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NCERT Solutions For Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2

Here, Below you all know about NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2 Question Answer. I know many of you confuse about finding this Chapter 16 Playing with Numbers Ex 16.2 Of Class 8 NCERT Solutions. So, Read the full post below and get your solutions.

TextbookNCERT
BoardCBSE
CategoryNCERT Solutions
ClassClass 8
SubjectMaths
ChapterChapter 16
ExerciseClass 8 Chapter 16 Playing with Numbers Exercise 16.2
Number of Questions Solved6
NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2

NCERT TEXTBOOK EXERCISES

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

Solution.

Since 21y5 is a multiple of 9, its sum of digits 2 + 1+ y + 5 = 8+ y isa multiple of 9; so 8 + y is one of these numbers: 0, 9, 18, 27, 36, 45,… .
But since y is a digit, it can only be possible that 8 + y = 9. Therefore, y = 1.

Question 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 to the last problem. Why is this so?

Solution.

Since 31z5 is a multiple of 9, its sum of digits 3 + 1 + z + 5 = 9 + z isa multiple of 9; so 9 + z is one of these numbers: 0, 9, 18, 27, 36, 45, … .
But since z is a digit, it can only be possible that 9 + z = 9 or 18. Therefore, z = 0 or 9.

Question 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 xis 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.)

Solution.

The solution is given with question.

Question 4. If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?

Solution.

Since 31z5 is a multiple of 3, its sum of digits 3 + 1 + z + 5 = 9 + z is a multiple of 3; so 9 + z is one of these numbers: 0, 3, 6, 9, 12, 15, 18, … .
But since z is a digit, it can only be possible that 9 + z = 9 or 12 or 15 or 18. Therefore, z = 0 or 3 or 6 or 9.

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Exercise 15.2 PDF

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