BLOG 4 PLAYING WITH NUMBERS Ex 16.2

BLOG 4 PLAYING WITH NUMBERS Ex 16.2

GOOD MORNING BOYS ,

LEARNING OUTCOMES 

 By the end of this lesson we will be able to  find the " DIGIT" to the " LETTER"  in  any given number using DIVISIBILITY TEST  and hence  solve the puzzle 

Exercise 16.2 : Solutions of Questions on Page Number : 260 

 CLICK ON THIS LINK TO UNDERSTAND EX 16.2 

(NCERT)

Q.1 If 21y5 is a multiple of 9, where yis a digit, what is the value of y?  

Answer : 

If a number is a multiple of 9, then the sum of its digits will be divisible by 9.  

Sum of digits of 21y5 = 2 + 1 + y + 5 = 8 + y

Hence, 8 + y should be a multiple of 9.  

This is possible when 8 + y is any one of these numbers 0, 9, 18, 27, and so on …  

However, since y is a single digit number, this sum can be 9 only. 

Therefore, y should be 1 only.  

Q2 :   

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?  

Answer : 

If a number is a multiple of 9, then the sum of its digits will be divisible by 9.  

Sum of digits of 31z5 = 3 + 1 + z + 5 = 9 + z

Hence, 9 + z should be a multiple of 9.   

This is possible when 9 + z is any one of these numbers 0, 9, 18, 27, and so on …  

However, since z is a single digit number, this sum can be either 9 or 18. 

Therefore, z should be either 0 or 9.  

Q3 :   

If 24x is a multiple of 3, where x is a digit, what is the value of x?  

Answer : 

Since 24x is a multiple of 3, the sum of its digits is a multiple of 3.  

Sum of digits of 24x = 2 + 4 + x = 6 +x

Hence, 6 + x is a multiple of 3.   

This is possible when 6 + x is any one of these numbers 0, 3, 6, 9, and so on … 

Since x is a single digit number, the sum of the digits can be 6 or 9 or 12 or 15 and thus, the value of x comes to 0 or 3 or 6 or 9 respectively. 

Thus, x can have its value as any of the four different values 0, 3, 6, or 9.  

Q4 :   

If 31z5 is a multiple of 3, where z is a digit, what might be the values ofz?  

Answer : 

Since 31z5 is a multiple of 3, the sum of its digits will be a multiple of 3.  

That is, 3 + 1 + z + 5 = 9 + z is a multiple of 3.  

This is possible when 9 + z is any one of 0, 3, 6, 9, 12, 15, 18, and so on …  

Since z is a single digit number, the value of 9 + z can only be 9 or 12 or 15 or 18 and thus, the value of xcomes to 0 or 3 or 6 or 9 respectively.  

Thus, z can have its value as any one of the four different values 0, 3, 6, or 9.  

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