Arithmetic

1. Looking at the two procedures to find the cube root of two and three digit numbers in previous post (Cube roots Part 1). We can follow the following procedure to find the cube root of any number. This method is called the division method.

1. Group the digits from right into triples. If the number has a decimal part, make triples to both sides of decimal. If the decimal part has one or two digit left, add one or two zero correspondingly to it.

2. Find  a number whose  cube is less than or equal to the first triple or the remaining digits after forming triple. Take it as divisor and quotient.

3. Subtract the cube of divisor  from the first triple or the remaining digits after forming triples.

4. Bring down the next triple to the right of the remainder. This is the new dividend.

5. Take the thrice of quotient below on the left column of the new dividend.

6. The new divisor is obtained by annexing the thrice of the quotient by a digit. The digit is such that [(10×(the thrice of quotient)×(quotient annexed  by new digit)×(new digit)) + (new digit)3] is less than or equal to the new dividend.

7. Annex the new digit to the top quotient.

8. Subtract the number obtained by [(10×(the thrice of quotient)×(quotient annexed  by new digit)×(new digit)) + (new digit)3]  from the dividend.

9. Repeat the process.

10. In case of taking quotient to decimal, add zeros to right of remainder in triples.

11. Cube root of 15625 is 25.

12. Cube root of 2352637

13. Cube root of 2352.637

14. Cube root of 2.352637

15. Cube root of 2 is 1.259...

Supplements

Read Basics of the above kind of mathematics.

Arithmetic

Cube

1. When we multiply a variable with itself twice it is called a cube of the variable. The cube of a is a3.

2. (a + b)3 = a3 + 3a2b + 3ab2 + b3

3. (a + b + c)3 = a3 + b3 + c3 + 3a2b + 3ab2 + 3a2c + 3ac2 + 3b2c + 3bc2 + 6abc

4. We can write a two digit number as (10a + b).

The cube of this number is (10a + b)3 = 1000a3 + 100×3a2b + 10×3ab2 + b3.

It can be written as (a3)(3a2b)(3ab2)(b3).

Read Basics of the following kind of mathematics.

Suppose we want to find the cube of 13 then 133 is (1)(9)(27)(27)
or (1)(9)(29)(7)
or (1)(11)(9)(7)
or (2)(1)(9)(7)
or 2197
and cube of 25 is (8)(60)(150)(125)
= (8)(60)(162)(5)
= (8)(76)(2)(5)
= (15)(6)(2)(5)
= 15625.

5. We can write a three digit number as (100a + 10b + c).

The cube of this number is (100a + 10b + c)3 = 1000000a3 + 100000×3a2b + 10000×3ab2 + 10000×3a2c + 1000×b3 + 1000×6abc + 100×3ac2 + 100×3b2c + 10×3bc2 + c3

It can be written as (a3) (3a2b) (3ab2 + 3a2c) (b3 + 6abc) (3ac2 + 3b2c) (3bc2) (c3)

Suppose we want to find the cube of 133 then 1333 is
(1)(9)(36)(81)(108)(81)(27)
or (1)(9)(36)(81)(108)(83)(7)
or (1)(9)(36)(81)(116)(3)(7)
or (1)(9)(36)(92)(6)(3)(7)
or (1)(9)(45)(2)(6)(3)(7)
or (1)(13)(5)(2)(6)(3)(7)
or (2)(3)(5)(2)(6)(3)(7)
or 2352637

and cube of 255 is (8)(60)(210)(425)(525)(375)(125)
= (8)(60)(210)(425)(525)(387)(5)
= (8)(60)(210)(425)(563)(7)(5)
= (8)(60)(210)(481)(3)(7)(5)
= (8)(60)(258)(1)(3)(7)(5)
= (8)(85)(8)(1)(3)(7)(5)
= (16)(5)(8)(1)(3)(7)(5)
= 16581375

6. We can write a three digit number with decimal point before the last digit as (10a + b + 10-1c).

The cube of this number is (10a + b + 10-1c)3 = 1000a3 + 100×3a2b + 10×3ab2 + 10×3a2c + b3 + 6abc + 10-1×3ac2 + 10-1×3b2c + 10-2×3bc2 + 10-3c3

It can be written as (a3) (3a2b) (3ab2 + 3a2c) (b3 + 6abc). (3ac2 + 3b2c) (3bc2) (c3).

Suppose we want to find the cube of 13.3 then 13.33 is
(1)(9)(36)(81).(108)(81)(27)
or (1)(9)(36)(81).(108)(83)(7)
or (1)(9)(36)(81).(116)(3)(7)
or (1)(9)(36)(92).(6)(3)(7)
or (1)(9)(45)(2).(6)(3)(7)
or (1)(13)(5)(2).(6)(3)(7)
or (2)(3)(5)(2).(6)(3)(7)
or 2352.637

and cube of 2.55 is (8)(60)(210).(425)(525)(375)(125)
= (8)(60)(210).(425)(525)(387)(5)
= (8)(60)(210).(425)(563)(7)(5)
= (8)(60)(210).(481)(3)(7)(5)
= (8)(60)(258).(1)(3)(7)(5)
= (8)(85)(8).(1)(3)(7)(5)
= (16)(5)(8).(1)(3)(7)(5)
= 1658.1375

Cube Root

1. The number which when multiplied by itself twice gives the cube of the number, the number is called the cube root of the cube. The cube root of a3 is a.

