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NUMBER SYSTEMS

NUMBER SYSTEMS

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NUMBER SYSTEMS

INTRODUCTION

PRIME NUMBERsA number is prime if it is not divisible by any prime number less than its 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.

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 rulesTest 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! Divisibility by Test7Subtract 2 x Last digit11Subtract 1 x Last digit13Add 4 x Last Digit17Subtract 5 x Last Digit19Add 2 x Last DigitHere 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.

Units Digit of a Number

Find the units digit of 71999 (7 to the power 1999)

Step 1: Divide the exponent by 4 and note down remainder1999/4 => Rem = 3

Step 2: Raise the units digit of the base (7) to the remainder obtained (3) 73 = 343

Step 3: The units digit of the obtained number is the required answer. 343 => Ans 3

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

Note: For bases with units digits as 1,0,5,6 the units digit for any power will be the 1,0,5,6 itself. Ex: Units 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


a) Number ending with 1 : Ex : Find the last 2 digits of 31786Now, multiply the 10s digit of the number with the last digit of exponent31786 = 3 * 6 = 18 -> 8 is the 10s digit.

Units digit is obviously 1

So, last 2 digits are => 81


b) Number Ending with 3, 7 & 9Ex: Find last 2 digits of 19266

We need to get this in such as way that the base has last digit as 119266 = (192)133 = 361133

Now, follow the previous method => 6 * 3 = 18So, last two digits are => 81


b) Number Ending with 3, 7 & 9

Remember : 34 = 81 74 = 240192 = 81

Last two digits of a numberEx 2: Find last two digits of 33288Now, 33288 = (334)72 = (xx21)72Tens digit is -> 2*2 = 04 -> 4So, 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 numberc)Ending with 2, 4, 6 or 8Here, we use the fact that 76 power any number gives 76.

We also need to remember that,242 = xx76210 = xx2424even = xx7624odd = xx24

Last two digits of a numberEx: Find the last two digits of 25432543 = ((210)54) * (23) = ((xx24)54)* 8 = ((xx76)27)*8

76 power any number is 76

Which gives last digits as => 76 * 8 = 608So last two digits are : 08

Highest powerHighest 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 shown60/5 = 12, 12/5 = 2 (omit remainders) 2/5 = 0 12+2+0 = 14So highest power of 5 that divides 60! is 14.

Highest powerEx: 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 = 44/5 = 0 So, 20 + 4 + 0 = 24For 3 : 100/3 = 33 33/3 = 11 11/3 = 33/3 = 11/3 = 0

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

Now, the smallest number of these is taken which will be 24.

Number of zeroesEx: Find the number of zeroes in 75!This means highest power of 10 which can divide 75!10 = 5*2If we consider highest power of 5 which can divide 75! , its sufficient.75/5 =1515/5 =3 3/5 =0 So, 15+3+0 = 18

So, there are 18 zeroes in 75!

Number of factors of a numberIf the number N can be expressed as a product of prime factors such thatN = (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 oddEven number => Divisible by 2Odd Number => Not Divisible by 2Important Results :

e x e = ee x o = eo x o = o

exerciseDownload the related exercise here

Exercise 1 - Number Systems

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presentsquantfor cat 2009

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