Playing with Numbers - Solutions 2

CBSE Class –VIII Mathematics
NCERT Solutions
CHAPTER - 16
Playing with Numbers (Ex. 16.2)
1. If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Ans. Since 21y5 is a multiple of 9.
Therefore according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.
 
   
Since 21y5 is a multiple of 9.
2. If 31z5  is a multiple of 9, where z is a digit, what is the value of z? You will find that there are two answers for the last problem. Why is this so?
Ans.  Since 31z5 is a multiple of 9.
Therefore according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.
 
    
 
If 
  
 
Hence 0 and 9 are two possible answers.
3. If 24x is a multiple of 3, where x is a digit, what is the value of x?
(Since 24x is a multiple of 3, its sum of digits 6 + x is a multiple of 3; so 6 + x is one of these numbers: 0, 3, 6, 9, 12, 15, 18, ... .But since x is a digit, it can only be that
6 + x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of (four different values.)
Ans.  Since  is a multiple of 3.
Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.
 
Since  is a digit.
     
     
    
    
Thus,  can have any of four different values.
4.     If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Ans. Since 31z5 is a multiple of 3.
Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.
Since  is a digit.
     
If  
   
If 
   
If  
    
Hence 0, 3, 6 and 9 are four possible answers.