SHREE BALAJI CLASSES (Rohit Sir)

We have to look at the numbers from 1 to 9.      Cube of 1 (1 3 )  =        1 Cube of 2 (2 3 )  =        8 Cube of 3 (3 3 )  =      27 Cube of 4 (4 3 )  =     64 Cube of 5 (5 3 )  =  125 Cube of 6 (6 3 )  =  216 Cube of 7 (7 3 )  =  343 Cube of 8 (8 3 )  =  512 Cube of 9 (9 3 )  =  729      Cube of 10 (10 3 )=1000 Some important things:- 1.      The sum of the digits of a perfect cube number can only be 1, 8 and 9, so if the sum of the digits is any other than these three, then that number cannot be a perfect cube. 2.      The numbers to the right of which zeros are not multiples of three are incomplete cube numbers like 100, 10000, 2160000 etc. 3.      As many groups of 3–3 digits are formed in a perfect cube number, there will be as ma...

1.     We have to look at the numbers from 1 to 9.      Square of 1 (1 2 )=    1 Square of 2 (2 2 )=    4 Square of 3 (3 2 )=    9 Square of 4 (4 2 )= 16 Square of 5 (5 2 )= 25 Square of 6 (6 2 )= 36 Square of 7 (7 2 )= 49 Square of 8 (8 2 )= 64 Square of 9 (9 2 )= 81 2.      From this we can infer that Square root of any number which ends with 1 will end with 1 or 9 (1+9=10). Square root of any number which ends with 4 will end with 2 or 8 (2+8=10). Square root of any number which ends with 9 will end with 3 or 7 (3+7=10). Square root of any number which ends with 6 will end with 6 or 4 (6+4=10). If the square ends in         1        4        9        6         5   ...

Complement of a Number    Octal and Hexadecimal Number    Convert Decimal to Binary Subtraction of numbers by using complement Subtraction with r’s Complements The direct method of subtraction taught in elementary schools uses the borrow concept. In this method, we borrow a 1 from a higher significant position when the minuend digit is smaller than the corresponding subtrahend digit. This seems to be easiest when people perform subtraction with paper and pencil. When subtraction is implemented by means of digital components, this method is found to be less efficient than the method that uses complements and addition as stated below.      The subtraction of two positive numbers (M - N), both of base r, may be done as follows: 1.        Add the minuend M to r’s complement of the subtrahend N. 2.        Inspect the result obtained in step 1 for an end carry: a.  ...

     This article is used to find out the square of  a number in a short period of time compare to normal maths. It increase the speed of your calculations.  The answer obtained by multiplying a number by itself is the square of that number. Example :- 6×6= 36 which is the square of 6.                          8×8= 64 which is the square of 8.      1.   Square of numbers ending in 5 :- Example 1-: 35 2 =   ? Step 1:- Multiply the first part by the first part plus 1: 3 ×(3+1) =3×4 =12 Step 2:- Write the number 25 next to the result from step 1: 35 2 = 12_25 = 1225 Example 2-: (in 3 digits) 125 2 =   ? Step 1:- Multiply the first part by the first part plus 1: 12 ×(12+1) =12×13 =156 Step 2:- Write the number 25 next to the result from step 1: 125 2 = 156_25 = 15625 Example 3-: (in decimal numbers) 9.5 2 =     ? Step 1:- M...

Subtraction of Binary Numbers  ;  Octal and Hexadecimal Number  ;  Convert Decimal to Binary Complements Complements are used in digital computers for simplifying the subtraction operation and for logical manipulations. There are two types of complements for each base-r system: (1) the r’s complement and (2) the (r-1)’s complement. When the value of the base is substituted, the two types receive the names 2’s and 1’s complement for binary numbers,  10’s and 9’s complement for decimal numbers. The r’s Complement Given a positive number N in base r with an integer part of n digits, the r’s complement of N is defined as r n   - N for N is not equal to 0 and 0 for N = 0. The following numerical example will help clarify the definition. The 10’s complement of (52520) 10 is 10 5 – 52520 = 47480 The number of digits in the number is n= 5 The 10’s complement of (0.3267) 10 is 1 – 0.3267 = 0.6733 No integer part, so 10 n = 10 0 = 1 The 10’s...