Reasoning and aptitude part 2

                                 Factorials

In this chapter we will discuss about number of zeros in the end. And also discuss higher power of exponent.

And also discuss divisibility rule we will elaborate one by 1.

First of all learnt the value of 0!=1.

Value of factorial zero is 1.

Rule to find out zeroes in the end:-

We take minimum power in the pair of (2^n)×(5^n). Sign ^ is indicate power ie 2 race to the power n and 5 race to the power n.

This rule valid only for prime number.

Finally we will find the zeros of any given factorial number.

So we use short trick to find out zeroes.

Short trick:-

Ex:-1 find the zeroes of 12!.

Solution:- 

First of all 12 divide by 2

12/2 then reminder will be zero and quesend will be 6. This quesend once again divide by 2. then we get quesend 3. And this quesend divide by 2 then we get quesend 1 Now these quesend added.

6+3+1=10

So we will write like 2^10.

Now 12 divide by 5

12/5 then quesend will be 2 and this quesend divide by 5 we get 0. Now we will add.

2+0=2

So we will write like 5^2.

Now we will see the min power of these two value. So the power of 5 is minimum ie 2. So final answer will be 2.

This was the 1st example u can solve all the questions of prime number by use this short trick method.

Ex:-2. Find the zeroes of 31!
Solution:- 
31 divide by 2
31/2= 15
15/2= 7
7/2  = 3
3/2  = 1
15+7+3+1= 26
2^26
Now divide by 5
31/5= 6
6/5  = 1
6+1 = 7
5^7
In both the pair minimum power is 7.
So zeroes will be 7.

Ex:-3. Find the zeroes of 100!.
Solution:-
100/2=50
50/2=25
25/2=12
12/2=6
6/2=3
3/2=1
50+25+12+6+3+1=97
2^97
Now divide by 5
100/5=20
20/5=4
20+4=24
5^24
So minimum power is 24.
Zeroes will be 24.

                       Exponent
Ex:- highest power of 2 that can divide 100!.
Solution:- 
100!/2^x= 2^97×...../2^97
X=97
Higher power of 2 is 97.
Note:- this trick valid for prime number.

For non prime number :-
Ex:- 100!/4^x   =   2^97×..(quesend)/(2^2)^x
2x=97
x=48
So highest power of 2 is 48.

Ex:- How many zero's will come in the end of 18!+19!.
Solution:-
18!+19!=18!+19×18!
             = 18!{1+19}
             = 18!×20
             = 5^3×5^1
             = 5^4
So zeroes will be 4.

                      Divisibility rule

Rule No. 1:- Divisible by 2  (N/2).
If last digits zero or even number then value will be divisible by 2.
Ex:- 2,4,6....

Rule No. 2:- Divisible by 3  (N/3).
If sum of the digit of N is divisible by 3.
Ex:- 654=6+5+4=15
        15/3=divisible.

Rule No. 4:- Divisible by 4  (N/4).
If last two digits will be zero or divisible by 4. Then number will be divisible by 4.

Rule No. 5:- Divisible by 5  (N/5).
If last digit 0 or 5. Then whole number will be divisible by 5.

Rule No. 6:- Divisible by 6  (N/6).
If number is divisible by 2 and 3 Then it divisible by 6.

Rule No. 7:- Divisible by 8  (N/8).
If last three digits divisible by 8 or last three digits will zero.
Ex:- abc/8 or 000.

Rule No. 8:- Divisible by 9  (N/9).
If sum of the digit of N is divisible by 9 then number will divisible by 9.

Rule No. 9:- Divisible by 11  (N/11).
Let a number 'abcd'
Now we check this number divisible by 11 or not .
N=abcd
Now start from right side.
So we will write like  (d+b)-(c+a)
If value will become zero or multiple of 11. Then number is divisible by 11.

Note:- 1001 is always divisible by 7,11,13.

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Reasoning and aptitude part 3

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