the remainder the remainder will be 1 so the answer when 4 raised to the power one is 4.So the unit digit in this case is 4.

For checking whether you have learned the topic, think of any number like this, calculate its unit digit and then check it with the help of a calculator.

* Note* :

*If the remainder becomes zero in any case then the unit digit will be the last digit of*

**a**^{cyclicity number }*where a is the given number and cyclicity number is shown in above figure.*

Lets solve another example:**The digit in the unit place of the number 7 ^{295} X 3^{158} is**

A. 7

B. 2

C. 6

D. 4

**Solution**

The Cyclicity table for 7 is as follows:

7^{1 }=7

7^{2 }=49

7^{3} = 343

7^{4} = 2401

Let’s divide 295 by 4 and the remainder is 3.

Thus, the last digit of 7^{295} is equal to the last digit of 7^{3} i.e. 3.

The Cyclicity table for 3 is as follows:

3^{1} =3

3^{2} =9

3^{3 }= 27

3^{4 }= 81

3^{5 }= 243

Let’s divide 158 by 4, the remainder is 2. Hence the last digit will be 9.

Therefore, unit’s digit of (7^{925} X 3^{158}) is unit’s digit of product of digit at unit’s place of 7^{925} and 3^{158} = 3 * 9 = 27. Hence option 1 is the answer.

A factorial is a non-negative number which is equal to the multiplication of numbers that are less than that number and the number itself. It is denoted by (!)**Let’s take an example to understand this **

What will be the value of 5!

So in the above definition we discussed that the multiplication of the numbers which all are less than that number and the number itself. Hence number less than 5 are 1,2,3,4 and 5 is number itself so

5! = 5 x 4 x 3 x 2 x 1 = 120

Always remember we define the value of 0! =1

**Lets take some more examples for this **

Value of 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720

Value of 4! = 4 x 3 x 2 x 1 = 24

Value of 3! = 3 x 2x 1= 6

We can also write n! = n x (n – 1)!

Concept of factorial is very important, there are few basic types in which this concept is used. We will discuss each type with the help of examples.** ****Type 1: Highest power of p (p is a prime number) which divides the q! **

Let’s take an example to understand this type** ****What is the highest power of 3 that divides 13! ?****Solution**:

13! = 13 x 12 x 11 x 10 x 9 x 8 x7 x 6 x 5 x 4 x 3 x 2 x 1

And we have to find highest power of 3 that can divide the above term

So first we need to know how many times 3 is multiplied in 13! .

So to find the number of 3’s we will divide 13 with 3.

13 divided by 3 gives 4 as quotient and 1 remainder. We will keep remainder aside and move further till the quotient cannot be divided further

Again we divide 4 by 3 we get quotient as 1 and again remainder is 1 .Now we stop at this stage because quotient 1 cannot be divided by 4.

Now add all the quotients

4+1 =5.

So the maximum power of 3 is 5, which can divide the 13! .

**We can also generalize the formula for calculating **

Maximum power of any prime p in any n! is calculated as

[np]+[np2]+[np3]+[np4]+………………">[np]+[np2]+[np3]+[np4]+………………[np]+[np2]+[np3]+[np4]+………………

Where [ ] represents integral part

Maximum power of 3 in 13!

[133]+[1332]+………..">[133]+[1332]+………..[133]+[1332]+………..

= 4+ 1 =5

**Other method**

Since the number 13! is not very big number in-fact we can write and check maximum power of 3

13 x 12 x 11 x 10 x 9 x 8 x7 x 6 x 5 x 4 x 3 x 2 x 1

13 x **3 x4** x 11 x 10 x **3 x 3** x 8 x 7 x **3** x 2 x **3** x 2 x 1

Here the number of 3’s is 5 so we can cancel it with 3^{5}. So maximum power of 3 is 5, which can divide 13!

**Let’s take one more example like this ****What is the highest power of 7 that exactly divides 49! **

To find the highest power of 7 that exactly divides 49! . We need to know the number of 7’s in the 49!

