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NUMBER SYSTEMS

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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NUMBER SYSTEMS

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  1. http://nov15.wordpress.com/ Presents QUANT For CAT 2009 NUMBER SYSTEMS

  2. NUMBER SYSTEMS

  3. INTRODUCTION

  4. 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

  5. DIVISIBILITY RULES

  6. 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

  7. 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

  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.

  9. 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

  10. 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

  11. 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

  12. 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

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