Playing with Numbers Class 8

 Playing with Numbers

Numbers in Generalised form

A number is said to be in a generalised form if it is expressed as the sum of the products of its digits with their respective place values.

Two - digits Numbers

       Consider a two digit number having a as the ten digit and b as the units digit. Then number may be written as (10a + b), where a can be whole number from 1 to 9 ; b can be any whole number from 0 to 9.

Examples : 43 = 4*10 +3 3 

1. In a two digit number the units digit is four times the tens digits and the sum of the digits is 10.      Find the number.  

Ans : Let the tens digit be x. then unit digit = 4x.

          so, x + 4x = 10

                     5x = 10

                       x = 10/5

                       x = 2

              so, tens digit =x=2 

                    and unit digit = 4x =4*2 = 8

          Hence, required number = 28

2.  The units digit of a two digit number is 3 and seven times the sum of the digits is the number itself. Find the number.              

Ans. Unit digit = 3 (given)

         let the tens digit be x 

         so, the number is written as 10x +3

        According to the question,

       10x + 3 = 7(x+3)

       10x + 3 = 7x + 21

       10x - 7x = 21 - 3

                3x  = 18

                  x = 6

       Hence, required number = 63

  3. In a two digit number, the digit at the units place is thrice the digit in the tens place. The number          exceeds the sum of its digits by 27 . Find the numbers.

      Let the ten digits be x , then unit digit = 3x.

      so, number is written as 10x+3x = 13x

       and sum of number = x+3x

        according to question,

       13x = x+3x+27

       13x = 4x + 27

       13x - 4x = 27

                9x = 27

                  x = 27 /9

                  x = 3

       the ten digits be x = 3 , then unit digit = 3x = 3*3 = 9

       Hence, required number = 39

4.  A two digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, its digits are reversed. Find the number.

      condition 1 : The number is 3 more than 4 times the sum of its digits.

       so, we can write the number as 10a+b and sum is a+b

       according to condition 1

$\mathrm{ \; \; \; \; 10}a+b= 4(a+b)+\mathrm{ 3}\mathrm{<br>}$

        10a +b = 4a + 4b +3

           6a -3b =3

           2a - b = 1 

            b = 2a -1

        condition 2 : If 18 is added to the number, its digits are reversed. so 

               10a + b +18 = 10b + a

                       9a -9b = -18

                         a - b =  -2

  put b = 2a -1

            a - (2a -1 ) = -2

             a -2a +1 = -2

               -a = -2-1

                -a = -3

                  a = 3

  and b = 2a -1 = 2*3 - 1 = 6 - 1 = 5

so, required number is 35.

5. The sum of the digits of two-digits number is 13. The number obtained by interchanging its digits exceeds the given number by 9. Find the original number.

Let the digits of two-digits number be a and b.

so the number is written as 10a + b

And sum is a + b = 13 so, b = 13 - a

The number obtained by interchanging its digits will be 10b + a

so, according to the question,

10b +a = 10a + b +9

10b - b + a -10a =  9

9b -9a = 9

b - a = 1

put value of b = 13 - a

13 - a - a = 1

13 - 2a = 1

-2a = 1 - 13

-2a = -12 

a = -12 /-2

a = 6

so, b = 13 -a = 13 - 6 = 7

hence required number is 67

6 . In a 3-digit number, the hundreds digit is twice the tens digit while the units digit is thrice the tens digit. Also, the sum of its digits is 18. Find the number.

Let the tens digit be x.

then, hundreds digit = 2x, units digit = 3x.

so, 2x + x + 3x = 18

                    6x = 18

                      x = 18/6

                      x = 3

so, hundreds digit = 2x = 2*3 = 6 , units digit = 3x = 3*3 = 9

hence, required number = 639.