So when we divide 49 with 7, the quotient will be 7 and there is no remainder, since the quotient can further divided by 7 we will divide 7 with 7 with quotient 1 and remainder 0 .

Now add all quotients and get the answer as 8

So the highest power which will divide the 49! will be 8.

**Type : Highest power of p which divides the q! ,where p is not a prime number**

The approach for this type is same as that for calculating maximum power of prime in any factorial buthere first we will break p into product of primes.

Lets take an example to understand this

__Example 1__**What will be the maximum power of 6 that divides the 9!**

In order to find maximum power of 6 we will first write as product of 2 and 3.

Maximum power of 2 in 9!

[92]+[922]+[923]+…..= 4 + 2 + 1 + 0=7Maximum power of 3 in 9![93]+[932]+…..= 3+ 1+0=4">[92]+[922]+[923]+…..= 4 + 2 + 1 + 0=7Maximum power of 3 in 9![93]+[932]+…..= 3+ 1+0=4[92]+[922]+[923]+…..= 4 + 2 + 1 + 0=7Maximum power of 3 in 9![93]+[932]+…..= 3+ 1+0=4

And there are seven 2’s and four 3’s are there, so we need to make the pairs of 2 and 3.

Now we know that 6 = 2 x3. So we need equal number of 2 and equal number of 3 in 9!

Hence there are maximum four pairs that can make 6.So the maximum power of 6 that

can divide 9! is 4.

**Other method**

Since the number 9! is not very big number in-fact we can write and check maximum power of 3

9! = 9x8x7x6x5x4x3x2x1 =**3×3**x**2x2x2**x7x**2×3**x5x**2×2**x**3**x**2**x1

So there are four pairs of 2 x 3, which can be formed

So the maximum power of 6 that can divide the 9! is 4 .

__Example 2__**What will be the highest power of 12 that can exactly divide 32!**

We can write 12 = 2x2x3 i.e. we need pair of 2^{2} x 3

Maximum power of 2 in 32!

[322]+[3222]+[3223]+[3224]+[3225]+…..= 16 + 8 + 4 + 2 +1=31Maximum power of 3 in 32![323]+[3232]+[3233]+…..= 10+ 3+ 1= 14">[322]+[3222]+[3223]+[3224]+[3225]+…..= 16 + 8 + 4 + 2 +1=31Maximum power of 3 in 32![323]+[3232]+[3233]+…..= 10+ 3+ 1= 14[322]+[3222]+[3223]+[3224]+[3225]+…..= 16 + 8 + 4 + 2 +1=31Maximum power of 3 in 32![323]+[3232]+[3233]+…..= 10+ 3+ 1= 14

Now in 32! Number of 2 are 31

And the number of three’s = 14

So therefore if we take 14 three’s then we need 28 two’s because for each three we need two 2’s so therefore we need 28 two’s from 31 two’s to make the equal pairs of 2^{2} x3

So the maximum power of 12 that can divide the 32! is 14

**Number of zeros at the end of a ****Problem Type: Number of zeros at the end of p!**

To solve such type of problems we see only pair of 2 x5 in p! .

Because if a number is divisible by 10 then it will have 0 in the end and 10= 2×5 so finding number of zeros is equivalent to finding maximum power of 10 in p! which is same as counting number of pairs of 2 and 5.

Lets take an example for this

__Example__**Find the number of zeros at the end of 45! ****Solution:**

Zero mainly comes from the combination of (5x 2) or by the presence of 10, and the number of zeros depends upon the number of times 10 is involved.So we check power of 2 and 5 first

Maximum power of 2 in 45!

[452]+[4522]+[4523]+[4524]+[4525]+…..= 22 + 11 + 5 + 2 +1= 41Maximum power of 5 in 45![455]+[4552]+[4553]+…..= 9+ 1= 10">[452]+[4522]+[4523]+[4524]+[4525]+…..= 22 + 11 + 5 + 2 +1= 41Maximum power of 5 in 45![455]+[4552]+[4553]+…..= 9+ 1= 10[452]+[4522]+[4523]+[4524]+[4525]+…..= 22 + 11 + 5 + 2 +1= 41Maximum power of 5 in 45![455]+[4552]+[4553]+…..= 9+ 1= 10

So in 45! Number of 2’s are 41

And the number of 5’s is 10

So maximum pair of 2 and 5 that can be made are 10 so the number of zeros at the end of the 45! is 10.

