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NUMBER SYSTEMS

      http://nov15.wordpress.com/
                Presents

         QUANT
          For
        CAT 2009
NUMBER SYSTEMS
INTRODUCTION
PRIME NUMBERS
      A number is prime if it is not divisible by
  
      any prime number less than it’s square
      root.

      Ex: Is 179 a prime number ?
  
                     179  13.3


      Prime Numbers less than 13.3 are
  
      2,3,5,7,11,13

      179 is not divisible by any of them, 179 is
  
      prime.
DIVISIBILITY RULES
SOME MORE DIVISIBILITY RULES
    Test for divisibility by 7: Double the last digit and subtract it

    from the remaining leading truncated number. If the result is
    divisible by 7, then so was the original number. Apply this rule
    over and over again as necessary.

    Example: 826. Twice 6 is 12. So take 12 from the truncated 82.

    Now 82-12=70. This is divisible by 7, so 826 is divisible by 7.


    Test for divisibility by 11: Subtract the last digit from the

    remaining leading truncated number. If the result is divisible by
    11, then so was the first number. Apply this rule over and over
    again as necessary.

    Example: 19151 --> 1915-1 =1914 –>191-4=187 –>18-7=11, so

    yes, 19151 is divisible by 11.
SOME MORE DIVISIBILITY RULES
    Test for divisibility by 13: Add four times the last digit to the remaining leading

    truncated number. If the result is divisible by 13, then so was the first number.
    Apply this rule over and over again as necessary.

    Example: 50661–>5066+4=5070–>507+0=507–>50+28=78 and 78 is 6*13, so

    50661 is divisible by 13.

    Test for divisibility by 17: Subtract five times the last digit from the remaining

    leading truncated number. If the result is divisible by 17, then so was the first
    number. Apply this rule over and over again as necessary.

    Example: 3978–>397-5*8=357–>35-5*7=0. So 3978 is divisible by 17.



    Test for divisibility by 19: Add two times the last digit to the remaining leading

    truncated number. If the result is divisible by 19, then so was the first number.
    Apply this rule over and over again as necessary.

    Example:101156–>10115+2*6=10127–>1012+2*7=1026–>102+2*6=114                    and

    114=6*19, so 101156 is divisible by 19.
REMEMBER IT!
Here is a table using which you can easily remember the
previous divisibility rules.

Read the table as follows :

For divisibility by 7 , subtract 2 times the last digit with the truncated
number.

           Divisibility by                               Test
                  7                             Subtract 2 x Last digit
                 11                             Subtract 1 x Last digit
                 13                               Add 4 x Last Digit
                 17                            Subtract 5 x Last Digit
                 19                               Add 2 x Last Digit
UNIT’S DIGIT OF A NUMBER
    Find the unit’s digit of 71999 (7 to the power 1999)



    Step 1: Divide the exponent by 4 and note down remainder

                           1999/4 => Rem = 3

    Step 2: Raise the unit’s digit of the base (7) to the

    remainder obtained (3)
                                   73 = 343

    Step 3: The unit’s digit of the obtained number is the required

    answer.
                                 343 => Ans 3

    If the remainder is 0, then the unit’s digit of the base is raised to 4 and

    the unit’s digit of the obtained value is the required answer.
                       If Rem = 0 , Then 74 = XX1 -> Ans 1

    Note: For bases with unit’s digits as 1,0,5,6 the unit’s digit for any

    power will be the 1,0,5,6 itself.
                                    453
LAST TWO DIGITS OF A NUMBER


    We will discuss the last two digits of numbers

    ending with the following digits in sets :

    a) 1



    b) 3,7 & 9



    c) 2, 4, 6 & 8

LAST TWO DIGITS OF A NUMBER
  a) Number ending with 1 :


 Ex : Find the last 2 digits of 31786

 Now, multiply the 10s digit of the number with
  the last digit of exponent
     31786 = 3 * 6 = 18 -> 8 is the 10s digit.

    Units digit is obviously 1




    So, last 2 digits are => 81

LAST TWO DIGITS OF A NUMBER

  b) Number Ending with 3, 7 & 9


 Ex: Find last 2 digits of 19266


    We need to get this in such as way that the

    base has last digit as 1
             19266 = (192)133 = 361133

  Now, follow the previous method => 6 * 3 =

  18
 So, last two digits are => 81
LAST TWO DIGITS OF A NUMBER

    b) Number Ending with 3, 7 & 9




  Remember :


 34 = 81

 74 = 2401

 92 = 81
LAST TWO DIGITS OF A NUMBER
    Ex 2: Find last two digits of 33288

              Now, 33288 = (334)72 = (xx21)72
               Ten’s digit is -> 2*2 = 04 -> 4
                So, last two digits are => 41

    Ex 3: find last 2 digits of 87^474



(872)*(874)118 => (xx69) * (xx61)118 (6 x 8 = 48)
                => (xx69)*(81)

                 So, last two digits are 89
LAST TWO DIGITS OF A NUMBER
 c)Ending with 2, 4, 6 or 8

Here, we use the fact that 76 power any
 number gives 76.