2. The cube root of a3 + 3a2b + 3ab2 + b3 is (a+b).

3. The cube root of a3 + b3 + c3 + 3a2b + 3ab2 + 3a2c + 3ac2 + 3b2c + 3bc2 + 6abc is (a + b + c).

4. We can represent a two digit number as (10a + b)or(a)(b). The cube of it is (1000a3 + 100×3a2b + 10×3ab2 + b3) or (a3)(3a2b)(3ab2)(b3). The figure describes how to find the cube root of a number whose cube root is a two digit number.

1. Take cube of a, (a3) and subtract it from the number left after making group of three from right.
2. Bring (3a) below.
3. Suffix (b) to it.
4. Main step: Multiply ten times of (3a) in column 1 row 2 ((3a) from (3a)(b)) leaving the new digit (b) to the number (a)(b) in the column 2 row 1 and new digit (b) to get [(3a2b)(3ab2)(0)], add the cube of new number (b) to it to get (3a2b)(3ab2)(b3). The expression looks like this [(3a)(0) ×(a)(b)×(b) + (b3)].
5. For detailed method read the points in the topic below as arithmetic.

5. We can represent a two digit number as (100a + 10b + c) or (a)(b)(c). The cube of it is 1000000a3 + 100000×3a2b + 10000×3ab2 + 10000×3a2c + 1000×b3 + 1000×6abc + 100×3ac2 + 100×3b2c + 10×3bc2 + c3 or (a3) (3a2b) (3ab2 + 3a2c) (b3 + 6abc) (3ac2 + 3b2c) (3bc2) (c3). The figure describes how to find the cube root of a number whose cube root is a three digit number.

1. Take cube of a and subtract it from the digit left after making triples.
2. Bring (3a) below in column 1.
3. Suffix (b) to (3a) to get (3a)(b). Bring next triad below.
4. Main step: Multiply ten times of (3a) in column 1 row 2 ((3a) from (3a)(b)) leaving the new digit (b) to the number (a)(b) in the column 2 row 1 and new digit (b) to get [(3a2b)(3ab2)(0)] add the cube of new number (b3) to it to get (3a2b)(3ab2)(b3). The expression looks like this [(3a)(0)×(a)(b)×(b)+ (b3)].
5. Subtract and bring the next triad (a group of three digits) below.
6. Suffix (3b) to (3a) to get (3a)(3b).
7. Main step: Multiply ten times of  (3a)(3b) in column 1 row 2 ((3a)(3b) from (3a)(3b)(c)) leaving the new digit (c) to the number (a)(b)(c) in the column 2 row 1 and new digit (c) to get [(3a2c)(6abc)(3ac2+3b2c)(3bc2)(0)] add the cube of new number (c3) to it to get [(3a2c)(6abc)(3ac2+3b2c)(3bc2)(c3)]. The expression looks like this [(3a)(3b)(0)×(a)(b)(c)×(c)+ (c3)].
8. Subtract this from above.
9. For detailed method read the points in the topic below as arithmetic.

Saturday, April 23, 2016

A two digit number

You are aware of two digit numbers. A two digit number is written in the form ab where ab = 10×a + b. Similarly as we would write 21 = 2×10 + 1. In this post I will teach you how to play with two digit numbers. I will teach fast tricks to do multiplication and finding squares.

Multiplication

You are aware of the formula (a + b)(c + d) = a×c + a×d + b×c + b×d. If you have two digit numbers kl and mn. Then you can represent kl as 10×k + l and mn as 10×m + n.

The product of kl and mn can be represented as
(10k + l)(10m + n) = 100×k×m + 10(k×n + l×m) + l×n

Example

21 × 65
= 100×2×6 + 10(2×5 + 1×6) + 1×5
= 1200 + 10(16) + 5
= 1200+160+5
= 1365

Steps to multiply two numbers
1. Multiply the digits at unit place
2. Multiply the two sets of digits at unit place and digit at tens place. Add. Multiply the result by 10.
3. Multiply the two digits at tens place. Multiply the result by 100.
4. Add the result obtained in steps 1,2 and 3.

Square

You know that the square of a number is product of a number with itself. So, if the number is kl then the square can be represented as
(10k + l)(10k + l) = 100×k×k + 10(k×l + l×k) + l×l
= 100×k2 + 2×10×(k×l) + l2

You can obtain the above formula easily by using the expansion
(a + b)2 = a2 + 2ab + b2

Example
15×15
= 100×1 + 2×10×5 + 25
= 100 + 100 + 25
= 225

Steps to square a two digit number
1. Square the digit at unit place
2. Multiply the two digits at unit place and at tens place. Multiply the result by 20.
3. Square the digit at tens place and multiply by 100.
4. Add the result obtained in steps 1,2 and 3.

Squaring a two digit number ending with 5

When the number ends with 5 it can be written as k5.
Its square is 100×k2 + 2×10×(k×5) + 52
Which can be written as 100×k2 + 100×k + 25
= 100×(k2+k) + 25
= 100×(k(k+1)) + 25

Rule: Find the product of digit at tens place and (digit at tens place + 1). Place 25 after it.