__Example __**Find the number of zeros in 500!****Solution: **Zero mainly comes from the combination of (5x 2) or by the presence of 10, and the number of zeros depends upon the number of times 10 is involved.So we check power of 2 and 5 first

Maximum power of 2 in 500!

[5002]+[50022]+[50023]+[50024]+[50025]+[50026]+[50027]+[50028]…= 250 + 125 + 62 + 31 +15 +7 + 3 + 1= 494Maximum power of 5 in 500![5005]+[50052]+[50053]+[50054]+…..= 100 + 20 + 4= 124">[5002]+[50022]+[50023]+[50024]+[50025]+[50026]+[50027]+[50028]…= 250 + 125 + 62 + 31 +15 +7 + 3 + 1= 494Maximum power of 5 in 500![5005]+[50052]+[50053]+[50054]+…..= 100 + 20 + 4= 124[5002]+[50022]+[50023]+[50024]+[50025]+[50026]+[50027]+[50028]…= 250 + 125 + 62 + 31 +15 +7 + 3 + 1= 494Maximum power of 5 in 500![5005]+[50052]+[50053]+[50054]+…..= 100 + 20 + 4= 124

Number of 5’s in 500! are 124

And the numbers of 2’s is 494

So the maximum possible pairs of 2 and 5 that can be made are 4 so the number of zeros in 500! are 124 .So the number of zeros in the end of the 500! are 124.

__Example__**What will be the number of zeros in (24!) ^{3! }?**

Now first calculate the number of zeros in 24!

Maximum power of 2 in 24!

[242]+[2422]+[2423]+[2424]+…..= 12 + 6 + 3 + 1= 22Maximum power of 5 in 24![245]+[2452]+…..= 4 + 0= 4">[242]+[2422]+[2423]+[2424]+…..= 12 + 6 + 3 + 1= 22Maximum power of 5 in 24![245]+[2452]+…..= 4 + 0= 4[242]+[2422]+[2423]+[2424]+…..= 12 + 6 + 3 + 1= 22Maximum power of 5 in 24![245]+[2452]+…..= 4 + 0= 4

Number of 5’s in 24! is 4

And the numbers of 2’s are more than 4

So the maximum possible pairs of 2 and 5 that can be made are 4 so the number of zeros in 24! are 4

Now 24! also have power of 3! i.e. 6 . So the whole expression would have 4 x 6 = 24 zeros in the end.

**Extra Problems for ‘Number of Zeros’ Question Type**

**Example 1: Find the number of zeros in 2 ^{145} x 5^{234 }. **

**Solution**:When we see the question it looks like a very difficult question but this type of question involving number of zeros is very simple and can be solved in seconds. Here the question is asking for the number of zeros, now we have to look for the pairs of 2x 5, so the maximum pairs we can form are 145 because we have the maximum power of 2 is 145 , so the number of zeros in this case are 145.

**Example 2**: **How many numbers of zeros are there in following expression**

2^{11} x 125^{3} x 7^{11}+ 3^{11} x 7^{11}x 2^{10 }x 5^{1}+ 11^{12} x 2^{13} x 5^{125}**Solution**: To explain the solution we must know something about the nature of zeros, before doing this lets have a simple example, the number of zeros in the end of

10+100+1000+……………..+ 10000000000000000000 are 1 because we can take 10 common out of this expression and write it as 10(1+10+100+……………..+ 1000000000000000000) And clearly the term present inside the bracket has unit digit 1 which when multiplied with 10 results into 1 zero at its end.Thus number of zeros in any such expression will depend on the least zero holder in the expression.