We also need to remember that,
 242 = xx76
 210 = xx24
 24even = xx76
 24odd = xx24
LAST TWO DIGITS OF A NUMBER
Ex: Find the last two digits of 2543
               2543 = ((210)54) * (23)
                    = ((xx24)54)* 8
                    = ((xx76)27)*8

           76 power any number is 76

  Which gives last digits as => 76 * 8 = 608


 So last two digits are : 08
HIGHEST POWER
    Highest power of a number that divides the

    factorial of another number.
    What is the highest power of 5 that divides

    60!(factorial)
    Note: N! = N*(N-1)*(N-2)*(N-3)….(2)*(1)


    Now, Continuously divide 60 with 5 as shown

                       60/5 = 12,
                        12/5 = 2 (omit remainders)
                        2/5 = 0 <- stop at 0
    Now add up all the quotients => 12+2+0 = 14

    So highest power of 5 that divides 60! is 14.

HIGHEST POWER
    Ex: Find Highest power of 15 that divides 100!



    Here, as 15 is not a prime number we first split 15 into prime factors.

                                     15 = 5 * 3
    Now, find out highest power of 5 that divides 100! and also highest power of 3

    that divides 100! .
    For 5 :                          100/5 =20

                                       20/5 = 4
                                        4/5 = 0
    So, 20 + 4 + 0 = 24
    For 3 :                          100/3 = 33

                                       33/3 = 11
                                       11/3 = 3
                                        3/3 = 1
                                        1/3 = 0

So, 33 + 11+ 3 + 1 + 0 = 48

    Now, the smallest number of these is taken which will be 24.

NUMBER OF ZEROES
  Ex: Find the number of zeroes in 75!

 This means highest power of 10 which can divide 75!
                         10 = 5*2
 If we consider highest power of 5 which can divide
  75! , it’s sufficient.
                         75/5 =15
                          15/5 =3
                           3/5 =0
So, 15+3+0 = 18

    So, there are 18 zeroes in 75!

NUMBER OF FACTORS OF A NUMBER
    If the number N can be expressed as a product of prime

    factors such that
                    N = (pa)*(qb)*(rc)
    where,
    p,q,r = prime factors
    a,b,c = powers to which each is raised


    Then,

No. of factors of N (including 1, N) = (a+1)*(b+1)*(c+1)*….
EVEN AND ODD

  Even number => Divisible by 2


 Odd Number => Not Divisible by 2

 Important Results :

                exe=e
                exo=e
                oxo=o
EXERCISE


   Download the related exercise here

      Exercise 1 - Number Systems
LET ME KNOW!!!

If you liked this presentation, do comment
                      on

  http://nov15.wordpress.com
                    or

          write to Nicky at

  nickyswetha20@yahoo.com
THANK YOU!

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Number Systems Theory for CAT 2009 Quant

