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Transcript
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
114=6*19, so 101156 is divisible by 19.
and
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.
Ex: Unit’s digit of 3589494856453 = 6
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
