NUMBER SYSTEMS

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# NUMBER SYSTEMS - PowerPoint PPT Presentation

http://nov15.wordpress.com/ Presents QUANT For CAT 2009. NUMBER SYSTEMS. NUMBER SYSTEMS. INTRODUCTION. 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 ? Prime Numbers less than 13.3 are 2,3,5,7,11,13

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### INTRODUCTION

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 ?

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

179 is not divisible by any of them, 179 is prime.

PRIME NUMBERs

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.

Some more divisibility rules
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.

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

Unit’s Digit 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

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

Last two digits of a number

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.

Highest power

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 zeroes

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)*….

Number of factors of a number

Even number => Divisible by 2

Odd Number => Not Divisible by 2

Important Results :

e x e = e

e x o = e

o x o = o

Even and odd

Exercise 1 - Number Systems

exercise

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