Factorial

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Definition 

In mathematics, the factorial of a positive integer n, denoted by n!, is the product of all positive integers less than or equal to n.

            n!=1×2×3.....×n

Fact 0!=1

We will soon discuss the reason behind it.

NOTE

2n!=2ⁿn![1*3*5....(2n-1)]

Illustrations

Do it yourself:

Factorise 

Illustration 1

Find the value of 5!

Answer:- 120

Solution 

5!=1×2×3×4×5=120

Illustration 2

n is a number less than 14 how many values can n take if n+1 is a factor of n!

Answer-5

Solution 

First let us use trial or error.

For 2!, 3 is not a factor.

For 3!, 4 is not a factor.

For 4!, 5 is not a factor.

........................................................

Now a prime number has only two factors 1 and itself.

So n+1 cannot be prime

Prime numbers smaller than or equal to 14 are 13,11,7,5,2

So n cannot be 12 ,10, 6, 4, 1.

Now let us think from 7.

If n is 7

7! has a factor 8

Since 7!=1×2×3×4×5×6×7=1×3×5×6×7×2     ×4=1×3×5×6×7×8

9 is not a factor of 8!

10 is a factor of 9!

12 is a factor of 11!

14 is a factor of 13!

So we have 5 possible cases.

Illustration 3

Find the maximum value of a, such that 2ᵃ divides 100!

Solution 

[100/2]+[100/2²]+[100/2³]+[100/2⁴]+[100/2⁵]+[100/2⁶]+[100/2⁷]........

Important note- this method is valid for prime numbers like 2,5,7,11,......

[*] Denotes greatest integer function

=97

Similar questions 

1) Find the maximum value of n such that 5ⁿ is a factor of 100!.

Answer-24

Illustration 4

Find the number of consecutive zeros at the end of  100!

                    Or

Find the maximum value  of n  such that 10ⁿ is a factor of 100!

Answer-24

Solution 

Number of zeros can be find out by 2ˣ × 5ˣ= 10ˣ

From our earlier observation we have the highest power of 2 and 5

2⁹⁷ × 5²⁴=2⁷³×(2²⁴×5²⁴)=2⁷³×10²⁴

So there are 24 zeros 

Incorrect solution 

[100/10]+[100/10²]+[100/10³]+...=11

But the above solution is incorrect. 

Since it is valid for only prime number 

Problems 

1)

Find the value of n

Answer-8

2)  let the value of the below given expression is k where k is an integer

Find the largest possible value of n

Answer-28

Solution 

5!= 120 = 5×4×3×2×1 = 5×2³x3

 

The above expression is equal to 

120!/120ⁿ

We can find the factors which comes as 120²⁸

3)Find the value of 

Answer-1

4) Find the unit digit of 

(1!)³+(2!)³+(3!)³....+(2020!)³

Answer-9

Solution 

From 5! Onwards we have unit digit as 0

5!=1×2×3×4×5=1×3×4×(2×5)=1×3       ×4×10

6!=6×5!

7!=7×6×5!

n!=k×5!    (If n>5)

So unit digit of cube of

5! is 0

6! is 0

7! is 0

8! is 0

.......

(1!)³=1

(2!)³=2³=8

(3!)³=6³ unit digit is 6

(4!)³=24³ unit digit is 4

So unit digit of the expression 

1+8+6+4=19

Is 9

Q)Show that (kn)! is divisible by (n!)ᵏ

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