Chapter 16Playing with Numbers, important questions for Class 8 Maths are available with solutions here. Students who are appearing for the final of class 8Â can practice these questions and score better marks. All these problems have been prepared as per NCERT(CBSE) curriculum by our subject experts.
In this chapter, students will come across expressing the numbers in a generalized form and normal or usual form. Also, they will learn to check the divisibility of a number by 2, 3, 9, 10, etc. Get important questions for all the chapters of class 8 Maths here at BYJU’S.
Also Check:
 Important 2 Marks Questions for CBSE 8th Maths
 Important 3 Marks Questions for CBSE 8th Maths
 Important 4 Marks Questions for CBSE 8th Maths
Important Questions with Solutions For Class 8 Maths Chapter 16 (Playing with Numbers)
Question 1: Express 3458 in a generalizedÂ form.
Ans: 3458 will be written in a generalized way in the following manner,
3458 = 3 x 10 ^{3} + 4 x 10^{2} + 5 x 10^{1 } + 8 x 10^{0}
Question 2: Express the given number in a normal form: (2 x 1000) + (2 x 10)
Ans: (2 x 1000) + (2 x 10) will be written in a normal form in the following way:
(2 x 1000) + (2 x 10) = 2000 + 20 = 2020
Question 3: If the number 1220 is divided by 13, find the quotient and remainder.
Solution : dividend = pq + r
1220 = 13 Ã— 93 + 11
Quotient = 93
Remainder = 11
Question 4: What is the least natural number which is larger than 100 and which leaves the remainder, R = 12 when it is divided by 19.
Ans: 100 = 19 Ã— 5 + 5
As 5 + 7 = 12, and the required number is 100 + 7 = 107
Question 5: What is the smallest number you have to add to 100000 to get a multiple of 1234?
Ans: 100000 = 1234 Ã— 81 + 46
Thus, 1234 – 46 = 1188.
Hence, the required number is 1188.
Question 6: How many numbers from 1001 to 2000 are divisible by 4?
Ans: 1004, 1008, 1012, 1016… are the numbers between 1001 and 2000 which are divisible by 4.
Therefore, there will be (20001004) /4 + 1 = 250 numbers between 1001 and 2000, and these will be divisible by 4.
Question 7: Show that if a palindrome is 4 digits, it would be divisible by 11?
Ans: Let us assume abba is the fourdigit palindrome.
So, abba = a x 1000 + b x 100 + b x 10 + a
= 1001a +110b
= 91 x 11a + 110b
= 11 x (91a + 10b)
Hence Proved.
Question 8: Make a 5 digit number using each of the digits 4,5,6,7,8. Also, see that the number made shall be divisible by 132.
Ans: 132 can be written as 11 x 12 = 3 x 4 x 11
As 4 + 5+ 6+ 7+ 8 = 30, the numbers formed by the 4,5,6,7,8 are all divisible by 3.
Letâ€™s check the divisibility by 11 and 4.
If the number is divisible by 4, then the last 2 digits of the number formed are divisible by 4.
Similarly, if the difference between the sum of digits placed at even and the sum of digits placed at odd places is divisible by 11, then the number is divisible by 11.
i.e. 67584 and 57684
Question 9. By which number 345111 is divisible among the given options – 15, 12, 3, 9.
Ans: Add all the digits in the number given – 345111
345111 + 3+4+5+1+1+1 = 15, which is divisible by 3. So, 3 is the correct option.
Question 10: Can you tell a 5 digit number which will be divisible by 11 and it should have digits 2,3,4,5,6?
Ans:
So now, 2 – 4 + 3 – 6 + 5 = 0, so 24365 is the least 5 digit number which is also divisible by 11.
Consider the number 24365 formed by the digits 2, 3, 4, 5 and 6
Here, 2 âˆ’ 4 + 3 âˆ’ 6 + 5 = 0
Thus, 24365 is the smallest number divisible by 11.
Question 11:Â Check the divisibility of 21436587 by 9.
Ans:Â The sum of the digits of 21436587 is 2 + 1 + 4 + 3 + 6 + 5 + 8 + 7 = 36.
This number is divisible by 9 (for 36 Ã· 9 = 4).
Hence, 21436587 is divisible by 9.
Check:Â 21436587/9 =Â 2381843
Question 12:Â Check the divisibility of 2146587 by 3.
Ans:Â
The sum of the digits of 2146587 is 2 + 1 + 4 + 6 + 5 + 8 + 7 = 33.
This number is divisible by 3 (for 33 Ã· 3 = 11).
Hence, 2146587 is divisible by 3
Question 13:Â 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?
Ans:
Given, 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.
3 + 1 + z + 5 = 9 + z
Therefore, 9 + z is a multiple of 9.
This is only possible when 9 + z is any one of these numbers: 0, 9, 18, 27, and so on.
This implies, 9 + 0 = 9 and 9 + 9 = 18
Hence 0 and 9 are two possible answers.
Question 14:Â If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Ans:Â
Given, 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.
That is, 3 + 1 + z + 5 = 9 + z
Therefore, 9 + z is a multiple of 3.
This is possible when the value of 9 + z is any of the values: 0, 3, 6, 9, 12, 15, and so on.
At z = 0, 9 + z = 9 + 0 = 9
At z = 3, 9 + z = 9 + 3 = 12
At z = 6, 9 + z = 9 + 6 = 15
At z = 9, 9 + z = 9 + 9 = 18
The value of 9 + z can be 9 or 12 or 15 or 18.
Hence 0, 3, 6 or 9 are four possible answers for z.
Extra Questions for Class 8 Maths Chapter 16


 If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
 If 24x is a multiple of 3, where x is a digit, what is the value of x?
 Check the divisibility of the following numbers by 3.
 108
 616
 294
 432
 927
 If the division of N+5 gives a remainder of 1, calculate the one’s digit of N?
(a) either 7 or 2
(b) 1
(c) 6
(d) 5Â Â Â Â Â Â Â Â Â 5.Â Is 6582 divisible by 4?

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