Now using the same logic we can say least number of zeros will be contributed by second term i.e. 3^{11} x 7^{11}x 2^{10 }x 5^{1} as it contains only single power of 5 .Hence it contains only 1 zero, so the number of zeros for whole expression is 1 .

**Example 3**:**Find the number of zeros at end of ****5 x 10 x 15 x 20 x 25 x 30 x 35…………………………………… x 240 x 245 x 250 ****Solution**: We can take 5 common out of this expression and write it as

5^{50} (**1 x 2 x 3 x 4…………………………………… x 49 x 50)**

5^{50} (**50!)**

To find number of zeros we first find

Maximum power of 2 in 50!

[50/2]+[50/2^2]+[50/2^3]+[50/2^4]+[50/2^5]+…..= 25 + 12 + 6 + 3 +1= 47

Maximum power of 5 in 50![50/5]+[50/5^2]+[50/5^3]+…..= 10 + 2= 12

Thus maximum power of 5 present in given expression is 62 and maximum power of 2 present in given expression is 47. Hence number of zeros will be 47.

**Assignment: **__Questions:__

1 : Find the number of zeros in 75!

a) 16

b) 18

c) 20

d) 21

Ans: b

Solution:

Maximum power of 5 in 75!

[75/5]+[75/5^2]+[75/5^3]+…..= 15 + 3= 18

Hence number of zeros will be 18.

2: Find the number of zeros in 255!

a) 63

b) 52

c) 62

d) 65

Ans: a

Solution:

Maximum power of 5 in 255!

[255/5]+[255/5^2]+[255/5^3]+[255/5^4]+…..= 51 + 10 + 2= 63

Hence number of zeros will be 63.

3: Find the number of zeros in 135!

a) 25

b) 30

c) 33

d) 32

Ans: c

Solution:

Maximum power of 5 in 135!

= 27 + 5 + 1

= 33

[135/5]+[135/5^2]+[135/5^3]+…..= 27 + 5 + 1= 33

Hence number of zeros will be 33.

4: Find the number of zeros in n! where n is number between 66 to 69

a) 12

b) 14

c) 15

d) 16

Ans: c

We pick any value of n between 66 and 69. Number of zeros will be same for any value we pick between 66 and 69 say 68

Maximum power of 5 in 68!

= 13 + 2

= 15

[685]+[6852]+[6853]+…..= 13 + 2= 15">[685]+[6852]+[6853]+…..= 13 + 2= 15[685]+[6852]+[6853]+…..= 13 + 2= 15**Hence number of zeros will be 15.**

5: Find the number of zeros in 350!

a) 84

b) 85

c) 86

d) 87

Ans: c

Solution:

Maximum power of 5 in 350!

= 70 + 14 + 2

= 86

[350/5]+[350/5^2]+[350/5^3]+[350/5^4]+…..= 70 + 14 + 2= 86* So the number of zeros in the end of the 350! are 86*.

**Remainders: Part-1**

When a particular number is divided by any another number(divisor) we get quotient and remainder.

For example: On dividing 9 with 4 we get 2 as the quotient and 1 as the remainder. Here 9 is dividend and 4 is divisor.

**Tool tip:**The value of the remainder is always less than the divisor.**Find the remainder when 39 is divided by 6 ?**

We can write 39 = 6(6) + 3. Here quotient is 6 and 3 is remainder. Note that remainder 3 is less than divisor 6.

**Find the remainder when 79 is divided by 11 ?**

We can write 79 = 11(7) + 2. Here quotient is 7 and 2 is remainder. Note that remainder 2 is less than divisor 7.

**If divisor is same then it hardly creates any difference whether we add numbers and then divide by divisor to calculate remainder or we separately calculate remainders and then add.**

Let’s understand first of all whether we deal numbers separately while calculating remainders or we deal together it doesn’t create any difference. Lets understand meaning of above with help of example

Suppose we have 53 sticks in a bundle and we are told to divide it in groups of 5. So we all know that while dividing 53 in groups of 5, three sticks will be left.

