MATH/P/7:WHOLE NUMBERS

                                               WHOLE NUMBER

LESSON 1:

FORMING NUMERALS USING GIVEN DIGITS

Examples:

LESSON  2:

PLACE VALUES OF DIGITS IN NUMERALS

LESSON  3:

READING AND WRITING VALUE IN WORDS TO (100 MILLION)

Writing in words

LESSON  4:

READING AND WRITING NUMERALS IN FIGURES

Examples:

LESSON  5:

EXPANDED NOTATION

Expanding numerals using:

LESSON  6:

FINDING THE EXPANDED NUMBERS (SHORT FORM)

Examples:

LESSON  7:

STANDARD FORM/SCIENTIFIC NOTATION

Writing whole numbers in Scientific notation

LESSON  8:

ROUNDING OFF WHOLE NUMBERS

Examples:

LESSON  9:

ROMAN NUMERALS; CONVERTING HINDU ARABIC NUMERALS TO ROMAN NUMERALS

LESSON  10:

ROMAN NUMERALS (CONVERSION OF ROMAN NUMERALS TO  HINDU ARABIC)

Example

LESSON  11:

OPERATION ON ROMAN NUMERALS

Example

LESSON  12:

BASES (CHANGING FROM DECIMAL BASES TO NON-DECIMAL BASES)

LESSON  13:

BASES (CHANGING FROM NON DECIMAL BASES TO DECIMAL BASES)

LESSON  14

BASES (CHANGING FROM NON DECIMAL BASES TO NON DECIMAL BASES)

LESSON  15:

BASES (OPERATION ON BASES – ADDITION)

LESSON 16:

BASES (SUBTRACTION OF BASES)

Examples:

• 671nine – 285nine

Solution:

5 6 10

6 7 1 _nine                 9 + 1 = 10

– 2 8 5 _nine                9 + 6 = 15

___________

3 7 5 _nine

LESSON 17:

MULTIPLICATION OF BASES

Example:

1 2 1 three               1 x 2 = 2

X 2 three                 2 x 2 = 4

__________            4 ÷ 3 = 1 rem 1

1012 three               1 x 2 = 2 + 1

__________            3 ÷ 3 = 1 rem o

3   3

(ii)            345 six                                     5 x 2 = 20

X 14 six                                    20 ÷  6 = 3rem2

__________                            4 x 4 = 16 + 3

2312                                        10 ÷ 6 = 3rem 1

345                                        3 x 4 = 12 + 3

__________                            15 ÷ 6 = 2 rem3

10202 six

LESSON18:

DIVISION OF BASES

Examples:

• 204 _five ÷14 _five.

Solution:

204 _five – base ten

(2 x 52) + (0 x 51) + (4 x 50)

(2 x 5 x 5) + (0 x 5) + (4 x 1) + (10 x 5) + 0 + 4

50 + 4

54 _ten

14 _five

(1 x 51) + (4 x 50)

(1 x 5) + (4 x 1)

5 + 4

= 9 _ten

5 ÷ 9 _ten

6 ten

6ten – base five

11 five

448 nine ÷  17 nine (answer in Septenary base)

LESSON 19:

FINDING THE UNKNOWN BASE (MISSING BASE)

Examples:

• If 44p = 35 nine

Solution:

(4 x 01) + (4 x p0) = (3 x 91) + (5 x 90)

(4 x p) + (4 x 1) = (3 x 9) + (5 x 1)

4p + 4 = 27 + 5

4p + 4 = 32

4p + 4 – 4 = 32 – 4

4p + 0 = 28

4p  =  28

4         4

P = 7