Page 305 - Start Up Mathematics_8 (Non CCE)
P. 305

19                                                 Playing with Numbers











                    In this chapter we will learn about numbers in more detail and deduce the divisibility test rules. Also, we will
                    learn how to solve number puzzles.


                    Numbers in Generalized Form
                    A number is considered to be in generalized form if it is written as the sum of the product of its digits
                    with their respective place values. On the other hand, you have already learnt that a natural number can be
                    expressed in exponential form by utilizing exponents of 10 and the digits of the number. For example, 86
                                                              1
                                                                       0
                    in exponential form can be written as 8 × 10  + 6 × 10 .
                    Let’s consider a 2-digit number ab having a and b as the digits at tens and ones places respectively. This
                    can be expressed in the generalized form as 10a + b, where a (π 0) and b are digits from 0 to 9. Similarly a
                    3-digit number abc can be expressed in the generalized form as 100a + 10b + c, where a (π 0), b and c are
                    digits from 0 to 9.


                        The number ab and abc does not mean a × b or a × b × c.

                    Reversing the Digits—2-Digit Number
                    Let there be a 2-digit number ab, where a (π 0) and b are digits at tens and ones places. On reversing the digits,
                    we get ba. In the generalized form, ab = (10 × a) + b and ba = (10 × b) + a.
                    On adding, we get,
                       ab + ba = (10a + b) + (10b + a)
                    fi ab + ba = 11a + 11b = 11(a + b)
                       ab ba+
                    fi         =11
                        ab+
                    So, it can be deduced that ab + ba is completely divisible by a + b and the quotient is 11. Also, ab + ba is
                    completely divisible by 11 and the quotient is a + b.
                    Now, let’s subtract ab and ba (a > b).
                      ab – ba = (10a + b) – (10b + a)
                    fi ab – ba = 9a – 9b = 9(a – b)

                       ab ba-
                    fi         = 9
                        ab-
                    So, it can be deduced that ab – ba is completely divisible by a – b and the quotient is 9. Also, ab – ba is
                    completely divisible by 9 and the quotient is a – b.
                        If on the other hand, b > a, then find ba – ab.


                    Example 1:      Without actual calculations, write the quotient when the sum of 68 and 86 is divided
                                    by (i) 14, (ii) 11.
                    Solution:       Here 68 and 86 are numbers that can be obtained by reversing the digits of the other.
                                    Sum of numbers = 68 + 86 = 154
                                    Sum of digits = 6 + 8 = 14
   300   301   302   303   304   305   306   307   308   309   310