Let’s consider another bundle of 21 sticks and we are told to divide it in groups of 5. So we all know that while dividing 21 in groups of 5, one stick will be left.

Now if we add the above two bundles we have total of 53 + 21=74 and again divide it in groups of 5

Then we will be left with 4 sticks.

Hence instead of mixing we can treat them by two groups as well and we can make groups of 5 separately and total numbers of remaining sticks are simply 4(3+1)i.e. addition of remaining sticks in the two groups separately.

Since 4 is lesser than 5 we cannot make another group of 5 from remaining sticks

But if instead of 21 sticks we had 22 sticks then the remainder would be 2 and the total remaining sticks would have been 5, and then we can make a new group of 5 with the remaining sticks with no remnant left.

**Remainders: Part-2****Type: Difference between divisor and remainder is same****On dividing by p, q and r the number gives different remainders a, b and c where ****p-a=q-b=r-c**

Let’s take an example to understand this

__Example__**What is the minimum two-digit number which when divided by 3, 4, 5 leaves 1, 2, 3 as the remainder respectively?****Solution:**

Let P be the required two digit number .As given in the question

P = 3 d + 1 =3 (d + 1) -2

P = 4 e + 2= 4( e + 1) -2

P = 5 f + 3= 5 (f+1) – 2

In each of above cases we observe difference between divisor and remainder is same.

Hence we have same remainder -2 when divided by 3,4,5

Thus number N must be in the form L.C.M. (3,4,5) k – 2 =60 k – 2

Since we are looking for two digit number putting k =1 we get 58 as minimum two-digit number which when divided by 3, 4, 5 leaves 1, 2, 3 as the remainder respectively.

__Example__**What is the minimum three-digit number which when divided by 2, 5, 6 leaves 1, 4, 5 as the remainder respectively?****Solution:**

Let P be the required three digit number .As given in the question

P = 2 d + 1 =2 (d + 1) -1

P = 5 e + 4 = 5( e + 1) -1

P = 6 f + 5 = 6 (f+1) – 1

In each of above cases we observe difference between divisor and remainder is same.

Hence we have same remainder -1 when divided by 2,5,6

Thus number N must be in the form L.C.M. (2,5,6) k – 1 =30 k – 1

Since we are looking for two digit number putting k =34 we get 1019 as minimum three-digit number which when divided by 2, 5, 6 leaves 1, 4, 5 as the remainder respectively.

*Assignment: ***QUESTIONS:****1.** What is the remainder when 72 + 51 is divided by 8 ?

a)1

b)2

c)3

d)5

Answer – c**Solution : **Since 72 is multiple of 8 hence it leaves remainder zero and 51 leaves remainder 3 when divided by 8.Hence 72 + 51 leaves remainder 3 when divided by 8.

**2.** What is the remainder when 111 + 68 + 48 is divided by 11 ?

a)5

b)7

c)9

d)None

answer – b**Solution:**

111 leaves remainder 1 when divided by 11

68 leaves remainder 2 when divided by 11

48 leaves remainder 4 when divided by 11

Thus 111 + 68 + 48 leaves remainder 1+2+4 =7 when divided by 11**3.** What is the remainder when 42×56 is divided by 8 ?

a)0

b)1

c)2

d)3

answer – a**Solution :**

56 is a multiple of 8 hence remainder is zero.

**4.**What is the remainder when 48×65 is divided by 7 ?

a)1

b)2

c)3

d)5

Answer – 5

48 leaves remainder 6 and 65 leaves remainder 2 so 48 65 divided by 7 gives 6×2=12 but 12 is greater than 7 hence we further divide by 7 and we obtain 5 as remainder.

**5.** What is the remainder of 84x71x41 by 9?

a)2

b)3

c)6

d)8

answer-b

84 leaves remainder 3, 71 leaves remainder 8 and 41 leaves remainder 5 when divided by 9. So 84x71x42 divided by 9 gives 3 8 =120 but 120 is greater than 9 hence we further divide by 9 and we obtain 3 as remainder.