  • 1. NUMBER SYSTEMS http://nov15.wordpress.com/ Presents QUANT For CAT 2009
  • 4. PRIME NUMBERS A number is prime if it is not divisible by  any prime number less than it’s square root. Ex: Is 179 a prime number ?  179 13.3 Prime Numbers less than 13.3 are  2,3,5,7,11,13 179 is not divisible by any of them, 179 is  prime.
  • 6. SOME MORE DIVISIBILITY RULES Test for divisibility by 7: Double the last digit and subtract it  from the remaining leading truncated number. If the result is divisible by 7, then so was the original number. Apply this rule over and over again as necessary. Example: 826. Twice 6 is 12. So take 12 from the truncated 82.  Now 82-12=70. This is divisible by 7, so 826 is divisible by 7. Test for divisibility by 11: Subtract the last digit from the  remaining leading truncated number. If the result is divisible by 11, then so was the first number. Apply this rule over and over again as necessary. Example: 19151 --> 1915-1 =1914 –>191-4=187 –>18-7=11, so  yes, 19151 is divisible by 11.
  • 7. SOME MORE DIVISIBILITY RULES Test for divisibility by 13: Add four times the last digit to the remaining leading  truncated number. If the result is divisible by 13, then so was the first number. Apply this rule over and over again as necessary. Example: 50661–>5066+4=5070–>507+0=507–>50+28=78 and 78 is 6*13, so  50661 is divisible by 13. Test for divisibility by 17: Subtract five times the last digit from the remaining  leading truncated number. If the result is divisible by 17, then so was the first number. Apply this rule over and over again as necessary. Example: 3978–>397-5*8=357–>35-5*7=0. So 3978 is divisible by 17.  Test for divisibility by 19: Add two times the last digit to the remaining leading  truncated number. If the result is divisible by 19, then so was the first number. Apply this rule over and over again as necessary. Example:101156–>10115+2*6=10127–>1012+2*7=1026–>102+2*6=114 and  114=6*19, so 101156 is divisible by 19.
  • 8. REMEMBER IT! Here is a table using which you can easily remember the previous divisibility rules. Read the table as follows : For divisibility by 7 , subtract 2 times the last digit with the truncated number. Divisibility by Test 7 Subtract 2 x Last digit 11 Subtract 1 x Last digit 13 Add 4 x Last Digit 17 Subtract 5 x Last Digit 19 Add 2 x Last Digit
  • 9. UNIT’S DIGIT OF A NUMBER Find the unit’s digit of 71999 (7 to the power 1999)  Step 1: Divide the exponent by 4 and note down remainder  1999/4 => Rem = 3 Step 2: Raise the unit’s digit of the base (7) to the  remainder obtained (3) 73 = 343 Step 3: The unit’s digit of the obtained number is the required  answer. 343 => Ans 3 If the remainder is 0, then the unit’s digit of the base is raised to 4 and  the unit’s digit of the obtained value is the required answer. If Rem = 0 , Then 74 = XX1 -> Ans 1 Note: For bases with unit’s digits as 1,0,5,6 the unit’s digit for any  power will be the 1,0,5,6 itself. 453
  • 10. LAST TWO DIGITS OF A NUMBER We will discuss the last two digits of numbers  ending with the following digits in sets : a) 1  b) 3,7 & 9  c) 2, 4, 6 & 8 
  • 11. LAST TWO DIGITS OF A NUMBER a) Number ending with 1 :   Ex : Find the last 2 digits of 31786  Now, multiply the 10s digit of the number with the last digit of exponent 31786 = 3 * 6 = 18 -> 8 is the 10s digit. Units digit is obviously 1  So, last 2 digits are => 81 
  • 12. LAST TWO DIGITS OF A NUMBER b) Number Ending with 3, 7 & 9   Ex: Find last 2 digits of 19266 We need to get this in such as way that the  base has last digit as 1 19266 = (192)133 = 361133 Now, follow the previous method => 6 * 3 =  18  So, last two digits are => 81
  • 13. LAST TWO DIGITS OF A NUMBER b) Number Ending with 3, 7 & 9  Remember :   34 = 81  74 = 2401  92 = 81
  • 14. LAST TWO DIGITS OF A NUMBER Ex 2: Find last two digits of 33288  Now, 33288 = (334)72 = (xx21)72 Ten’s digit is -> 2*2 = 04 -> 4 So, last two digits are => 41 Ex 3: find last 2 digits of 87^474  (872)*(874)118 => (xx69) * (xx61)118 (6 x 8 = 48) => (xx69)*(81) So, last two digits are 89
  • 15. LAST TWO DIGITS OF A NUMBER c)Ending with 2, 4, 6 or 8  Here, we use the fact that 76 power any number gives 76. We also need to remember that,  242 = xx76  210 = xx24  24even = xx76  24odd = xx24
  • 16. LAST TWO DIGITS OF A NUMBER Ex: Find the last two digits of 2543 2543 = ((210)54) * (23) = ((xx24)54)* 8 = ((xx76)27)*8 76 power any number is 76 Which gives last digits as => 76 * 8 = 608   So last two digits are : 08
  • 17. HIGHEST POWER Highest power of a number that divides the  factorial of another number. What is the highest power of 5 that divides  60!(factorial) Note: N! = N*(N-1)*(N-2)*(N-3)….(2)*(1)  Now, Continuously divide 60 with 5 as shown  60/5 = 12, 12/5 = 2 (omit remainders) 2/5 = 0 <- stop at 0 Now add up all the quotients => 12+2+0 = 14  So highest power of 5 that divides 60! is 14. 
  • 18. HIGHEST POWER Ex: Find Highest power of 15 that divides 100!  Here, as 15 is not a prime number we first split 15 into prime factors.  15 = 5 * 3 Now, find out highest power of 5 that divides 100! and also highest power of 3  that divides 100! . For 5 : 100/5 =20  20/5 = 4 4/5 = 0 So, 20 + 4 + 0 = 24 For 3 : 100/3 = 33  33/3 = 11 11/3 = 3 3/3 = 1 1/3 = 0 So, 33 + 11+ 3 + 1 + 0 = 48 Now, the smallest number of these is taken which will be 24. 
  • 19. NUMBER OF ZEROES Ex: Find the number of zeroes in 75!   This means highest power of 10 which can divide 75! 10 = 5*2  If we consider highest power of 5 which can divide 75! , it’s sufficient. 75/5 =15 15/5 =3 3/5 =0 So, 15+3+0 = 18 So, there are 18 zeroes in 75! 
  • 20. NUMBER OF FACTORS OF A NUMBER If the number N can be expressed as a product of prime  factors such that N = (pa)*(qb)*(rc) where, p,q,r = prime factors a,b,c = powers to which each is raised Then,  No. of factors of N (including 1, N) = (a+1)*(b+1)*(c+1)*….
  • 21. EVEN AND ODD Even number => Divisible by 2   Odd Number => Not Divisible by 2  Important Results : exe=e exo=e oxo=o
  • 22. EXERCISE Download the related exercise here Exercise 1 - Number Systems
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